How Can I Populate A Listbox With The Result Of Mysql Query?
I would like to populate a listbox with the result of a mySQL Query.
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Populate Result Query
I query database and it produces rows of results. I want to populate it so that the first 10 rows go to the 1st page, the next 10 rows to the 2nd page and so on. The last page will be at most 10 rows and it can be less than that. How do I do it? Or do anybody know if there is any sample code that I can refer to?
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Code To Populate Listbox
I'm trying to create a php script that creates a listbox and populates the list with values from a table named Inventorytable. I only need the values from the ItemName field to be populated in the listbox. Here is what I have so far.. <? $conn=mysql_connect("127.0.0.1", "odbc", "") ; mysql_select_db("php012",$conn) or die ("error cannot connect"); $sql = "select itemname from inventorytable;"; $result = mysql_query($sql,$conn); while ($array = mysql_fetch_array($result)) { print $array[ItemName];} <html> <SELECT > Â <OPTION VALUE="01">Option 1 Â <OPTION VALUE="02">Option 2 Â <OPTION VALUE="03">Option 3 Â </SELECT> </html ?>
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Mysql Query Result
I have successfully selected 4 rows using a mysql SELECT query. However, I would like to random pick a row from this result so that every page refresh will yield something different.
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Same Mysql Query Doesn't Always Return A Result
We have no access to a mysql NG on my provider's server, so we ask here: We have a long query (long in text) with a UNION between 2 select. We have been informed that some times the query doesn't return any result. We have tried on our server and we always get a result. BUT, trying on the hosting server, many times the query doesn't return any result and doesn't get any error. Any idea ? does Union have any problem ? how to check if the query failed ? My hoster said that sometimes the table can be locked (if the server is overloaded) and then mysql doesn't return any result, but this seems to me an aberration. the query takes about 0.0050 sec to execute when it doesn't return any result, and 0.030 when I get results
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MySQL Query Result Rows
Still learning the ropes of PHP/MySQL here so pardon if this is something from a 101 class. I am wanting to know if/how I can query a MySQL database, pull two rows of information and assign them individual variable names. I know this isn't too clear. Let me try to explain what I want to do. Let's say I have something similar to this: Code: $query = mysql_fetch_assoc(mysql_query("SELECT entry_id, user_id, date FROM my_entries WHERE user_id = '$user_info[user_id]' ORDER BY date DESC LIMIT 2")); What I would like to be able to do is assign separate variables for the info on the two rows returned by the query. I want to have the following variables: $first_date = $query[date]; <= from the first row $second_date = $query[date]; <= from the second row I can then use the two variables to compare to each other, etc. Can this be done with a single query? Or are two queries required (if so how)?
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Processing MySQL Query Result
after executing the query, say it returns 1000 results. if i just want to do some sampling and like to have a look on 4 of them, how can i refer to these 4 random records out of 1000? i'm working on processing the result of $res_array which has 1000 records. query = select * from test; result = mysql_query(query); records1000 = db_result_to_array($result); //how to select/remove records1000 to get say, records4? function db_result_to_array($result) { $res_array = array(); for ($count=0; $row = $result->fetch_assoc(); $count++) { $res_array[$count] = $row; } return $res_array; }
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How To Get Mysql Query Result Into Temp. Txt File
I did a lite search in the archive but didn't find anything specific to my problem. Here is what I am trying to do: I've got a table with about 25k entries of company addresses. I need to be able to pull specific data out of the table and put it in a temporary text file and then display the result in the browser for the user to save locally. The query part and the format of the data is all simple enough but I am having trouble figuring out how to get the data from the query into a temporary text file and then displaying that file. here is what I whipped up but it obviously doesn't work: Code: <? $tmpfname = tempnam("/path/to/file/", "FOO"); $list = mysql_query("select * from FH_LIST where State='AK'"); while ($list_results = mysql_fetch_array($list)) { $data = blah blah blah; fwrite($tmpfname, ".$data "); } fopen($tmpfname, "r"); fpassthru($tmpfname); ?> I get an error message saying: Warning: Supplied argument is not a valid File-Handle resource (referring to $data).
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Can I Put A Mysql Query Result Directly Into A Multidimensional Array?
I have two joined tables: Departments - which contains the department description and a key Positions - which contains position data What I'm attempting to achieve is to display the department title in one table cell and then list all the positions associated with that department in the cell under neither the title. I need to generate a table two column table with as many rows as required, depending on the number of departments, which is dynamic. Do this I'm trying to turn the result of my SQL query into multidimensional array e.g. dept = VioP positions = "designer", "engineer" etc, which I can do then use to generate the table. I've done a search on the forum and looked around the manual and I'm stumped on how to achieve this, a point in the right direction would be greatly appreciated. The query I'm using is below. PHP Code:
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Send Mysql Query Result Displayed In A Table
I have just created a shopping cart and I would like to send my user the receipt of there order by email. The receipt is generated by a mysql query and is displayed within a table. I have worked out how to send a html using simple HTML code as my “$message� variable: $success = mail($to, $subject, $message, $headers); But I can not get it to send my mysql query result displayed in a table. How can i create the result of my mysql query and make it a simple variable i.e $message.
