Displaying Multiple Items From Database
what i have is a page that has four sections. what i want to have is when the page opens, in that page i want the lastest two items for each section from a database to be displayed. the items to be displayed are an image, a title and a description. but i also want the description to be shortend to say about 25-30 words or about 100 chars, if thats easier?
so in the end i have 4 sections, each showing the lastest two items for each section.
would the best way to do this be to have a seperate table in the database for each of the sections? or would i be able to do it from one table?
would i have to creat a query for each of the sections?
e.g:
SELECT * FROM table WHERE section=section
View Complete Forum Thread with Replies
See Related Forum Messages: Follow the Links Below to View Complete Thread
Multiple Cart Items
I've created a shopping cart application using php 4.06 and postgres. a problem i had was that when i added the same item twice it was adding an extra item as opposed to adding the quantities together. i did manage to sort this out to an extent but my only solution means using a block of code for each item which has made my script very long and i'm pretty sure has made the script somewhat unreliable. I would appreciate it greatly if someone could offer a solution to this, this is my code for adding items to the cart: PHP Code:
Multiple Items Of Same Name In Session?
I have a shopping cart where I can add tshirt designs to the cart. The designs are added to a php session object called cart. The object 'cart' consists of the design name, color and the quantity. The probem is that if someone adds the same design to the cart again then it just increments the quantity of the existing design with the same name instead of adding another instance of it to the cart. How can I have multiple instances of the same design with the same name in the cart at the same time. One thing I thought of was to append some sort of incrementing id to the design name before it is added to the cart. Thus the next time the design is added it would have a new id appended and the resulting name would be different.
Multiple Items In Foreach()
Can someone tell me what im doing wrong with the code bellow please? I get print_ticket called lots of times, with 1 as the hash,name,ticket_no responce. (i have chequed that the query is returning the right results). $return = pg_query($pg_connection, $query); $data = pg_fetch_array($return); foreach ( $data as $temp ) { echo print_ticket ($temp['hash'],$temp['name'],$temp['ticket_no']); }
Updating Multiple Items With One Form
I am using a do while loop to run through a list of items on my page. Say I have 5 items on the page and each has one field that the user can input information to. Is there a way to do an update where it will update all records on the page with one submit form?
Adding Multiple Listbox Items To Mysql
I want to be able to select multiple items from a listbox in a form and send the result to mysql blob. I only get one item to the bd. Here is part of the script. Item: <select name="itm" align="bottom" multiple size="2"> <option value="item1">item1</option> <option value="item2">item2</option> <option value="item3">item3</option> <option value="item4">item4</option> <option value="item5">item5</option> </select> "Insert into xyz set itm = '$itm'";
Count Of How Many Items Are In Database
I am wanting to create an select menu for displaying the order of the item on a page. I am guessing that I need to get the count of how many items are in the db and them put them into a select menu. Does anyone know how to do this or can point me in the right direction?
Adding List Items To A Database
Not sure if this is javascript or php so i put it in both newsgroups, sorry if it's the wrong group. I wrote (with help) a code to put several items in a list object (select) and I would like to add the items in this list to a database. I know how to add a selected item to the database but how can i put al the items tot the database. I need somekind of loop wich go's through the list.
How To Pull Only Unique Items From Database?
I have a table like this: date, author, title, content now I want to grab the 3 latest items from this table from 3 DIFFERENT authors. I want this: -article by jim -article by sally -article by john Not this: -article by jim -article by jim -article by sally even if the two latest articles happen to be by jim, lets only grab the first one and move on. So basically I need (in pseudo-sql/plaintext): select title from contenttable where author=unique order by date limit 3 I've got everything code-wise except the "author=unique" part, how do I do that?
Display Items From A Mysql Database In A Table
I am trying to display items from a mysql database in a table but I need it to start a new table row each time multiples of 3 rows of the database table have been displayed like the following example: item 1 item 2 item3 item 4 item 5 item 6 Does anybody know how I can make this work using php.
Preventing The Addition Of Duplicate Items To A Database
I'm trying to stop my file upload system adding the same files to my database. The script below is a much simplified version of what I'm using. It's getting frustrating as I can't get this to work. I have a page previosu to this that posts up the details form a previous form. The form and everything works fine apart from the duplicate check. Code:
Passing Multi Items To Database - Shopping Cart.
I'm developing/designing a shopping cart with a flash front end and php/MySQL. The products information in the basket are stored in Objects which are stored in an array - so the information to be passed to the database is an array of objects and each object describes a product in the basket. The problem I have is how do I pass this array of objects to the database considering they will all be related to one customer. Is it possible to pass an array to php ?
Displaying Results On Multiple Pages
I have a search query via PHP and MySQL that has list of results, but I only want to show 10 or 15 on a page. Much like Yahoo! or even here on the Devshed forums. I have been working on a system that cycles through all the results with a while loop and a counter. It only displays results >= 11 or <=20 as an example. I am passing that data (the results start point through the url) I am just worried that if the results are in the thousands this system would get very slow.Is there an easier/better way that I am not seeing?
Uploading And Displaying Multiple Images
I am building a website where people can post jobs that need to be done and have laborers bid on them.... Im done with pretty much everything on the site but im having a hard time figuring out how to upload and display multiple images for each job.... I am able to upload and display one image but i can upload multiple images.... Everytime I try to do so, It only displays multiple copies of the first image that i upload. Code:
Displaying Select Multiple Box And Using Mysql Join
i have an edit page where the query lists all the information from the database about a business. each business can have multiple types so i made a many-to-many database and now i need to be able to show all the types that are in the db for each business on the edit page. here is what i have so far but it's only selecting the last type of business in the result. PHP Code:
Displaying From Database!
I'm trying to display products that are pulled form a database in a certain way using only DIV tags I would like the products to display across the page in rows of 4, however when i use div tags it is just putting one on each row.. whats the best way to display in a row of 4? here is an example script of what im using to experiment with: Code:
Retrieving,Sorting & Displaying Multiple Results From Mutliple Tables
I need to know how to run a php / mysql query to do the following. I have 1 database (lets call it "database" for now) within this database there are multiple tables ("News","Feature","Header_feature") For NEWS I need only the 6 latest results pulled from that table and displayed in a table. The "News" table has the following columns "ID","Title","Date","Body","Image" For Feature I need only 1 result retrieved from the database and displayed under the variable $ticker as it will display as a scrolling marquee within a table. The "Feature" table has the following columns "ID","Ticker". For Header_feature I need only 1 result retrieved from the database and displayed in the following manner <a href="$link"><img src"$header_feature">[/url] The "Header_feature table has the following columns "ID","image","link" How would I code this all into 1 page? I cant seem to get the query to do everything I want it to and display the results properly. You can see what I have so far at www.getlivemedia.com The $ticker is the scrolling text near the top of the page and the news is obviously the bulk of the page and $feature_image is the broken image.
Displaying Content From Database
need some simple help with displaying content from a mySQL database using PHP. i want to know how to write a php script to display the latest entry in a table to an html page. and also how to adapt the script to display the latest x amount of entries.
