comparing Query results

Question

$query1= mysql_query("select r.nid from ....");     

$query2= mysql_query("select t.nid from....");

Both these queries return a nid. How to compare the 2 returned nid are equal.. I'm just a beginner.

why do need it? you can compere then in query. Do you know how to get data from database using php? do you know how to compare 2 variables in php? — El', May 09 '12 at 07:48
where do you like to make that compare? in php code or in sql? — AKZap, May 09 '12 at 07:55
If you're going to continue doing mysql using PHP (and just started learning) I highly suggest learning PDO instead of using mysql_* . Suggested Reads: http://stackoverflow.com/questions/866860/mysql-vs-pdo http://stackoverflow.com/questions/13569/mysqli-or-pdo-what-are-the-pros-and-cons http://www.php.net/manual/en/book.pdo.php — Event_Horizon, May 10 '12 at 05:24

score 1 · Accepted Answer · answered May 09 '12 at 07:28

1

$row1 = mysql_fetch_row($query1);
$row2 = mysql_fetch_row($query2);
if($row1[0] == $row2[0])
{
//something
}

answered May 09 '12 at 07:28

arun

3,637
3
27
53

score 1 · Answer 2 · answered May 09 '12 at 07:30

1

You could do it in pure sql. Like this:

select 
    r.nid 
from 
    ....
WHERE EXISTS
(
    select 
        NULL 
    from
        ....
    WHERE
        t.nid = r.nid 
)

answered May 09 '12 at 07:30

Arion

29,817
9
65
81

score 1 · Answer 3 · answered May 09 '12 at 07:34

If you are certainly sure that the query really returns one id, you can speed up checking it by:

$query1 = mysql_query("select r.nid from ...."); 
$query2 = mysql_query("select t.nid from ...."); 
if(mysql_fetch_field($query1, 0) === mysql_fetch_field($query2, 0))
{
    //do something
}

comparing Query results

3 Answers3

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