更简单地用一行SQL语句完成:

SELECT * FROM information_schema.columns WHERE column_name = 'column_name';

SELECT DISTINCT TABLE_NAME, COLUMN_NAME  
FROM INFORMATION_SCHEMA.COLUMNS  
WHERE column_name LIKE 'employee%'  
AND TABLE_SCHEMA='YourDatabase'

获取数据库YourDatabase中所有列为columnna或ColumnB的表:

SELECT DISTINCT TABLE_NAME 
    FROM INFORMATION_SCHEMA.COLUMNS
    WHERE COLUMN_NAME IN ('columnA','ColumnB')
        AND TABLE_SCHEMA='YourDatabase';

information_schema的问题是它可能非常慢。使用SHOW命令速度更快。

选择数据库后，首先发送查询SHOW TABLES。然后对每个表执行SHOW COLUMNS。

在PHP中是这样的

    $res = mysqli_query("SHOW TABLES");
    while($row = mysqli_fetch_array($res))
    {   $rs2 = mysqli_query("SHOW COLUMNS FROM ".$row[0]);
        while($rw2 = mysqli_fetch_array($rs2))
        {   if($rw2[0] == $target)
               ....
        }
    }

使用这一行查询。将desired_column_name替换为列名。

SELECT TABLE_NAME FROM information_schema.columns WHERE column_name = 'desired_column_name';

对于那些与此相反的搜索，即寻找不包含特定列名的表，这里是查询…

SELECT DISTINCT TABLE_NAME FROM information_schema.columns WHERE 
TABLE_SCHEMA = 'your_db_name' AND TABLE_NAME NOT IN (SELECT DISTINCT 
TABLE_NAME FROM information_schema.columns WHERE column_name = 
'column_name' AND TABLE_SCHEMA = 'your_db_name');

当我们开始缓慢地实现InnoDB特殊的ai_col列的使用，并需要找出200个表中哪些表还没有升级时，这真的很方便。