php - mysqli_query 有问题，返回 NULL 但查询在 phpmyadmin 中有效

Question

当用户从 iOS 应用程序发送登录命令时，我尝试从数据库检索信息。为了测试这个功能，我手动启动我的 php 页面(例如 http://www.testdatabase.com/LoginFunctions.php )并以编程方式强制用户名。

问题是 mysqli_query 返回 NULL 值。如果我使用“or die(mysql_error()”，什么也不会发生。即使我使用 mysqli_num_rows 返回 1，但 $result 仍然为空。因此，当执行 mysql_fetch_assoc 时，程序崩溃，但没有显示任何错误。任何想法？谢谢

<?php
    // Create connection
    $con=mysqli_connect("localhost","super","super","testdb");

    // Check connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $action = "login";
    $username = "Peperoncino";
    $response = array();

    if ($action == "login")
    {
        $query = "SELECT psw AS pswrd,id FROM Activities WHERE nome = 'Peperoncino' LIMIT 1";

        if ($result = mysqli_query($con, $query))
        {
            $values = mysql_fetch_assoc($result);
            $password = $values['pswrd'];
            $response["password"] = $password;
            $response["message"] = "Get information from db";
        }
        else
        {
            echo "err";
        }

        echo json_encode($response);
    }

    // Close connections
    mysqli_close($con);
?>

Answer 1

您正在使用已弃用的 mysql_fetch 函数。请使用新函数

<?php
    // Create connection
    $con=mysqli_connect("localhost","super","super","testdb");

    // Check connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $action = "login";
    $username = "Peperoncino";
    $response = array();

    if ($action == "login")
    {
        $query = "SELECT psw AS pswrd,id FROM Activities WHERE nome = 'Peperoncino' LIMIT 1";

        if ($result = mysqli_query($con, $query))
        {
            $values = mysqli_fetch_assoc($result);
            $password = $values['pswrd'];
            $response["password"] = $password;
            $response["message"] = "Get information from db";
        }
        else
        {
            echo  mysqli_error($conn);  
        }

        echo json_encode($response);
    }

    // Close connections
    mysqli_close($con);
?>