我有2-3个不同的列名，我想在整个数据库中查找，并列出所有具有这些列的表。有没有简单的脚本?
当前回答
SELECT DISTINCT TABLE_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE COLUMN_NAME LIKE '%city_id%' AND TABLE_SCHEMA='database'
其他回答
SELECT DISTINCT TABLE_NAME, COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE column_name LIKE 'employee%'
AND TABLE_SCHEMA='YourDatabase'
SELECT TABLE_NAME, COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE COLUMN_NAME LIKE '%wild%';
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)
....
}
}
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。
更简单地用一行SQL语句完成:
SELECT * FROM information_schema.columns WHERE column_name = 'column_name';
