php - 尝试查找用户时 MYSQL "Impossible WHERE noticed after reading const table"

Question

当我尝试从 php 运行以下查询时，我没有得到任何结果(即使有一个用户与给定的信息相匹配。

SELECT * FROM `virtual_users` WHERE `email`='$u', `password` = ENCRYPT('".$p."', CONCAT('$6$', SUBSTRING(SHA(RAND()), -16)))

在另一个脚本中，以下查询工作正常:

"UPDATE `virtual_users` SET `password`=ENCRYPT('".$sNewPassword."', CONCAT('$6$', SUBSTRING(SHA(RAND()), -16))) WHERE `email`='". $oAccount->Email () ."'"

在困惑了一会儿之后，我尝试使用 PHP my Admin 中的 explain 运行查询:

EXPLAIN SELECT * FROM `virtual_users` WHERE `email`='fakeuser@fake.fake' AND `password` = ENCRYPT('fakepass', CONCAT('$6$', SUBSTRING(SHA(RAND()), -16)))

返回以下内容

+---+---+---+---+---+---+---+---+---+---+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +---+---+---+---+---+---+---+---+---+---+ | 1 | SIMPLE | NULL | NULL | NULL | NULL | NULL | NULL | NULL | Impossible WHERE noticed after reading const tables | +---+---+---+---+---+---+---+---+---+---+

表结构如下: Table Structure Image

任何帮助将不胜感激。签，迈克

Answer 1

WHERE 不使用逗号。我相信你想要的语法是 AND:

SELECT *
FROM `virtual_users`
WHERE `email` = '$u' AND `password` = ENCRYPT('".$p."', CONCAT('$6$', SUBSTRING(SHA(RAND()), -16)))