[QTortZ]

Hello everyone..i have this problem..i want to know how to search from a form to a database
on my form.php i have this code:

<b>Search by : </b>
<form method="post" action="test8.php">
<select name="searchList">

<option value="own"name="owner" >Owner</option>

<option value="sec"name="section">Section</option>

<option value="ipad"name="ipaddress">IP Address</option>
</select>
<input type="text" name="frm_search" size="30">
<input type="Submit" value="Search" align="MIDDLE">
</form>


and on my test8.php:

<?php
$hostname = "10.18.1.78"; // Usually localhost.
$username = "root"; // If you have no username, leave this space empty.
$usertable = "ipa"; // This is the table you made.
$dbName = "ipadd"; // This is the main database you connect to.
mysql_connect($hostname, $username) or die("Unable to connect to database");
@mysql_select_db( $dbName) or die( "Unable to select database");

?>
<?
//error message (not found message)
$XX = "No Record Found";
$query = mysql_query("SELECT * FROM ipa WHERE $searchList LIKE '%$frm_search%' limit 0, 30");
if ($row = mysql_fetch_array($query))
{
echo "<table border=1>\n";
echo "<tr><td>Owner</td><td>Section</td><td>IP Address</td></tr>\n";
do {
printf("<tr><td>%s</td><td>%s</td><td>%s</td></tr>\n", $row["owner"], $row["section"], $row["ipaddress"]);
} while ($row = mysql_fetch_array($query));
echo "</table>\n";
}
if (!$variable1)
{
print ("$XX");
}

?>

but it has this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\apache\apache2\htdocs\TMP9qjlrwi7z2.php on line 14
No Record Found

can anyone help me?

[NeverMind]

this means that the query has somethign wrong with it !
change the query line to:

PHP Code:

$query = mysql_query("SELECT * FROM ipa WHERE $searchList LIKE '%$frm_search%' limit 0, 30")or

die(mysql_error()); 

if any error was found, shall show it ..
also don't depend on globals !
add these lines at the top of your test8.php file:

PHP Code:

$searchList = $_POST['searchList'];

$frm_search = trim(addslashes($_POST['frm_search'])); 

[QTortZ]

thanks for your reply but i still can't get it..
when i change the code above..it prompt out error like:

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'LIKE '%%'' at line 1

thanx again...

[Tim]

It is probably due to using the mysql_fetch_array function
twice. Try changing:
if ($row = mysql_fetch_array($query))
to
if ($row = !$query)

It will accomplish the same thing without the extra function call.

[QTortZ]

ermmm...still the same problem..how to solve it?


thanks..

[Infinite_Hackers]

it seems to me that $frm_search is empty... check if that's the right variable or w/e that holds the keyword.

[ron_barber]

NEVER MIND!! i could kiss you, im just learning php mysql dah dah, geez i been trying to query a search and rahh you just saved me from havign to turn the kettle on for the 5th time this morn haha, MWAHH,