Hallo All,

I m a bit new to PHP. I am having a mySQL database and I want to get a field value of this database to a combobox. How can I manage to do it? Can anyone help me?

Your attitude determines your altitude

__________________
Your attitude determines your altitude

You need a few different things. First you need to connect to the database, then you need to query the database and get the array you want to put into the combo box. Then you loop through your array and populate the combo box. Post how much of this you have already done so we know which piece you need answered.

Do you mean something like this (assuming you've got the connect piece down)?

echo "<select>\n";
  $query = "SELECT t.val,t.label FROM t1 AS t ORDER BY t.label ASC";
  $res = @mysql_query($query);
  while($row = mysql_fetch_array($res)) {
    printf("<option value=\"$row[val]\">$row[label]</option>\n");
  }
  mysql-free_result($res);
echo "</select>";

Yes that was it what I tried to do.

So I am having another problem.

First of all I am having a database named fernschule. It has 3 tables which are:

t_kurse
t_institute
t_ins_kurse

t_kurse includes course names and their IDs like KURSE_NAME KURSE_CODE
t_institute includes institute names which give these courses like INS_NAME INS_CODE
t_ins_kurse includes the foreign keys KURSE_CODE INS_CODE

What I am trying to do is to build a php page which takes the course names into a combobox and their IDs as OPTION value and search the institute(s) which includes the selected course.

So here is my first php page

<?php
/*
* Created on 12.11.2006
*
* To change the template for this generated file go to
* Window - Preferences - PHPeclipse - PHP - Code Templates
*/

$link = mysql_connect('localhost','root','admin');
$db = mysql_select_db("fernschule");
if (!$link) {
die('Could not connect: ' . mysql_error());
}



$sqlquery = "select * from t_kurse ORDER BY KURSE_NAME";
$res = mysql_query($sqlquery);

echo "<FORM ACTION = 'trial1.php' METHOD = 'POST'>";
echo "<SELECT NAME ='t_kurse'>";
while($dsatz = mysql_fetch_assoc($res)) {
$value = $dsatz["KURSE_CODE"];
echo "<OPTION VALUE=".$value.">".$dsatz["KURSE_NAME"]."</OPTION>";
}
echo "<INPUT TYPE = 'SUBMIT' VALUE = 'Suchen'>";
echo "<input type='reset' VALUE = 'Abbrechen'>";

mysql_close($link);
?>

So the trial1.php is

<?php
/*
* Created on 11.12.2006
*
* To change the template for this generated file go to
* Window - Preferences - PHPeclipse - PHP - Code Templates
*/

$link = mysql_connect('localhost','root','admin');
$db = mysql_select_db("fernschule");
if (!$link) {
die('Could not connect: ' . mysql_error());
}
$sqlquery = "SELECT t_institute.INS_NAME
FROM t_kurse INNER JOIN (t_institute INNER JOIN t_ins_kurse ON t_institute.INS_CODE=t_ins_kurse.INS_CODE) ON t_kurse.KURSE_CODE=t_ins_kurse.KURSE_CODE
WHERE (((t_kurse.KURSE_CODE)=".$_POST['value']."))";
$res = mysql_query($sqlquery);
while($dsatz = mysql_fetch_assoc($res)) {
echo "<p>".$dsatz["INS_NAME"]."</P>";
}
?>

So it gives error in SQL code. I tried to include the $value variable in the second file as a posted value but it doesn't accept it and gives this error.

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\Programme\xampp\htdocs\trial1.php on line 18

So how should I write this SQL query so that it takes the OPTION VALUE $value from trial.php and compares it in trial1.php? Can anyone help me?



Your attitude determines your altitude