资源ID＃4 PHP MYSQL

    $result = mysql_query("SELECT indvsum.sum1 + indvsum.sum2
    FROM (SELECT SUM(Cash) AS sum1,
                 SUM(Bank) AS sum2
          FROM players) indvsum");

    echo $result; 

For some reason this is returning Resource id #4.

How do I get the results of sum1 + sum2 returned?

This is the expected behavior.

Please check out the manual for some example about how to fetch rows:

mysql_query

This is the signature:

resource mysql_query ( string $query [, resource $link_identifier = NULL ] )

For getting the rows you should use

mysql_fetch_array

or

mysql_fetch_assoc

for example.

Because that's the standard output of mysql_query function. It returns the identifier related to that query. To get selected rows use mysql_fetch_array($result) or mysql_fetch_row($result)

Why dont you just do

SELECT SUM(Cash) AS sum1,
       SUM(Bank) AS sum2
FROM players;

Assign the two values in php and then add them.

or

Use mysql_fetch_row

Resource id #4 is being returned because $result is an array.

As an example:

$q_example = "SELECT indvsum.sum1 + indvsum.sum2 AS `aSUM`
                  FROM (SELECT SUM(Cash) AS sum1, SUM(Bank) AS sum2 FROM players) indvsum";
$rsexample = mysql_query($q_example, $DB) or die(mysql_error());
$row_rsexample = mysql_fetch_assoc($rsexample);

echo $row_rsexample['aSUM'];

...should get you what you are looking for.

use mysql_fetch_array

$result = mysql_query("SELECT indvsum.sum1 + indvsum.sum2
FROM (SELECT SUM(Cash) AS sum1,
             SUM(Bank) AS sum2
      FROM players) indvsum");

$arr = mysql_fetch_array($result);

print_r($arr); // echo $result; 

$values = mysql_fetch_array($result);                      
var_dump($values);