Php Script failing to continue after a for(;;) loop

Question

I have a PHP script that queries a database and fills out some <option> tags inside a <select> tag. See code below:

$stmt = sqlsrv_query($dbc, $tsql, $params, $dbcOptions);
if($stmt === false) {
        die ( print_r (sqlsrv_errors(), true));
}
$rows = sqlsrv_num_rows($stmt);
#echo $rows.'rows';
echo '<!--'.$rows.'-->';
echo '<select onchange="getNamesByDep(this.value)">';
echo '<option value="">Select a Department. . .</option>';

for ($i = 1; i <= $rows; $i++)
{
        if(sqlsrv_fetch($stmt) !== false)
        {
                $DepName = sqlsrv_get_field($stmt,0);
                echo '    <option row="'.$i.'" value="'.$DepName.'">'.$DepName.'</option>'."\r\n";
        }
}
echo '</select>';
echo 'Debugging';

It never echos </select> or Debugging to the HTML of the page.

I know that I configured $dbc, $tsql, $params, and $dbcOptions correctly because I am getting the desired results from my query.

Do I have a syntax error that my web server (WIMPServer) isn't catching?

Also, make sure there are no `"`, `<`, or `>` in any values of `$DepName`. — elixenide, Jul 25 '14 at 17:13
@EdCottrell I found this in the error logs: Use of undefined constant i. Now I have a PHP plugin to shutdown before it fill my hard drive with errors being generated by an infinite loop. — Joel Trauger, Jul 25 '14 at 17:18
@Fred-ii- It's a bit of debugging to see if any rows are being skipped. I will remove it in the final version. — Joel Trauger, Jul 25 '14 at 17:24
I understand. However, I take (posted) code as being literal. Anything out of the ordinary, I question ;-) — Funk Forty Niner, Jul 25 '14 at 17:26

score 3 · Accepted Answer · answered Jul 25 '14 at 17:16

3

I think I found it:

for ($i = 1; i <= $rows; $i++)

You forgot the dollar sign before the second i. It should be like this:

for ($i = 1; $i <= $rows; $i++)

answered Jul 25 '14 at 17:16

Mariano D'Ascanio

1,202
2
16
17

1

I was just about to comment on that lol @EdCottrell however am questioning ` – Funk Forty Niner Jul 25 '14 at 17:19
@EdCottrell Edited my comment above. Seems like invalid HTML. – Funk Forty Niner Jul 25 '14 at 17:22
@Fred-ii- That's also true, but it seems the syntax error is what's killing the output entirely. An invalid attribute shouldn't break the output like this. – elixenide Jul 25 '14 at 17:24
It filled my error log. 256MB and climbing...I hate infinite loops. – Joel Trauger Jul 25 '14 at 17:26

score 1 · Answer 2 · answered Jul 25 '14 at 17:21

1

you missed a dollar sign:

for ($i = 1; i <= $rows; $i++)

should be

for ($i = 1; $i <= $rows; $i++)

answered Jul 25 '14 at 17:21

Asterisk

96
2

score 0 · Answer 3 · answered Jul 25 '14 at 17:23

0

Discovered the error.

Instead of:

for ($i = 1; i <= $rows; $i++)
{
        etc
}

I needed to have:

for ($i = 1; $i <= $rows; $i++)
{
        etc
}

answered Jul 25 '14 at 17:23

Joel Trauger

720
1
4
24

2

*"Discovered the error."* So did a few others before posting this. You should accept the first answer given, mainly Mariano's – Funk Forty Niner Jul 25 '14 at 17:23
I accepted it. I discovered the error after Ed asked me to check the error logs. I thought I was posting an answer before anyone else. – Joel Trauger Jul 25 '14 at 17:30

Php Script failing to continue after a for(;;) loop

3 Answers3