SQL INTO将错误作为INTO表 - undefined

Question

我正在尝试复制1个表选定列INTO第2个表选定列,使用此:

$sql = "SELECT (p.col1, p.col2)
INTO tblColors(b1,b2)
FROM tblPuppies as p
WHERE p.col3=1
";

但它给了我错误:未定义的变量tblColors.对我来说很奇怪.任何猜测PLZ,哪里出错？先谢谢你.

Answer 1

只是用

INSERT INTO tblColors (b1, b2)
SELECT col1, col2 FROM tblPuppies WHERE b3 = 1