Question

因此，我在PHP中编写此应用程序，其中用户具有“学生姓名”，每个用户都有一个唯一的学生姓名。所以，在我进一步解决我的问题之前，这里是代码 *注意我已经阻止了SQL注入

    function hello($username123) {
       // Connect to Database //
$host3     = "db"; 
$username3 = "db"; 
$password3 = "db"; 
$db3       = "db";
$con3 = mysqli_connect($host3,$username3,$password3,$db3) or die("Can not connect to Server.");
$query3 = mysqli_query($con3,"SELECT student1 FROM users WHERE username = '$username123'");
$student1name = "$query3";
return $student1name;

因此，该人输入手前注册的用户名，每个用户都有一个学生姓名。我开始一个选择student1的查询，student1等于student1name。然后将学生1的名称定义为query3。当我全部测试时，我得到的是（null）..有谁知道这个问题？谢谢！

Answer 1

我怀疑你想要的是这样的：

function hello($username123) {
    // Connect to Database //
    $host3     = "db"; 
    $username3 = "db"; 
    $password3 = "db"; 
    $db3       = "db";
    $con3 = mysqli_connect($host3,$username3,$password3,$db3) or die("Can not connect to Server.");
    $query3 = mysqli_query($con3,"SELECT student1 FROM users WHERE username = '$username123'");
    while ($row = mysqli_fetch_array($query3))
    {
        $student1name = $row['student1'];
    }
    return $student1name;

这会将查询的最后返回行的内容“student1”列的内容放入变量$ student1name中，然后将其返回。

Answer 2

您没有从结果中获取数据。试试这个：

function hello($username123) {
    // Connect to Database //
    $host3     = "db"; 
    $username3 = "db"; 
    $password3 = "db"; 
    $db3       = "db";

    $con3 = mysqli_connect($host3,$username3,$password3,$db3)
    if (!$con3)
        throw new Exception("Connection error");

    $result = mysqli_query($con3,"SELECT student1 FROM users WHERE username = '$username123'");
    if ($result)
        return $result->fetch_object();
    else
        throw new Exception("Query error: " . mysqli_error($con3));
}

返回查询结果

2 个答案: