Question

Hello All我为我的android应用程序项目创建了一个php，它创建了一个包含数据库返回数据的json 但问题是除了＆＃34; []＆＃34;它没有返回任何东西。方括号。

FetchUserData.php

<?php

$con=mysqli_connect("mysql8.000webhost.com" , "====" , "passwordGoesHere" , "=====");

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

$statement = mysqli_prepare($con, "SELECT * FROM User WHERE username = ? AND password = ?");
mysqli_stmt_bind_param($statement, "ss" , $username, $password);
mysqli_stmt_execute($statement);

mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement ,$userID , $name , $age , $username , $password);

$user = array();

while(mysqli_stmt_fetch($statement)){
    $user[name] = $name;
    $user[age] = $age;
    $user[username] = $username;
    $user[password] = $password;
}

echo json_encode($user);
mysqli_stmt_close($statement);
mysqli_close($con);
?>

Answer 1

这是因为你正在使用php解析器解释为未定义的常量。你没有把你的钥匙包在报价单中。应该是：

while(mysqli_stmt_fetch($statement)){
    $user['name'] = $name;
    $user['age'] = $age;
    $user['username'] = $username;
    $user['password'] = $password;
}

Answer 2

检查你的sql查询是否返回任何结果。我认为您的查询没有给出任何结果。因此你得到空白数组。直接将查询运行到数据库＆amp;检查你是否得到任何东西

PHP没有传递正确的JSON结果

2 个答案: