On May 31, 2007, at 3:27 PM, Davi wrote:
Em Quinta 31 Maio 2007 16:25, Jason Pruim escreveu:
Hi Everyone, I am attempting to setup a search field on a database
application I'm dinking around with and running into problems that
I'm hoping someone might be able to shed some light on.
Here is the code I am using to display the results of the search:
echo ('<table border="1">');
echo "<tr><th>First Name</th><th>Author</th><th>Pages</th></tr>";
$result_row[] = mysql_query($query) or die(mysql_error());
echo "Result_row dump: $result_row <BR>";
$num=mysql_numrows($result);
echo "Num dump: $num <BR>";
$i= 0;
/*
while($i < $num) {
*/
$result=$result_row[0];
while($result_row = mysql_fetch_array($result) {
Worked perfectly after adding a closing ) Thanks for the tip!
echo "<tr><td>";
echo $result_row[0] . '</td><td>';
echo $result_row[1] . '</td><td>';
echo $result_row[2] . '</td></tr>';
$i++;
}
echo ("</table>");
the problem is instead of printing out the actual results it prints
out 6 fields that say: "Resource id #5"
the query I'm using is: SELECT FName, LName, Add1, Add2 FROM current
WHERE FName like '%jason%' which works in MySQL
I can't find any errors in my log files to even give me a hint as to
what is going on.
If someone could take a look I would greatly appreciate it.
HTH
--
Davi Vidal
davividal@xxxxxxxxxxxxxxxx
davividal@xxxxxxxxx
--
"Religion, ideology, resources, land,
spite, love or "just because"...
No matter how pathetic the reason,
it's enough to start a war. "
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