the problem is that your query is not returning a result i would start by
changing
"$Link = mysql_connect ($Host, $User, $Password);"
AND
"$Result = mysql_db_query ($DBName, $Query, $Link);"
To
"$Link = mysql_connect ($Host, $User, $Password) OR die("could not
connect");"
AND
"$Result = mysql_db_query ($DBName, $Query, $Link) OR die("could
query");"
i would also echo the variables
$Host, $User, $Password, $DBName, $Query, $Link
and check they contain what you expect
hope that helps
nick
>>I think that the following code looks fine, but I get an error (Warning:
>>Supplied argument is not a valid MySQL result resource) where it says "while
>>($Row = mysql_fetch_array ($Result)) {" any ideas?
>>
>>
>><?php
>>$Link = mysql_connect ($Host, $User, $Password);
>>
>>
>>$title = trim($title);
>>
>>$title2 = ereg_replace (" ", " %' || like '% ", $title);
>>
>>if ($category) {
>> $Query = "SELECT * from $TableName WHERE ((category)=$category)";
>> print "Your search query of <b> $category </b> returned the following
>>results:<br><br>";
>>} else { if ($title) {
>> $Query = "SELECT * from $TableName WHERE title LIKE '%$title2%' ";
>> print "Your search query of <b> $title </b> returned the following
>>results:<br><br>";
>> }
>>}
>>
>>if ($category = "") {
>> print "";
>>}
>>
>>$Result = mysql_db_query ($DBName, $Query, $Link);
>>
>>while ($Row = mysql_fetch_array ($Result)) {
>> print "<a href=item.php?p=$Row[id]>";
>> echo ( $Row[title] );
>> print "</a><br>";
>>}
>>
>>
>>mysql_close ($Link);
>>?>
>>
>>
>>
>>
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