Question

php + mysql 查询结果作为条件再查另一个表，如何弄？

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 } ?>

Answer 2

你这里是不能用do while的，结果都没遍历出来，不可能有值的
<?php while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?>

<tr>
<td><?php echo $row_Recordset1['name']; ?></td>
<?php
$sql="select * from 表B where userid=".$row_Recordset1['userid'];
$Recordset2=mysql_query($sql);
$row_Recordset1 = mysql_fetch_assoc($Recordset2)

?>
<td><?php echo $row_Recordset2['mobile']; ?></td>
</tr>

<?php } ?>

Answer 3

你用联合查询
$sql="select * from 表A，表B where 表A.userid=表B.userid";
$re=mysql_query($sql);
while($row=mysql_fetch_assoc($re)){
$class[]=$row;

};
一条sql语句就搞定了

Answer 4

联合查询：$sql = "select a.name as name,b.mobile as mobile from a,b where a.userid=b.userid";
$res = mysql_query($sql);
while($row = mysql_fetch_assoc($res)) {
echo "<tr>";
echo "<td>{$row['name']}</td><td>{$row['mobile']}</td>"

echo "</tr>";

}

Answer 5

你把表A的数据查出来,然根据id update进去表B不就可以了么