SELECT COUNT和SELECT错误

Question

在我的代码中,我想从不同的表中创建多个SELECT.我有一个SELECT COUNT和一个普通的SELECT.我想在一个声明中写这些.当我这样做时,我得到以下错误:

mysqli_error()期望参数1为mysqli,给定布尔值

这是我发表的声明:

$check = "SELECT 
            (SELECT COUNT(bloopp) FROM bloopp WHERE email='$email') AS bloopp, 
            (SELECT first_name AS first_name, last_name AS last_name 
             FROM user WHERE email ='$email')";

Answer 1

问题是您的第二个select语句,您试图在第二个选择中返回多个列.您应该使用类似于以下内容的内容:

SELECT 
    (SELECT COUNT(bloopp) FROM bloopp WHERE email='$email') AS bloopp, 
    first_name AS first_name, 
    last_name AS last_name 
FROM `user` 
WHERE email ='$email';

当您以编写的方式使用select语句时,它一次只能返回一列.