assigning value from a list box to a variable

[wombatwoo]

Hi I am very new to php and mysql, so I'm sorry if I sound incredibly thick!

I am trying to assign the selected value from a list box to a variable, so that I can retrieve the related value from a database using another query statement. I have populated a list box with stock codes, so that the user can select one. When this is selected, I would like the related stock description to appear in an adjacent text box, or something similar.

This is the code I have used so far for the list box:

<?php

$link = mysql_connect("localhost", "root");
mysql_select_db("invoice");

echo "<table border=0 width=50% align=center cellpadding=2>
<TR>\n
<TD align=center><B>Stock Code</B></TD>";

echo "<TD> <SELECT name=item>";

$query = "SELECT stockcode FROM stockitem";
$result = mysql_query($query);

while ($line = mysql_fetch_array($result))
{
foreach ($line as $value)
{
print "<OPTION value='$value'";
}
print ">$value</OPTION>";

} "</TD> </TR>";

mysql_close($link);

print "</SELECT>";

echo "</table>";

?>

Thanks for any help you can offer.

😕

[DeltaRho2K]

I may be missing something, but I don't see your <form> tag. Above your while loop (like above your table tag), add something like:

echo "<form action=\"page_you_are_posting_variables_to.php\" method=\"GET\">";

Then after your closing table tag, add

echo "</form>";

[Jona]

[font=arial]Well, you might want to try...[/font]

foreach ($line as $value) 
{ 
print "<OPTION value='$value'>$value</OPTION>"; 
}
echo "</TD></TR>";

[font=arial]Instead of what you have--because if it's not in the loop, it only prints once. Thus a very odd looking page. 😉

Hope that helped.[/font]

[J]ona

[DeltaRho2K]

It sounds like his drop down list is filling in fine (based on the first post), but rather he is having troubles passing the variable... I could be mistaken. (Wouldn't be the first time ;-)

[Jona]

Originally posted by DeltaRho2K
It sounds like his drop down list is filling in fine (based on the first post), but rather he is having troubles passing the variable... I could be mistaken. (Wouldn't be the first time ;-)

[font=arial]Well, actually you saw what I missed--and I saw what I believe to be causing an error, although I could be wrong. I never tried just puting "</TD></TR>"; and expecting it to echo it. :rolleyes: [/font]

[J]ona

[wombatwoo]

I've got slightly further with my quest!! I've managed to populate the listbox and post the results to a new page, displaying certain items from the database that relate to the result.

However, I want to display these results on the same page as the original listbox. It won't let me use include or require with the new page, because of reusing class objects.

Any brainwaves?

Thanx
Wombat

[wombatwoo]

Sorry, forgot to post the code!

$query = "SELECT stockcode FROM stockitem";
?>
<form action="menu.php?link=coderesults.php" method = "post">
<?
listbx($query);
?><input type=submit value="get">
</form><?

//function
function listbx($query)
{
?><select name = "item"><?

global $result, $value;
$result = mysql_query($query);

while ($line = mysql_fetch_array($result))
{
foreach ($line as $value)
{
print "<OPTION value='$value'";
}
print ">$value</OPTION>";
}
return $value;

}

//coderesults.php page

$query = "select stockcode, stockitem, price from stockitem where stockcode='$item' ";
$result = mysql_query($query);

while ($row = mysql_fetch_array($result))
{
echo "stock code = {$row['stockcode']}<br>";
echo "stock item = {$row['stockitem']}<br>";
echo "price = {$row['price']}";
}