mysql_fetch_array() expects parameter 1 to be resource, boolean given. Any Ideas?

Question

I'm trying to make a page where you search an IGN then it returns stats relevant to that player stored in the database.

This is my search.php page

<?php
mysql_connect ("185.14.187.33", "lightpvp","Notsayingpassword")  or die (mysql_error());
mysql_select_db ("lightpvp_test");

$term = $_POST['term'];

$sql = mysql_query("select * from Stats where IGN='%$term%'");

while ($row = mysql_fetch_array($sql)){
    echo 'Player Username: '.$row['IGN'];
    echo '<br/> Number of Kills: '.$row['Kills'];
    echo '<br/> Number of Deaths: '.$row['Deaths'];
    echo '<br/><br/>';
    }

?>

However it returns the error:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/content/51/11301251/html/mockup/search.php on line 9

Help? Thanks In Advance.

What does `var_dump($sql);` give you? It is possible that the sql query failed so `$sql` is `false`. — Jonathan, Feb 13 '14 at 19:11
Keep in mind, that this piece of code is vulnerable to SQL Injection! — Chris, Feb 13 '14 at 19:12
@Hell_Yeah96. Add after the `$sql = mysql_query("select * from Stats where IGN='%$term%'");` this: `var_dump($sql);` It should print out the value of `$sql`. — Jonathan, Feb 13 '14 at 19:17
Please, Please, PLEASE look up MySQLi and use it instead of these deprecated mysql_* functions! — Andrew McGivery, Feb 13 '14 at 19:23
@Bonzo error is still there but now also says above it bool(false) — Hell_Yeah96, Feb 13 '14 at 19:24
PS: You're getting this error because you have some kind of error in your SQL. mysql_query returns FALSE if it fails to run. — Andrew McGivery, Feb 13 '14 at 19:26
do an echo of mysql_error() before your while loop to find out what the error is. — Andrew McGivery, Feb 13 '14 at 19:27
@Hell_Yeah96 It's probable that `mysql_select_db("lightpvp_test");` is returning false then and so all queries afterwards will fail. Remove the `var_dump($sql);` and add `echo mysql_error();`. This will print out the error you are getting. — Jonathan, Feb 13 '14 at 19:29

echo_Me · Answer 1 · 2014-02-13T19:30:12.010

1

You may use LIKE instead of =. Try this:

    $sql = mysql_query("SELECT * FROM Stats WHERE IGN LIKE '%$term%'") or die (mysql_error());

edited Feb 13 '14 at 19:30

answered Feb 13 '14 at 19:16

echo_Me

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Same issue with that code. – Hell_Yeah96 Feb 13 '14 at 19:21
@Hell_Yeah96 edited my answer , try that and tell what error you got . – echo_Me Feb 13 '14 at 19:30

mysql_fetch_array() expects parameter 1 to be resource, boolean given. Any Ideas?

1 Answers1