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
– elixenide, Commented Jul 25, 2014 at 17:13
<option row? and you're using $rows <option row which btw, is invalid HTML. — Funk Forty Niner
– Funk Forty Niner, Commented Jul 25, 2014 at 17:17
@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
– Joel Trauger, Commented Jul 25, 2014 at 17:18

Mariano D'Ascanio · Accepted Answer · 2014-07-25 17:16:59Z

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, 2014 at 17:16

Mariano D'Ascanio

1,2002 gold badges16 silver badges17 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Funk Forty Niner Over a year ago

I was just about to comment on that lol @EdCottrell however am questioning <option row

Funk Forty Niner Over a year ago

@EdCottrell Edited my comment above. Seems like invalid HTML.

elixenide Over a year ago

@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.

Joel Trauger Over a year ago

It filled my error log. 256MB and climbing...I hate infinite loops.

Asterisk · Accepted Answer · 2014-07-25 17:21:01Z

1

you missed a dollar sign:

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

should be

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

answered Jul 25, 2014 at 17:21

Asterisk

962 bronze badges

Comments

Joel Trauger · Accepted Answer · 2014-07-25 17:23:00Z

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, 2014 at 17:23

Joel Trauger

7321 gold badge4 silver badges24 bronze badges

2 Comments

Funk Forty Niner Over a year ago

"Discovered the error." So did a few others before posting this. You should accept the first answer given, mainly Mariano's

Joel Trauger Over a year ago

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.

Collectives™ on Stack Overflow

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

3 Answers 3

4 Comments

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

4 Comments

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related