有效的SQL命令不能在PHP中运行

Jaa*_*älä 0 php mysql

我在PHPAdmin的终端中运行了以下代码:

SELECT COUNT( * )
FROM User
WHERE email = 'ijp' 
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并得到结果0.

然后我的部分PHP程序如下:

$email = clean_input($_POST['email']);
$query = "SELECT COUNT(*) FROM user WHERE '.$email.'=email"; 
echo $query;
$result = mysqli_query($link, $query);
echo $result;
mysqli_close($link);
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我已经检查过我登录数据库的情况很好.输出仍然是

SELECT COUNT(*) FROM User WHERE email='ijp'
Catchable fatal error: Object of class mysqli_result could not be converted to string   in .../html/register.php on line 49
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如何在PHP中运行MySQL命令?

Xin*_*nt0 5

您正在传递期望字符串的$result资源echo.您需要使用其中一个mysqli_fetch_*函数从中读取值$result然后打印它.

mysqli_fetch()已弃用,使用mysqli_fetch_arraymysql_fetch_assocmysql_fetch_all,例如:


$row = mysqli_fetch_array($result);
if($row) {
    echo $row[0];
}
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