Question

PHP+MYSQL 查询结果作为条件再查另一个表，怎么弄？

表A
userid name
1 张三
2 李四
3 赵五

表B
id userid mobile
1 3 13899999999
2 2 15566667777

查询表B后如下列表
<?php do { ?>
<tr>
<td><?php echo $row_Recordset1['user_id']; ?></td>
<td><?php echo $row_Recordset1['mobile']; ?></td>
</tr>
<?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?>

如何把查询结果列表中的user_id替换成表A中的name

Answer 1

方法一：很简单，再查询一次就好了，代码如下

<?php

while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)) {
$sql = "select * from 表A where userid={$row_Recordset1['user_id']} limit 1";
$query = mysql_query($sql);
$user = mysql_fetch_assoc( $query );

?>

<tr>
<td><?php echo $user['name']; ?></td>
<td><?php echo $row_Recordset1['mobile']; ?></td>
</tr>

<?php } ?>

方法二： 用联合查询,直接查出两张表的数据
<?php

$sql = "select * from tableA,tableB where tableA.userid = tableB.userid limit 10";
$query = mysql_query($sql);
$user = mysql_fetch_assoc( $query );
while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)) {

?>

<tr>
<td><?php echo $user['name']; ?></td>
<td><?php echo $user['mobile']; ?></td>
</tr>

<?php } ?>

好像刚才解答了一个类似的问题，不会是你的同学吧？

Answer 2

怎么最近老问 多表联查的问题 简单点就是 select a.name,b.mobile from a,b where a.userid=b.userid

当然还有很多方法，熟悉后你会发现 条条大道通罗马

Answer 3

select a.name b.mobile from a as a left join b as b where a.userid=b.userid group by a.userid order by b.id desc limit 0,10;