Forum Moderators: coopster
when i try to run this file
<html>
<?php
$open = mysql_connect("localhost");
$data = mysql_db_query("video1, SELECT * FROM video1 WHERE namn LIKE 'sok' ORDER BY namn");
while ($rad = mysql_fetch_array ($data)) {
print "<B>";
print $rad["namn"];
print "nummer";
print $rad["nummer"];
print "</B><BR>";
print "Tid:";
print $rad["tid"];
print "<P>";
}
?>
</html>
is there anybody who knows whats wrong pleas help me
[edited by: jatar_k at 8:52 pm (utc) on Dec. 3, 2002]
[edit reason] no sigs please [/edit]
the like won't match anything as steffan mentioned, if there is no variable part of the match it will work just like equals.
take a look at String Comparison Functions [mysql.com]
you could also try to add an or die [php.net] to your query to get the error from mysql.
like so
$data = mysql_db_query($somequery) or die(mysql_error());
What does the "video1," before the SELECT do? I've never written a query like that before.
mysql_db_query -- Send a MySQL query
Description
resource mysql_db_query ( string database, string query [, resource link_identifier])
mysql_db_query("video1", "SELECT * FROM video1 WHERE namn LIKE 'sok' ORDER BY namn");
instead of:
mysql_db_query("video1, SELECT * FROM video1 WHERE namn LIKE 'sok' ORDER BY namn");
which is what he is doing... perhaps this is why it is causing a mysql error?
mysql_fetch_array(): supplied argument is not a valid MySQL result resource
$result = mysql_query("select * from db where field like '$somevar'");
if (mysql_numrows($result)>0) {
echo stuff;
} else {
echo "No results";
}
