比较查询结果

Question

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

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

这两个查询都返回一个nid。如何比较2返回的nid是相等的..我只是一个初学者。

Answer 1

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

Answer 2

你可以在纯sql中完成它。像这样：

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

Answer 3

如果你确定查询确实返回了一个id，你可以加快检查：

$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
}