Re: Challenge: MySql result to drop down box

Jim S wrote:
I have made the two changes you suggested but am still not receiving
a result set. Let me repost my modified code.

<?php
$res=mysql_query("SELECT DISTINCT job_name FROM
oats_jobs_users_laborCode where user='$username' order by job_name");
echo "<select name=jobname> <option default='default'>Choose
One</option>";
for ($i=0; $row=mysql_fetch_row($res); $i++)
{
echo "<option value='$row[0]'>$row[0]</option>";
}
echo "</select>";
I think I have narrowed it down to the for loop causing the problem...

Well, let's build some error handling into it.

<?php
$username = mysql_real_escape_string($username);
$res=mysql_query("SELECT DISTINCT `job_name`
FROM `oats_jobs_users_laborCode`
WHERE `user`='$username'
ORDER BY `job_name`");
if($res===false){
echo 'Error in query: '.mysql_error();
} else {
if(mysql_num_rows($res)<1){
echo 'Query did not match any records.';
} else {
echo "<select name=jobname><option selected='selected'>Choose
One</option>";
while($row=mysql_fetch_row($res)){
echo "<option value='{$row[0]}'>{$row[0]}</option>";
}
echo "</select'>";
}
}
?>

--
Rik Wasmus


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