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How To Use Selected Data In The Listbox For My Query..
I have a multiple select listbox and I pass the data using array in php. How can I use these records to my query, or in one select statement. For an example: I have a listbox of item codes; the user selects an item code/s to view it's inventory. The selected itemcodes will be pass on php using array and these records will be used as a query or search items on my database.
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Create A Query In Order To Dynamicly Popular A Listbox
I am trying to create a query in order to dynamicly popular a listbox however I am trying to fill the listbox with years such as 2007 etc based on if they are stored in the data base. The problem is that two tables have to be checked that have no relation to eachother but this caues the problem of duplication so if both tables have 2007 it will appear in the list more than once. Code:
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Listbox Contents Dependant On Selection In Previous Listbox
I need to create a form so that when an option is selected in listbox1, the contents of listbox2 change depending on what option has been selected. I'm afraid I don't know where to begin with this one, I can create the listboxes so they populate dynamically by data from a database - but thats it!
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Parse A Query String To Populate Form...
I have created a Content management System that writes content to flat files. I have a single PHP page that calls/opens each flat file and places its content all around the page. I want to send a URL to people such as (www.??????.com/whatever.php?flatfile&firstname=mike&lastname=jones) where the "whatever.php is the landing page, the "flatfile" is the name of the text file that supplies the content, the "firstname" and lastname" will populate a form found on the php page. Can this be done? And if so, can you help me out? Code:
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Turn Select Query Result Into Hyperlink To Other Query
I have a query which gives results of selecting coursenames from a table called trainingtopics but this does so in a continuous bulk of text without any breaks between each record. my question how do I create line breaks between each record and also how can I force each query result to become a hyperlink which when clicked runs another query that gives details of that course. <?php $user = "root"; $host = "localhost"; $password = ""; $connection = mysql_connect($host, $root, $password) or die ("Couldn't connect to server."); $database = "courses"; $db = mysql_select_db($database) or die ("Couldn't select database."); $sql = "SELECT coursename FROM trainingtopics"; echo $sql."=sql<br>"; $result = mysql_query($sql) or die(mysql_error()); echo $result."=result<br>"; while($row=mysql_fetch_array($result)) { // NOTE this one ABOVE the echo echo "result found!"; echo $row[0]; } ?>
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Listbox And Multiple Php/mysql Queries
i've run into a bit of a problem which i can't seem to solve. Basically i'm making a forum and i'm working on a section that enables you to view information about the other users signed up the forum in one big table. I have 6 columns and a listbox that enables you to sort the data regarding the selected item chosen (ie. if postcount is chosen the table will be sorted by post count) Anyway, here's my code for the listbox: Code:
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Speeding Up Query/code (query Within Result Set)
I'm looking for ideas on how to speed up this script. Basically it finds all the zip codes in a zipcode table, then looks for all the records in another table with those zip codes. Right now it finds all the zips then within that WHILE, it looks for a record in another table with that zip: PHP Code:
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Adding Multiple Listbox Items To Mysql
I want to be able to select multiple items from a listbox in a form and send the result to mysql blob. I only get one item to the bd. Here is part of the script. Item: <select name="itm" align="bottom" multiple size="2"> <option value="item1">item1</option> <option value="item2">item2</option> <option value="item3">item3</option> <option value="item4">item4</option> <option value="item5">item5</option> </select> "Insert into xyz set itm = '$itm'";
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Multiple Select From Listbox And Execute A Sql Query Select Statement
Supposing i have a html form with 2 listbox and i name it status and resolution. So it goes <select name='status[]'>... <select name='resolution[]'> Then i have a another php page. $status = $_POST['status']; $resolution = $_POST['resolution']; Now my question is, how do i do the sql query statement? mysql_query("SELECT * from bugs where status=$status and resolution=$resolution"); The problem is, i select 3 options form the status listbox eg. new, closed, duplicate. And from the resolution, i chose fixed, invalid.
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Use Query Result Field As Query Key For New Query..
