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| alexjaquet@gmail.com 2006-05-25, 3:58 am |
| Hi
I do the following code but I got this error : PHP Warning:
mysql_num_rows(): supplied argument is not a valid MySQL result
resource in C:\\Apache2\\htdocs\\recordz\\php\\main.php on line 36
Any idea ?
<?php
$lang = $_GET['lang'];
$link = mysql_connect("localhost", "root", "xxxx") or die ("Could not
connect");
mysql_select_db ("recordz",$link) or die ("Could not select
database");
$query="SELECT nom FROM category_$lang";
$result=mysql_query($query,$link);
$num = mysql_num_rows($result);
for ($i = 0; $i > $num;$i++) {
print "<option>$result[$i]</option>";
}
?>
:
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| Usenet 2006-05-25, 6:58 pm |
| Probably something wrong with your query (since it should die if
there's something wrong with the database connection).
I'd check that the substitution for $lang is working properly.
Mark Stanton
One small step for mankind...
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| Colin McKinnon 2006-05-25, 6:58 pm |
| alexjaquet@gmail.com wrote:
> I do the following code but I got this error : PHP Warning:
> mysql_num_rows(): supplied argument is not a valid MySQL result
> resource in C:\\Apache2\\htdocs\\recordz\\php\\main.php on line 36
>
> Any idea ?
Well first, you're machine seems to be running Microsoft Windows which is
always a handicap.
>
> <?php
> $lang = $_GET['lang'];
> $link = mysql_connect("localhost", "root", "xxxx") or die ("Could not
> connect");
> mysql_select_db ("recordz",$link) or die ("Could not select
> database");
> $query="SELECT nom FROM category_$lang";
> $result=mysql_query($query,$link);
try
print mysql_error();
here
> $num = mysql_num_rows($result);
> for ($i = 0; $i > $num;$i++) {
> print "<option>$result[$i]</option>";
> }
> ?>
> :
C.
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