我试图使用一个选择语句从某个MySQL表中获得除一个以外的所有列。有什么简单的方法吗?
编辑:在这个表格中有53列(不是我的设计)
我试图使用一个选择语句从某个MySQL表中获得除一个以外的所有列。有什么简单的方法吗?
编辑:在这个表格中有53列(不是我的设计)
当前回答
只做
SELECT * FROM table WHERE whatever
然后用您最喜欢的编程语言php删除该列
while (($data = mysql_fetch_array($result, MYSQL_ASSOC)) !== FALSE) {
unset($data["id"]);
foreach ($data as $k => $v) {
echo"$v,";
}
}
其他回答
我使用这个工作,尽管它可能是“离题”-使用mysql工作台和查询生成器-
打开列视图 Shift选择所有你想在你的查询列(在你的情况下,所有但这是我所做的) 右键单击并选择发送到SQL编辑器->名称短。 现在你有了列表,然后你可以复制粘贴查询到任何地方。
您可以使用DESCRIBE my_table并使用其结果动态地生成SELECT语句。
也许我有一个解决Jan Koritak指出的矛盾的方法
SELECT CONCAT('SELECT ',
( SELECT GROUP_CONCAT(t.col)
FROM
(
SELECT CASE
WHEN COLUMN_NAME = 'eid' THEN NULL
ELSE COLUMN_NAME
END AS col
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME = 'employee' AND TABLE_SCHEMA = 'test'
) t
WHERE t.col IS NOT NULL) ,
' FROM employee' );
表:
SELECT table_name,column_name
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME = 'employee' AND TABLE_SCHEMA = 'test'
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
table_name column_name
employee eid
employee name_eid
employee sal
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
查询结果:
'SELECT name_eid,sal FROM employee'
(不要在大桌子上尝试,结果可能是……令人惊讶的!)
临时表
DROP TABLE IF EXISTS temp_tb;
CREATE TEMPORARY TABLE ENGINE=MEMORY temp_tb SELECT * FROM orig_tb;
ALTER TABLE temp_tb DROP col_a, DROP col_f,DROP col_z; #// MySQL
SELECT * FROM temp_tb;
DROP语法可能因数据库而异
如果它总是相同的一列,那么你可以创建一个不包含它的视图。
否则,不,我不这么认为。