$query  = "SELECT ProjID, UserID, ProjDesc, file, OrigProj, OrigUser, ProjDate FROM projects"; $result = mysql_query($query); I would like to take the field 'UserID', and utilise it for a Query statement to my users table in my database, to read and fill in information to my table, which looks like this: PHP Code:
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Using Query Result To Do Another Query In Loop
I have 2 tables, one called users2 and one called books. Books has a field called UserId, which is the ID of the user that added the book to the database. My problem, is that i need to take this ID from the book table, and use it to get some information from the user table. Here is my code to get the ID from the book table: Code:
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Populate Form From MySQL
I am kinda confused why the states select FORM will not populate from MySql. PHP Code: <tr>               <td class="label">State:</td>               <td class="field">                 <select name="state">                 <?php if (isset($_POST['state'])) { ?>                   <option selected value="<?php echo $_POST['state'] ?>"><?php echo $_POST['state'] ?>                 <?php } ?>                 <option value="">                 <?php                   $result = mysql_query("SELECT abbrv, states FROM wsd_states ORDER BY states");                   if ($result) {                     if ($frow = mysql_fetch_row($result)).
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Using PHP And MySQL To Populate (select) Menus
I am a newbie to php *gasp* and I am trying to make a <select> drop down menu, but I want the menus populated by information from a field in my database. There are a bunch of entries in the field MUD, that are identical......I would like the menu NOT to repeat identical names. EX... Each name only once, even though it is in the database many times.
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How To Populate HTML Form From PHP/MySQL
I have found many examples of how to access an HTML text field from PHP, but cannot find the reverse - how to load an HTML form from a MySQL database. Specifically, I have an HTML form which contains several text boxes that correspond to the fields in a MySQL file. What I want to do is call a PHP form that reads one record from MySQL and then populates all of the HTML fields on the calling form. The call to MySQL is not a problem - that works and I can use "Echo" to verify that I have successfully read a record. Now, how do I get that data back into the original HTML form?
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Populate Array With Mysql Table
What I would like to do is populate an array with some information pulled from a mysql table. as an example lets say I have a table that shows: id ¦ name ¦ nickname what I would like to do is have my array as follows: $name_ar = array( 'name1' => 'nickname1', 'name2' => 'nickname2', 'name3' => 'nickname3', /* and on and on */ ); Arrays are not my favourite things, some help?
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Populate Drop Down From Mysql With Php Part II
I have this working sort of, but I still can't figure out why it is skipping the "uniqid" field. Does it not work if the field is an auto_incrementing INT value or something? Also, where do I add the "form action" elements to this code to allow the user to actually submit after selecting? ....
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Populate MySQL With Multiple Text Box Entries
I have about 300 rows of mysql data which I have chosen to diplay with a text box at the end of each row, that is available for adding comments. I would like to be able to add comments to entries and then have an update scripte that would take the text box, named "MC" and place the data in its row identified by a hidden field which would corrrespond to the Primary Key of the MySQL Database. I've done something similar using check box arrays, ie. Checkbox name="MC[]" My question is whether I can write a script that would test for a value in the corresponding text box, and update the MySQL data accordingly. I have also provided in my display script that if the MC column is already populated in MySQL it will not display in the form.
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Populate A Calendar Page With MySQL Data
I'm trying to figure out the best way to populate a Calendar page with data from a database. I have the code that will generate the Table with Days of the week, but I'm at a loss on how to populate this with data from a database. Do I build the calendar and then for every day that I generate a cell, query the DB or should I load the DB in an Array and compare that to the date of the cell?
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Result From A Sql Query
Is there anyway I can exclude just one result from the database based upon the value of a field, for example i want to exclude the field with an ID of 1 but get all the other fields in the database. is this possible and can anyone tell me how? PS: I had thought about doing something like $query1 = mysql_query("SELECT * FROM table ORDER BY id ASC"); while ($row = mysql_fetch_array($query1,MYSQL_ASSOC)) { $id = $row['id']; if ($id == 1) { //DO NOT DISPLAY ITEM } else { //DISPLAY ALL OTHER ITEMS } }
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Result For Query
I dont know why this query is not returning a value, when I should be getting the value of ཋ'. PHP Code: mysql_select_db($dbname, $conn) or die (mysql_error()); $query = "SELECT MIN(q_id) FROM questions2 WHERE quiz_id='".$quiz_id."'"; echo $query.'<br />' $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($result) or die(mysql_error()); $first = $row['q_id']; echo 'first='.$first.'<br />'
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Linked Result From A Query
I'm trying to create a query that will return a linked record from a database that will take me back to an update page with that record so I can change things in the record. I'm having trouble finding simple code to do this.
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Echo A $result Of Query?