Problem Displaying Image From Database
I used the code below to pull out an image from a mysql database but nothing shows on the page and there was no error message to indicate any problem.What could be wrong.The image is actually in the database with all the details: <?php if(isset($_GET[loginid])) { $con=mysql_connect("localhost","name","pass"); if(!$con) { die('Could not connect to database:'.mysql_error()); } mysql_select_db("datab",$con); $id = $_GET[loginid]; $query = "SELECT name, type, size, content " . "FROM image WHERE id= '$id'"; $result = mysql_query($query) or die('Error, query failed'); $file=mysql_fetch_array($result); $name=$file['name']; $type=$file['type']; $content=$file['content']; header('Content-length: '.strlen($content)); header("Content-type: $type"); header("Content-Disposition: inline; filename=$name"); echo $content; } mysql_close($con) ?>
Displaying Values From A Database From An Include
I've got a URL going to a template, e.g. www.mydomain.com/product.php?id=product_a where the template is product.php the tag after the ? refers to the product that should be displyed in the template, in this case product_a Inside the template (product.php) I've got:- <? $url = $id.".htm"; include("$url"); ?> This works fine so far. How ever I've also got an include inside the the template (product_a.php), called product_info.php, which is another php file that displays the product version and the platforms that it sits on. so in the the include (product_info.php) I have:- <?php $db = mysql_connect("localhost", "username", "password"); mysql_select_db("DATABASE",$db); $result = mysql_query("SELECT version FROM table1 WHERE id = id",$db); do { if ($myrow[0] != "") { echo "$myrow[0]n"; } } while ($myrow = mysql_fetch_array($result));?> however this displays all the version numbers within the table (for every product) I just want it to display the version number of the value of $id (www.mydomain.com/product.php?id=product_a)
Displaying Image Stored In Database
The user can browse and upload an image and the image is stored successfully in a seperate binary table (with userid and bin_data, filesize, filename, filetype). The photo is displayed properly when I point my broswer to a script called fileshow.php: PHP Code:
Problem Displaying One Column Of One Row From A Database With PHP.
I have a small script with PHP that queries a MySQL database to pull out one row, where I want to be able to access each of the columns separately. I have tried several different variations and am able to get the entire row to print, but when I attempt to access the individual columns I get an error. Here is what I have so far: if (isset($_POST['memberNo'])): $link = mysql_connect('...','...','...'); mysql_select_db("..."); //Perform a test query $query = "SELECT * FROM users WHERE 'id' = " . $_POST['memberNo']; $result = mysql_query($query) or die(mysql_error()); $line = mysql_fetch_assoc($result); print $line; //Close connection mysql_close($link); The point of this code is to retrieve the user information to validate the login information sent in the previous form. I do not get an error with this code, but it also does not print anything. I know there is an entry in the database that matches data sent in the $_POST['memberNo'] variable, it is the only entry in the database. Changing the 'print $line;' to 'print $line["id"];' does not display anyting either.
Displaying Database And Getting Unwanted Results...
I want to display everyone in the database, but only if the 'level' field is larger than 0. Well, this works fine for most of the database... but if the last entry in the database is 0, then it prints it anyways. Is there anyway around this? I tried adding an if statement before the last bit where I echo out the table rows, but it was killing the entire table, or still showing the same results. Code:
Displaying Results From Database In A Particular Format
How do I display results from a database in a particular format. For example the following text in this format including the lines in stored in the database: Quote“Hello, Welcome to the news for 11/06/07. Thanks” When I echo the results using the usual, <? echo $row_news[‘news’]; ?> it doesn’t echo the information in the database in the format it is stored for example it will display like this. Quote“Hello, Welcome to the news for 11/06/07. Thanks” How do I get it to display in the format it in displayed in the database?
Displaying Database Information In Columns
I have a database with products in and I want iT to automatically show every product on the page. Now I am able to do this easy enough but what I am struggling to do is that I want the products to be displayed in 3 columns. I can get them to display one under the other easy enough but I want them to be in rows with 3 columns instead.
Displaying A List Of Categories From A Database In A Select Box
Could someone please tell me why this outputs nothing <select name=categories> <? $cat_array = get_categories(); foreach($cat_array as $this_cat) { echo "<option value=""; echo $this_cat["category_id"]; echo """; echo ">"; echo $this_cat["category_name"]; echo " "; } ?> </select> ========================================================== function get_categories() { //get the list of categories from the database $conn = mysql_pconnect("localhost", "user", "pwd"); $query = "select * from categories"; $result = mysql_query($query); if(!$result) return false; $num_cats = mysql_num_rows($result); if($num_cats == 0) return false; $result = db_result_to_array($result); return $result; } //A function that returns a query to the database as an array function db_result_to_array($result) { $res_array = array(); for($count=0; $row=@mysql_fetch_array($result); $count++) $res_array[$count] = $row; return $res_array; }
Reading And Displaying A Date From A MySql Database
I've just started using php, and although I am very impressed by it, there are some things with which I am confounded! What I'm trying to do is read in a Date from a mySql database and then display it on screen (the Date is in the usual yyyy-mm-dd format). If I use SQL on phpadmin SELECT Date FROM details WHERE details.Name = "Paul Lee" LIMIT 0 , 30 The result comes up 2005-01-01, so this seems to work. But if I use the following: $db = mysqli_connect("localhost","root",""); @mysqli_select_db($db, "personnel") or die ( "Unable to select database" ); $query = 'SELECT Date FROM details WHERE details.Name = "Paul Lee" LIMIT 0, 30' $result=mysqli_query($db, $query); echo "<br>"; echo $result; mysqli_close($db); I get "Object id #2" displayed on screen. I have tried to use the explode function to separate the months, year and day using the "-" as a delimeter, but this doesn't work, and I can't seem to get the other php date/time functions to work either.
Displaying Information From A Database In A Text Field,
I am trying to create an admin page, but before I get anywhere near that, I need to understand how to get data froma database to show up in appropriate form fields *text, textarea, etc...* I am able to get content from my database to show up in my template, now I want to create a page that allows a user to click on a link for which they want to edit the information (name, and bio) So I want that page to show up with a simple form, a text field displaying the name, and a textarea displaying the bio information, so that they can modify it and save it to the database. how do you get the information to show up in the form fields? Code:
Previous And Next Links And Displaying Database Information
I would like to display a listing of files on a web page as follows: If there is only one file: display the section name and then display the current file. If there is more than one file (for the first page): display the section name, the current file and a few archive files. If there is more than a page full (for each additional page): display the section name, and the next set of archive files. This is the code I have been working with (it's a bit ugly, suggestions are welcome). For some reason the second page of listings skips the first item. 1st page lists items 0 through 4, second page skips item 5 and starts displaying item 6. function display_table($sec_id, $num_item){ GLOBAL $link; $updir = "../uploads/"; //maybe this should be a GLOBAL variable??? if (isset($_GET['start'])){$start = $_GET['start'];}else{$start = 0;} //GLOBAL $start; //query DB $query_total = "SELECT file_id FROM files WHERE file_section_id =" . $sec_id ; $query = "SELECT file_id, file_meeting_date, file_meeting_time, file_title, file_modified, file_section_id, sec_name, sec_id, sec_disp_archive FROM files f, section s WHERE f.file_section_id =" . $sec_id . " AND f.file_section_id = s.sec_id ORDER BY file_modified DESC LIMIT " . $start . "," . $num_item; $result = mysqli_query( $link, $query ) or die(mysqli_error($link)); $result_total = mysqli_query( $link, $query_total ) or die(mysqli_error($link)); $num_rows = mysqli_num_rows($result_total); //print $query; //create hyperlinks and display table //display current file /* print "START: " . $start . "NUMBER: " .$num_item . " ROWS: " . $num_rows;*/ /*$obj = mysqli_fetch_object($result);*/ if ($num_rows > 1){ //Display current and archive files //only display current table if start is not set (not on page 1) $obj = mysqli_fetch_object($result);//Call DB $content = "<div id ='MainContent'><p><b>" . $obj->sec_name . "</b></p><table width=ď`%' border=Ƈ'>"; if (!isset($_GET['start'])||$_GET['start']==0){ //for 1st item display file info $content.= "<tr><td colspan=Ɖ' align='center'>Current</td></tr>"; $content.= "<tr bgcolor='#CCCCCC'><td width=ཝ%'>Meeting Date</td><td width=ཝ%'>Meeting Time</td><td width=ཞ%'>File</td></tr>"; $content .= "<tr><td>" . date("F jS, Y",strtotime($obj->file_meeting_date)) . "</td><td>" . date("g:i A",strtotime($obj->file_meeting_time)) . "</td><td><a href='" . $updir . $obj->file_title . "'>" . "View Current File" . "</td><tr>"; $content .="</table>"; print "START: " . $start . "NUMBER: " .$num_item . " ROWS: " . $num_rows; //DEBUG
Displaying Block Of Text From MySQL Database
I would like to create a details.php page that displays all my products. I understand how to get the title, price, description etc from mySQL. But my problem is how do I get a list of data? For example say I want to display the features of a product in point form (list). What is the best way to do this? How do most common websites do this? Code:
Displaying Data From Mysql Database In A Drop Down List On Form.
I'm trying to do is display data from two different mysql tables from the same database in a drop down list on a html form. I have a fixtures table with the player1(userid), player2(userid), gameid, game, score1 and 2, what I want is to use the userid to get the players first name and surname from the members table (as it is a unique id), I need to do this bit before displaying it in the drop down. I think i need 2 querys to do this but when I have tried it it just echo's a blank value or the userid not the forname and surname that I want. I'm using the fetch_array function but just can't see where I'm going wrong, Code:
Multiple Database Connections
I have a simple update form that creates a MySQL connection with a DB to display on the form all previous user preferences. I also need to dynamically populate various drop down lists (select objects) by running entirely different SQL commands based upon data in different tables in the same DB. I would need to set up more DB connections in order to do this. Q1. Can I use the same connection ID ? ie can I avoid re-connecting to the same DB ? Q2. Will not multiple connections like this degrade performance ? Q3. Is there an easy way of doing this ? (running multiple queries)
Preventing Multiple Database Entries.
I am trying to make my form processor detect if an item name submitted to the database already exists I would like to display a message saying the item is already in the db. I tried something like this if ($Item_Name == "$row["Item_Name"]") { print "That item is already in the database"; } Now I know this is wrong but if someone could point me in the right direction it....
Multiple Database Updates From 1 Form
I am writing a very simple script that pulls a resultset out of a database. These rows are then made into form variables where the user has the option to update multiple entries, hit submit and the system updates the appropriate fields. I am using the convention of "variable[rowid]" in the input form, it allows me to loop through all the results and update the database. This solution is working fine, but I hate the idea that I'm updating data that isn't changing. I've considered keeping state when a form is shown, then comparing the values between the original state, and if its different then update the database. I hate to re-invent the wheel, how is this handled in other applications? I've tried to divine it out of some other programs, but I'm having trouble finding an application that is doing something like what I am doing.
Multiple Database Crosstab MYSQL & PHP
I've got two databases and I'm trying to look with values from one table in one database and then look up a name from the other database, the two tables are linked where 'postedby' in my newsdb database refers to 'ID' in my tblemployees table in my basedb database. All I get is one name shown and 6 records where there are actually 7. Code Below: -------------------------------------------------------------- <?php require_once('Connections/newsdb.php'); require_once('Connections/basedb.php'); mysql_select_db($database_newsdb, $newsdb); $res_newsdb = mysql_query("SELECT * FROM tblnews ORDER BY newsdate DESC"); mysql_select_db($database_basedb, $basedb); $row_newsdb=mysql_fetch_object($res_newsdb); $res_basedb=mysql_query("SELECT * from tblemployees WHERE ID=$row_newsdb->postedby"); ?> <html> <link href="body01.css" rel="stylesheet" type="text/css"> <body> <table width="500" border="0" class="body01" style='BORDER-COLLAPSE: collapse'> <tr> <td>Latest News:</td> <td><div align="right"><a href="../phpbb" target="_MainFrame">Bulletin Board</a></div></td> </tr> </table> <p class="body01"> <?php echo ("<TABLE BORDER=Ƈ' WIDTH=450 style='BORDER-COLLAPSE: collapse'>"); $rows=1; while($rows <= 10) { $res_news = mysql_fetch_array($res_newsdb); $res_base = mysql_fetch_array($res_basedb); $ndate = $res_news['newsdate']; echo ('<TR><TD><FONT SIZE=-2><div align="center">' . $ndate . '</div></FONT></TD><TD>' . $res_news['news']) . '</TD><TD>' . $res_base['IDname'] . '</TD></TR>' $rows++; } mysql_free_result($res_newsdb); echo('</TABLE>'); echo("<BR /><P><A HREF='addnews.php'>Add news</A></P>"); ?> </p> <p><span class="body01">To email the website manager click </p> </body> </html> ---------------------------------------------------------------------
Multiple Page Gallery Without Database
I have 20 images and 20 thumbs. same filenames but in different folders. I have 20 thumbs per page, but if i want to have more, there has to be a link to 2nd page, which shows thumbs from 021.jpg till xxx.jpg. If possible also third and fourth pages. I dont want to use a database. The questions is how to achieve this?
