PHP登录脚本json错误

我的php脚本中有一个错误，我正在postman中执行该脚本，以便在android studio中使用。响应部分没有执行该脚本，并给出一个错误

请帮帮我

<?php
    $con = mysqli_connect("", "", "", "");

    $username = $_POST["username"];
    $password = $_POST["password"];


     $query = "SELECT * FROM user_info WHERE username = '$username' AND password ='$password'";
   $result=mysqli_query($con,$query);

    $response=array();


    if(mysqli_num_rows($result)>=1)
    {
        $data=mysqli_stmt_fetch($result)
        $response['success'] = 'true';  
        $response["name"] = $data['name'];
        $response["age"] = $data['age'];
        $response["username"] = $data['username'];
        $response["password"] = $data['password'];

    }

    if(mysqli_num_rows($result)<1){
        $response["success"] = 'false'; 

    }
    echo json_encode($response);
?>

您缺少分号；在这一行的末尾：

$data=mysqli_stmt_fetch($result)
                                ^

---

在SQL查询中将php变量与字符串连接时犯了一个小错误。请把鳍放在正确的下面

<?php
    $con = mysqli_connect("", "", "", "");

    $username = $_POST["username"];
    $password = $_POST["password"];


     $query = "SELECT * FROM user_info WHERE username = '".$username."' AND password ='".$password."'";
   $result=mysqli_query($con,$query);

    $response=array();


    if(mysqli_num_rows($result)>=1)
    {
        $data=mysqli_stmt_fetch($result);
        $response['success'] = 'true';  
        $response["name"] = $data['name'];
        $response["age"] = $data['age'];
        $response["username"] = $data['username'];
        $response["password"] = $data['password'];

    }

    if(mysqli_num_rows($result)<1){
        $response["success"] = 'false'; 

    }
    echo json_encode($response);
?>

---

没有按数据命名的数组。因此，$data['name']不可用

if(mysqli_num_rows($result)>=1)
{
    $data=mysqli_stmt_fetch($result);//semi colon was missing
    $response['success'] = 'true';  
    $response["name"] = $data['name'];//unable to find $data array
    $response["age"] = $data['age'];//unable to find $data array
    $response["username"] = $data['username'];//unable to find $data array
    $response["password"] = $data['password'];//unable to find $data array

}

---

我们不会为你做家庭作业。23号线在哪里，你试图解决什么问题等等……哎呀！您正在使用纯文本密码！！！！PHP提供并请使用它们。如果您在这一行中使用的是5.5$data=mysqli\u stmt\u fetch$result之前的PHP版本，那么这里有一些；。它应该$data=mysqli\u stmt\u fetch$result；

if(mysqli_num_rows($result)>=1)
{
    $data=mysqli_stmt_fetch($result);//semi colon was missing
    $response['success'] = 'true';  
    $response["name"] = $data['name'];//unable to find $data array
    $response["age"] = $data['age'];//unable to find $data array
    $response["username"] = $data['username'];//unable to find $data array
    $response["password"] = $data['password'];//unable to find $data array

}