Trying to have an echo display a result from a query. Not such a hard concept. Yet I am having trouble. What good is the data without analysis. Here is what I am using to try and grab result. Have been told it looks correct, must be missing something small. When this runs, alternate echo displays in browser. So you know: 'testdata' is the table 'score' & 'id' are columns or fields. Almost think it is missing a table structure to display in. Looking for player name column and score column (for the sake of example). Code:
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Paging Of Query Result
This script used to be working in splitting query result into pages. But now, PREV 1 2 3 4 NEXT links is not working anymore. Any idea how to fix this ? or what is wrong with this script ? Code:
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Filtering In Query Result With WHERE
I am trying to filter out records in a query using WHERE expressions. See example below. if ($league="All") {$leaguebool="1"} else {$leaguebool="League='$league'"}; $query = "SELECT League,HR FROM ranfranbattingsim WHERE ('$leaguebool') ORDER BY $sorder DESC LIMIT $limit"; My desire is to select all records if $league='All' and select only records where LEAGUE=$league if $league is anything other than 'All'. LEAGUE is one of the fields of the database. I hope that I explained that well enough. I am open to any solutions, and have considered that perhaps I should be using something other than the WHERE statement to filter the records.
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One Row Result Database Query
Most of the pages I write are multi-row queries. I have about 30 pages that will only yield one row of about 15 columns. What is the best way to design this page? Right now, I am using $row = mysql_fetch_object($results) - this is causing a minor issue with positive/negative signs with numbers.
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Query Not Returning First Result
I have a query that is returning the correct number of rows. However, when I try to print the rows out, it will not print the first row, but all after that. I can't figure out why it isn't printing out the first row, even though the query is returning the results. Here is my code below to print the results in a table: Any suggestions? while($row2 = mysql_fetch_array($result1)) { $affinity_name = $row2['affinity_name']; $affinity_detail = $row2['affinity_detail']; $affinity_no = $row2['affinity_name']; $track_id = $row2['id']; get_name($affinity_name); $affinity_details .=<<<EOD <tr> <td><a href="/m_t5.php?affinity_id=$affinity_no&q=$name">$name</a></td> <td>$affinity_detail</td> <td><a href="/delete_track.php?track_id=$track_id">Delete</a></td> </tr> EOD; }
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If Query Result Returns Nothing
i have a marquee which scrolls across news items, all this is database driven, however if no news items are returns from the query, my page errors, can someone help me write an else statement or something so that the page doesn't error. my code is below:- <?php require('database3.php'); if(!isset($news_item)) $news_item='' $query = "select * from monksc.news where DATE_SUB(CURDATE(),INTERVAL 14 DAY) <= startDate order by id"; $result = mysql_query ($query) or die (mysql_error()); if (mysql_num_rows($result)>0) ..............
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Displaying Result Of A Sum Query:
$sql = "SELECT sum(value) FROM `offers' WHERE sid='$sid'"; $result = mysql_query($sql); As you can see, I'm trying to select a column in my database which is named "value" I want to return the sum of the records which match the current user's subscriber id ($sid) I think the query is structured correctly. However, I'm having trouble placing the result of the query into a variable that php can echo to the page. In the database, the numbers are formatted as decimal. I just want to get the sum of the fields into a numeric php variable that I can add/subtract/divide etc and echo to the page. How do I go about doing this?
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Weird Query Result
I am trying to do a simple query, but instead of a numeric value, I get "Resource id #34". $newdodgemod1 = "select default_dodgemod from phpbb_classes where class='$newclass'"; $newdodgemod2 = $db->sql_query($newdodgemod1); Any idea what is wrong with the code? All field exist and the variables are set correctly. I don't understand why I am getting this error.
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Query Result To Array
i have a table with two field ID and CAPACITY. i want the ID to point to the capacity on an array $top = "SELECT * from topics"; $db = new mysqli('localhost','root',?'); $db -> select_db('testing'); $res_top = $db->query($top); $topics = array(); while ($row = $res_top->fetch_row()) { $topics[$row['id']] = $row['capacity']; }
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Query Result Values
i have two tables a customer and package table. a customer can have many packages. how i can i set a status in the customer table to 'complete' if all the packages in the package table that relate to that customer are set to 'paid'? i can do a simple select statement but how will i know if all the status rows returned from packages are set to 'paid'.
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Searching A Query Result
I have a simple query that selects everything from a database and limits the results to one row ordered by desc. The code does work. The next step is to search the query result to see is if a certain feild has a result. <?php Include("config.php"); $con; if (!$con) { die('Could not connect: ' . mysql_error()); } $result = mysql_query("SELECT * FROM photos ORDER BY photo_id DESC LIMIT 1"); while($row = mysql_fetch_array($result)) { echo $row['photo_id']; } ?>
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Query Doesn't Show First Result
I've created this query but it never shows the first result. I can have any number of results and order them however I like but whatever the first result in the list is it doesn't appear. Code:
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Query Result Display
PHP Code: <html> <head><title>Display Table</title></head> <body> <?php $database="temp"; /* DB name */ $host="localhost"; $user="root"; /* Set DB Username */ $pass=""; /* Set DB password*/ mysql_pconnect($host,$user,$pass); /* DB connect....*/
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