Searching My Database For Multiple Fields
I have a number of different tables containing a number of different fields and i want to give people on my site an advanced search. The easiest way to explain this is for you to think of it as an estate agent's website. People visit the site and search for houses based on a number of different criteria, for example they would have a number of diffent boxes they could narrow a search down such as 'number of rooms', 'type of property', 'price' etc... Now this is similar to my site, I want people to b able to choose to narrow there search down but not be forced to enter information into all boxes. So i want a search where a user would be able to select the number of rooms they want a property to have but if they choose not to enter that information then the intial value of '0' (or null) would be submitted to the search query. A user should be able to enter number of rooms AND/OR property type AND/OR price. IF they dont enter that information then a value of '0' (or null) would be submitted for each of the options to the query however if they do select a value then this would form the query. This is where im struggling. After racking my small brains and trying to think logically about how i would do this, my logic tells me (and correct me if im wrong) that if i choose too insert each variable 1 at a time into a search query such as $search_result = mysql_query("SELECT * FROM tablename WHERE rooms='$rooms' AND type='$type' AND price='$price'",$database); ...then if a user hadnt entered how many rooms but had entered 'flat' as the property type they wanted then the sql would be searching the 'rooms' field for 0 (which it obviously WONT find) AND 'type' for 'flat' and will return no result even if a 'flat' did exist in the database What I have considered is building up the query as i go, so for example. Code:
Multiple Record Form To Database
I have a form that has header information on the top such as dates, city, state, event, etc. The detail information of the form has several rows, 32 to be exact. Each row has 4 fields (name, gender and association). I named them 1name, 1 gender, 1 association for row 1. 2name, 2gender, 2association for row 2, etc etc etc. I need each row to be a separate record within the sql database but also carry in the header fields.
Multiple Select From Database Where Clause
I have a website with a database full of category names...I want to be able to chose lets say 5 of these categories and use a select from database clause where I could simply enter the category names I want to use but am not sure how I would go about doing it. I can get the list to display just the Aprilia category using the code below: $data1->q("SELECT * FROM categories WHERE cat_name = 'Aprilia' ORDER BY cat_name asc"); However...how could I get my site to display Aprilia, Ducati, Honda, Suzuki, Yamaha?
ADODB Multiple Database Connection
I'm using ADODB for a PHP project and am having problems with multiple (MySQL) databases. I've looked through all the docs and am stumped. I know how to use PHP and MySQL and I still haven't got a clue why this isn't working. Code:
Updating Multiple Rows In Database On The Same Form.
Im writting a content management system that generates menu's / pages etc. A feature i want is for people to be able to order the menus themselves.. ive created a page.. that looks thru all the records in the mysql database and lists them and also displays a text field called "ORDER" ie. Menu Header 1 [ 1 ] Menu Header 2 [ 2 ] Menu Header 3 [ 3 ] .... and so on.. ([ ]represents a text field...) what i want to do now.. is be able to press submit and all the rows in the same table update. PHP Code:
|
TRACKER
