Question

我正在尝试获取指定用户名的密码，将其分配给PHP变量，然后将其打印出来。这是它的代码 -

$user = $_POST["uname"];
$pass = mysqli_query($con, "SELECT pass FROM users WHERE username = '$user'");                                              
$result = mysqli_fetch_array($pass);

$specific = $result['pass'];

echo $specific;

问题是根本没有印刷品！甚至没有错误。我该怎么办？

Answer 1

$pass = mysqli_query($con, "SELECT pass FROM users WHERE username = '$user'") or die(mysqli_error($con));

这将在您的查询中显示正确的错误

Answer 2

转义您的值并检查错误：

$user = mysqli_escape_string( $con, $_POST["uname"] );
$pass = mysqli_query( $con, "SELECT pass FROM users WHERE username = '$user'");

# Error checking
if( $pass === false ) {
    echo 'Error: ', mysqli_error( $con );
}

# Check for no user with that password
if( mysqli_num_rows( $pass ) == 0 ) {
    echo 'No user with that username.';
}

# Use as associate arary              
$result = mysqli_fetch_assoc($pass);

$specific = $result['pass'];

echo $specific;

编辑：添加了无结果检查。

将MySQL数据库值分配给PHP变量

2 个答案: