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.
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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: " . $num_rows; //DEBUG
Displaying Columns With A Value
what is the most efficient way to display column names from a database that only have the value '1' in its fields?? I cant seem to work it out, could someone give me a code example??
Displaying Rows Of Data As Columns?
Does anyone know of an efficient way to retreive data from a mysql db and display the record rows as collumns? I have been doing this: <display headers> <while ($row = mysql_fetch_array($result)) { > <display row 1 of result> <display row 2... } > <end table> but I want to do this instead: <col 1 header > <row 1 col 1> <row 2 col1> <row3 col1> <col 2 header> <row 1 col 2> <row 2 col2> <row3 col2> <col 3 header> <row 1 col 3> <row 2 col3> <row3 col3>
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Is there an easy way to display in my browser window all of the tables and their corresponding column names for a selected mysql database?
Displaying Information In Prev/next Link?
I have Prev and next link in which it takes information from a Customer table in the database. It supposed display the information in a table but the information doesn't show. The Prev links show. PHP Code:
Displaying Information In A 1 Row/2 Column Table
I'm very new to php scripting and haven't quite figured out what I need to do here. I need to display information in a 1 row/2 column table. I have the MySQL query working fine, but the scripting part in php is killing me! In the first column a graphic about 70x70 will be shown (plus some additional padding). In the second column will be the description information. The second column will be around 400 pixels wide. How do I set this up in php? It's easy enough in HTML, but I've about killed myself trying to figure out how to properly lay this out in php!
Selecting And Displaying Information From 2 Tables
I have 3 tables in my database: The client table holds the client name, address, city, state, zip, sub_category and category (and a few other fields but that is not important). In the sub_category and category fields in the client table are just numbers which match the id numbers in the other 2 tables. the category table holds the id number and title the sub_category table holds the id number, title and category (which is just a number that matches the id number in the category table. Hopefully that all makes sense. Now what the problem I have run into is I have a table on my page that loads all businesses for a category. So if a user clicks on the Dining button on the home page a list of all restaurants will display in my table. Well I want the table to show the sub_category the business belongs to, however the field in the client table is just a number. How do I get it to show the text (title) of that sub_category? Code:
Displaying Information In A Update Form
I can display the information in a table. But now I need to display it in a form, so I can edit it and then upload it to the database. I have the form from my upload page, But I can not seem to get the information to appear back in it. Code:
Sending/displaying POST Information Without Using Forms
What I need to do is build onto an existing login form, using POST forms. I would like to have to have to access the existing code, so what I ultimately have to do is write a PHP script that will spoof a form POST. I have found 2-3 different examples, however none of these will send the post, and then go to that page. All that these scripts do is send the post, and keep you on the existing page. The return from the form cannot be displayed in the page with the script on it because of licensing issues. So the script needs to send the POST information to that page, just as a regular form.
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I am new to both php and xml. I have a school assignment to update an existing php-based website where I have to display search results from an online library. I can use a url search which returns an XML file that includes tags such as position, title, url, description, date, etc. I only want to display the title, url (preferably the title should be a link to the url), and description for all of the results on my website's search results (different template as well). What classes/functions are available to help me do this? I would appreciate any guidance as well as links to online tutorials etc. that would point me in the right direction. I have read about the general aspects of php, and I have experience with Java and C; but I don't know about xml at all. By the way, the website is running on the school server and has php version 4. I don't think I'll be able to install php 5 but I can probably upload some helper files or something.
Loading Information From Database
I need to load information from a database to a form to be modified. Below is the code to load the field title into a text field. Not working...... what is wrong with the code below? echo "Title:<input type="text" name="title" size="50" value=$row['title']"><br>";
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Now i know how to connect to access database & PHP. But i want to arrange the fields which i am retreiving from the access database... for e.g. I got 2 fields in the access database called name and namedetails. So I want to display name as hyperlink and below that name details . Now i am getting information from the databse but not in that way....how to do that?
Send Information To A MySQL Database
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I created a very simple form and a MySQL database to see if I could get the information submitted from the form to the database, and also a copy of the form submission emailed to me. I have managed to do that with the help of a few online tutorials. Now, I would like to collect the IP address and browser type of the person submitting the form and have this information transferred to my MySQL database, along with the other form field information. I would like this information to be submitted to columns in my MySQL database labeled "ip" and "browser" respectively. Code:
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How To Display Mysql Database Information In Multiple Pages?
how to display mysql database information in multiple pages? i would like to display all my member with consist of 400 people and i wan it list in mulitiple pages, and contain a "delete tick" for me,and when i click the member, can redirect me to member detail imformation. it just similiar to email pages of yahoo inbox.
Displaying From Database!
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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?
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Displaying A List Of Categories From A Database In A Select Box
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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 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:
Sortable Columns
Hi there, I am new to this Forums...I am currently learning PHP and I was wondering if any of you guys knows how to make columsn in a table sortable. What would I have to change in the code? Thanks!
UPDATE Columns With PHP
So you know I'm going through Kevins book. I'm updating data in a test table. I'm adding names to each joke, say I have 7 jokes and 3 of them go to the same person. I can update each one seperately but I can't find a way of bulk updating. In this case it's not a problem. I've looked though Kevins book and a couple of others but can't find anything. Tried several sites to but I may just be searching wrong. I can do this: mysql> UPDATE Jokes SET AID="1" WHERE ID=2; but I'd like to update WHERE ID equals 2,3,4,5 all at once.
Columns In Html
I am trying to make a table with two columns. I am really close. The problem is the table's first row has three TD's. ie one two there four five six seven PHP Code:
Sortable Columns
Does anybody know here how to make columns in a table sortable? I know that with a sql query I can accomplish that. PHP Code:
Grouping The Columns
Let's say I have a very simple Table: +---------------------+ | id | name | level | +---------------------+ | 1 | Adam | 1 | | 2 | Jack | 7 | | 3 | Bill | 1 | | 4 | Pete | 3 | | 5 | Ross | 7 | +---------------------+ but I want the PHP to output their id's and names, but group their Levels together like so: Level 1: Adam (id: 1) Bill (id: 3) Level 3: Pete (id: 4) Level 7: Jack (id: 2) Ross (id: 5) And then afterwards, If I create someone with a new level, PHP will automatically create a new heading (e.g. Level 10) and put the corresponding names under it. How should I approach this? Should my PHP contain two "while"s? Or do I use the SQL <group by> function?
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I've been using this code to split records from a db into columns when it displays the data. Items are shown with 3 on a row and then a new row starts. It works really well but if ever there is only one item on a row it causes me a problem in this instance. I have far left and far right columns which display a background image, which is a border. The problem is this only works if there are three columns visible in a row. Is it possible to manipulate the following code so that there are always three columns present, they're just empty if there's no more records to fill them? Code:
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I was wondering if there is a way to calculate the total of columns in a table? (something like $totalRows= mysql_num_rows($Recordset1); , but than for columns)
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I have several forms that display information from the database after users log in. I would like the column titles to be sortable so that when the user clicks on a column heading, the data re-displays in sorted order of that column. Next time they click on it, it toggles the order from ascending to descending, and then back to ascending again. Since I have many forms and each form has different columns, I was wondering if there was a method that would be better than a brute force approach. What are the best practices for this functionality?
Query Results In 2 Columns?
I'm struggling to figure out how to do this properly. I have done a mySQL query to extract a list of names, and done a mysql_num_rows to determine how many results I've gotten. I want to split the list in to two columns. I've created $percol = $mysql_num_rows/2 but I'm not quite sure how to "loop" through the first half of the list, echo </td><td> and then list the balance. Can anyone give me some guidance on how best to structure this?
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