2013-12-10 86 views
0

我想使用多个查询来计算多个值并在一个表中显示这些值。计数多个值查询

$resultnieuws = mysql_query("select gamer_int, count(gamer_int) as gamer_count_nieuws  FROM berichten WHERE gamer_int LIKE 'Kenny' AND soort LIKE 'nieuws'"); 
$resultvideo = mysql_query("select gamer_int, count(gamer_int) as gamer_count_video FROM berichten WHERE gamer_int LIKE 'Kenny' AND soort LIKE 'video'"); 

echo $resultnieuws['gamer_count_nieuws']; 
echo $resultvideo['gamer_count_video']; 

上面的回声给我没有结果。我究竟做错了什么?

回答

1

尝试像这个 -

$resultnieuws = mysql_query("select gamer_int, count(gamer_int) as gamer_count_nieuws  FROM berichten WHERE gamer_int LIKE 'Kenny' AND soort LIKE 'nieuws'"); 
    $resultvideo = mysql_query("select gamer_int, count(gamer_int) as gamer_count_video FROM berichten WHERE gamer_int LIKE 'Kenny' AND soort LIKE 'video'"); 

    $row1 = mysql_fetch_array($resultnieuws); 
    $row2 = mysql_fetch_array($resultvideo); 

    echo $row1['gamer_count_nieuws']; 
    echo $row2['gamer_count_video']; 

注:mysql_ *功能已被弃用,用起来不推荐。

+0

使用他们甚至**不鼓励**。 –

+0

这不行;由于某种原因,我的编辑被拒绝了,但是您不能从'$ row1 ['gamer_count_nieuws']'中引用该字段,您需要从结果的第一行中选择它。例如。 '$ row1 [0] ['gamer_count_nieuws']' – Ryan

+0

完美运作!谢谢! – user2959441