MySQL-a problem that calculates PHP results. Daniel, Seek Master

//从part_time数据库中查找$sql="select * from part_time where agents=6";$result=mysql_query($sql);while($row=mysql_fetch_assoc($result)){    $id=$row['id'];//循环出所有agents=2的id        $resu="select count(*) from userinfo where part_person=$id";    $re=mysql_query($resu);    $roo=mysql_fetch_assoc($re);    $number= $roo['count(*)']; //计算出userinfo中是相同兼职人员（part_time）的人数    echo $number;    echo "----";}//现在需要把$number 相加得到最终的数字。应该怎么做？求大牛解答

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//从part_time数据库中查找$sql="select * from part_time where agents=6";$result=mysql_query($sql);while($row=mysql_fetch_assoc($result)){    $id=$row['id'];//循环出所有agents=2的id        $resu="select count(*) from userinfo where part_person=$id";    $re=mysql_query($resu);    $roo=mysql_fetch_assoc($re);    $number= $roo['count(*)']; //计算出userinfo中是相同兼职人员（part_time）的人数    echo $number;    echo "----";}//现在需要把$number 相加得到最终的数字。应该怎么做？求大牛解答

whiledefine one in the front $number and then directly $number+=$row['count(*)'] .

In addition, I will give you the code to simplify the next ...

$sql = 'select count(*) as total from userinfo where part_person IN (select id from part_time where agents = 6)';$result = mysql_query($sql);$row = mysql_fetch_assoc($result);echo $row['total'];

Define a $sum=0;
Add $sum+= in while

+ = $roo [' Count (*) '];

Wow, in the loop you call the database so repeatedly, the access to the large estimate database will be unbearable. You can use join queries.

Try aggregate.

$number += $roo['count(*)'];

But this code is a little bit.

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