Mysql-php mysql_fetch_array()

Question

        $avg_pgt = mysql_query("SELECT avg(convert(custom_var_v1,unsigned)),min(convert(custom_var_v1,unsigned)), max(convert(custom_var_v1,unsigned) FROM `table_name` WHERE server_time BETWEEN '$date 00:00:00' AND '$date 23:59:59'");

        $row_all = mysql_fetch_array($avg_pgt);         
        $string_avg = (string)$row_all[0];      
        echo $string_avg;

It gives an error mysql_fetch_array() expects parameter 1 to be a resource boolean given in the code

If this is new code: Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. — Amelia
– Amelia, Commented Feb 8, 2013 at 8:35
That means there is some problem in your query, always use die(mysql_error());. — Rikesh
– Rikesh, Commented Feb 8, 2013 at 8:35

s.lenders · Accepted Answer · 2013-02-08 08:39:18Z

1

max(convert(custom_var_v1,unsigned)

Is missing a ) at the end, This would work:

    $avg_pgt = mysql_query("SELECT avg(convert(custom_var_v1,unsigned)),min(convert(custom_var_v1,unsigned)), max(convert(custom_var_v1,unsigned)) FROM `table_name` WHERE server_time BETWEEN '$date 00:00:00' AND '$date 23:59:59'");

    $row_all = mysql_fetch_array($avg_pgt);         
    $string_avg = (string)$row_all[0];      
    echo $string_avg;

answered Feb 8, 2013 at 8:39

s.lenders

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Hanky Panky · Accepted Answer · 2013-02-08 08:34:31Z

0

$avg_pgt = mysql_query("SELECT avg(convert(custom_var_v1,unsigned)),min(convert(custom_var_v1,unsigned)), max(convert(custom_var_v1,unsigned) FROM `table_name` WHERE server_time BETWEEN '$date 00:00:00' AND '$date 23:59:59'");

This query is failing and mysql_query is returning false. Check if your query is fine and returns results elsewhere. Is table_name really the name of your table?

answered Feb 8, 2013 at 8:34

Hanky Panky

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Gung Foo · Accepted Answer · 2013-02-08 08:35:13Z

You shouldn't be using the mysql_* functions since they are being deprecated since php5.

Use PDO or mysqli!

From the mysql_query() documentation page:

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

mysql_query() will also fail and return FALSE if the user does not have permission to access the table(s) referenced by the query.

Call mysql_error() to find out what is going wrong.

Prasanth Bendra · Accepted Answer · 2013-02-08 08:44:42Z

0

you are missing ) at max(convert(custom_var_v1,unsigned)

mysql_* functions are dprecated, use mysqli_ or PDO

answered Feb 8, 2013 at 8:44

Prasanth Bendra

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