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
实际上有一种方法，当然你需要有权限才能这样做…
SET @sql = CONCAT('SELECT ', (SELECT REPLACE(GROUP_CONCAT(COLUMN_NAME), '<columns_to_omit>,', '') FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME = '<table>' AND TABLE_SCHEMA = '<database>'), ' FROM <table>');
PREPARE stmt1 FROM @sql;
EXECUTE stmt1;
替换<表>，<数据库>和< columns_to_省略>
其他回答
你可以:
SELECT column1, column2, column4 FROM table WHERE whatever
没有得到列3，尽管您可能在寻找一个更一般的解?
您可以使用DESCRIBE my_table并使用其结果动态地生成SELECT语句。
视图在这种情况下工作得更好吗?
CREATE VIEW vwTable
as
SELECT
col1
, col2
, col3
, col..
, col53
FROM table
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'
我也想要这个，所以我创建了一个函数。
public function getColsExcept($table,$remove){
$res =mysql_query("SHOW COLUMNS FROM $table");
while($arr = mysql_fetch_assoc($res)){
$cols[] = $arr['Field'];
}
if(is_array($remove)){
$newCols = array_diff($cols,$remove);
return "`".implode("`,`",$newCols)."`";
}else{
$length = count($cols);
for($i=0;$i<$length;$i++){
if($cols[$i] == $remove)
unset($cols[$i]);
}
return "`".implode("`,`",$cols)."`";
}
}
所以它的工作原理是，你输入表格，然后是你不想要的列或在数组中:array("id"，"name"，"whatevercolumn")
所以在select中你可以这样使用它:
mysql_query("SELECT ".$db->getColsExcept('table',array('id','bigtextcolumn'))." FROM table");
or
mysql_query("SELECT ".$db->getColsExcept('table','bigtextcolumn')." FROM table");
