Using this extract from
http://ie.php.net/manual/en/control-structures.foreach.php
Amaroq
09-Mar-2008 06:40
Even if an array has only one value, it is still an array and foreach will run
it.
<?php
$arr[] = "I'm an array.";
if(is_array($arr))
{
foreach($arr as $val)
{
echo $val;
}
}
?>
The above code outputs:
I'm an array.
-------------------------------------
So if I use:
$query = "Select answer from answers where studentID='A123456789'";
$result = mysql_query($query,$connection) or die(mysql_error());
while($row = mysql_fetch_assoc($result))
{
foreach($row as $answer)
{
echo $answer . "\n";
}
}
I thought I would be able to print out each array element, but I am not getting
any output. Has anyone got a better idea?
--- Begin Message ---
Tried that as well and got the same result.
I tried "Select count(answer) as total from answers where studentID='A123456789'";
from the CLI and got total = 2 as a result.
So, we got rid of the Invalid Resource error and we know that the
student id you use occurs in both rows in your table and that your query
works fine.
Did you get rid of the semicolon @ line 15 "while($row =
mysql_fetch_assoc($result));", as Jason suggested? Also, an:
error_reporting(E_ALL);
at the top of your code might help in backtracing.
--- End Message ---
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
