Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in

Question

I have this problem with short php code I am using in to get suggestions for JQuery autocomplete. I have tried many things but I am unable to find solution:

//connect to your database
<?php require('Connections/con_1.php');?>
<?php

//retrieve the search term that autocomplete sends
$term = trim(strip_tags($_REQUEST['term']));

//save search queries to csv file
$file = fopen("searchqueries.csv","a");
fputcsv($file,explode(',',$term));
fclose($file);

//query the database for entries containing the term
$qstring ="USE database_2";"SELECT DISTINCT Pracodawca as value FROM Pensje WHERE Pracodawca LIKE '%".$term."%'";
$result = mysql_query($qstring) or die($result.mysql_error());

//loop through the retrieved values        
while ($row = mysql_fetch_array($result,MYSQL_ASSOC))
    {

    //build an array
    $row['value']=htmlspecialchars(stripslashes($row['value']));
    $row_set[] = $row['value'];
    $row['value']=
    $row_set[] = array('value' => ($row['value']));
    }

//format the array into json data
echo json_encode($row_set, JSON_UNESCAPED_UNICODE);
?>

I get error "Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in" and the result is NULL.

What I am doing wrong?

If you're writing new code, **_please_ don't use the `mysql_*` functions**. They are old and broken, were deprecated in PHP 5.5 (which is so old it no longer even receives security updates), and completely removed in PHP 7. Use [`PDO`](https://secure.php.net/manual/en/book.pdo.php) or [`mysqli_*`](https://secure.php.net/manual/en/book.mysqli.php) with _prepared statements_ and _parameter binding_ instead. See http://stackoverflow.com/q/12859942/354577 for details. — Chris, Mar 19 '17 at 12:26
First of all, use `mysqli_*` ... second... you can't send 2 queries at once like this.... you need to seperate them... 3rd... well fix the first ones and maybe someone will help ya. — Flash Thunder, Mar 19 '17 at 12:28

score 0 · Answer 1 · answered Mar 19 '17 at 12:29

Please don't use the mysql_* functions.

Change the following line:

$qstring ="USE database_2";"SELECT DISTINCT Pracodawca as value FROM Pensje WHERE Pracodawca LIKE '%".$term."%'";

to

$qstring = "SELECT DISTINCT Pracodawca as value FROM Pensje WHERE Pracodawca LIKE '%".$term."%'";

As there is no need to pass USE dbname when you are querying through php as database is already selected at the time of connecting with database.

and try again.

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in

1 Answers1