[Resolved] mysql help

[lukeurtnowski]

Here's the problem, I have two combo boxes, the first one yas a name of year and lists some years, once a selection is made from the list (onchange=submit()), the second combo box should be populated based on which year was selected, thanks a lot

[rjl]

<form method="post" action="<?= $_SERVER['PHP_SELF'] ?>">

...
<select name="combo1">
<option value="...">...</option>
...
</select>
<?
if (isset($POST['combo1']))
{
switch ($POST['combo1'])
{

case "1999": $combo2_values = array("...","..."); break;

case "...":
}
?>
<select name="combo2">
<?
foreach ($combo2_values as $value)
{
echo "<option>$vales</option>"
}
?>
</select>
<?
}
?>
</form>

[TheDefender]

Where is the data coming from to populate the data in the second list? It sounds like a MySQL database given the title you gave, so my example goes as follows:

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">

<select name="Year" onchange="submit()">
<option value="2001" />
<option value="2002" />
<option value="2003" />
<option value="2004" />
</select>

<select name="otherlist">
<option value="Please Select" />
<?
if(isset($_POST['year'])){
  // Insert DB Connection script here
  $getlist = mysql_query("SELECT field FROM table WHERE year = '".$_POST['year']."'"); // Change according to your needs
  while($row = mysql_fetch_array($getlist)){
    echo "<option value=\"".$row['field']."\" />";
  }
}
?>
</select>

[lukeurtnowski]

K, her'e the php code im using

$year = $_GET['year'];
$query1 = mysql_query("select Make from MAKE where
        ((yr1=='$year') || (yr2=='$year'))")
        or die(mysql_error());

Here's the html code to display the first combo box, and for it to submit itself when the onchange event is triggered
<select NAME="year" onchange="submit()">
<option value="">Select Vehicle Year</option>
<option VALUE="2005">2005
<option VALUE="2004">2004

Lastly, here is the php code to go into the mysql database and display the 2nd combo box (yr1, and yr2 are year fields in the MAKE table

<?php
				while ($result1 = mysql_fetch_array($query1)) { 
				printf('<option value="'.$result1['Make'] . '">'.$result1['Make'] . '</option>'); 
				}
			?> 

thanks

[TheDefender]

OK, so is this working for you? If not, let us know what the problem is so we can help. If so, be sure to mark this thread resolved.

[tsinka]

Hi,

if the method of your form is POST then you need to use $POST instead of $GET.

Two questions:

1) do you have a submit button in your form and what is it's name ?
2) does the form submit at all if you select a year ?

Thomas

[TheDefender]

Just a side note, you should look into coding for XHTML compliance too... Basically put, when a tag doesn't have a closing tag, or the closing tag is optional (like <option>), you should close that tag like so:

<option VALUE="2005" />

(Note the forward slash ( / ) before the >...

[lukeurtnowski]

This is the error I get when I run the page.
You have an error in your SQL syntax near '=='') || (yr2==''))' at line 2

[tsinka]

Hi,

don't use == in SQL queries, just =.

Thomas