Question

我正在使用以下SQL查询：

Select * from table1 as t1, table2 as t2 where t1.id = t2.col

但我的问题是两个表都有相同名称的字段place。那么如何在我的PHP代码中从place中选择名为table2的列？我想使用以下php代码

 while($row_records = mysql_fetch_array($result_records))
    {

            <?  echo $row_records['place']; ?>

     }

如何从特定表中获取字段？

Answer 1

永远不要使用......

Select * from ...

...在生产环境中 - 始终明确指定要返回的列。

因此，您可以将SQL修改为：

Select t1.Place as T1Place, t2.Place as T2Place
  from table1 as t1, table2 as t2 where t1.id = t2.col

所以在你的PHP中你会有：

 while($row_records = mysql_fetch_array($result_records))
 {

        <?  echo $row_records['T2Place']; ?>

 }

Answer 2

为什么不使用表别名和字段名称。例如，

    Select t1.place as t1_place, t2.place as t2_place 
      from table1 as t1, table2 as t2 where t1.id = t2.col

在PHP代码中，您可以使用

选择它

while($row_records = mysql_fetch_array($result_records))
    {
    echo $row_records['t1_place']; 
    echo '<br />';
    echo $row_records['t2_place']; 
    }

从表中选择特定字段？

2 个答案: