Question

我在我的Android应用程序上成功注册。但不知何故，我真的想用PHP更新我注册的用户信息。但是，我失败了。我工作的代码有什么问题。

<?php

    $host = 'example.com';
    $user = 'username';
    $pass = 'password';
    $db = 'database';

    $con = mysqli_connect($host, $user, $pass, $db);

    if (!$con) {
      printf("Connection failed: %s\n", mysqli_connect_error());
      exit();
    }

    $query ="UPDATE user_info VALUES (?, ?, ?, ?, ?, ?)"; // make this query into "update query"
    $statement = mysqli_prepare($con, $query);

    if($statement === FALSE){ die(mysqli_error($link)); }

    mysqli_stmt_bind_param($statement, "isssss" ,$UserID, $LastName, $FirstName, $Username, $EmailAddress, $Password);

    $LastName = $_POST["LastName"];
    $FirstName = $_POST["FirstName"];
    $Username = $_POST["Username"];
    $EmailAddress = $_POST["EmailAddress"];
    $Password = $_POST["Password"];

    mysqli_stmt_execute($statement);

    $response = array();
    $response["success"] = true;  

    echo json_encode($response);
?>

顺便说一句，我通过简化问题完成了这个问题。

Answer 1

您应该在绑定它们之前设置变量的值：

<?php
//I remove your credentials
$con = mysqli_connect($host, $user, $pass, $db);

if (!$con) {
  printf("Connection failed: %s\n", mysqli_connect_error());
  exit();
}

$query ="UPDATE user_info SET lastname = ? AND firstname = ? AND username = ? AND emailaddress = ? AND password = ? WHERE userid = ?"; // make this query into "update query"
$statement = mysqli_prepare($con, $query);

if($statement === FALSE){ die(mysqli_error($link)); }

$LastName = $_POST["LastName"];
$FirstName = $_POST["FirstName"];
$Username = $_POST["Username"];
$EmailAddress = $_POST["EmailAddress"];
$Password = $_POST["Password"];

mysqli_stmt_bind_param($statement, $LastName, $FirstName, $Username, $EmailAddress, $Password, $UserID);

mysqli_stmt_execute($statement);

$response = array();
$response["success"] = true;  

echo json_encode($response);

编辑 $UserID未定义。你的UPDATE语句是完全错误的

使用PHP更新用户信息（android app）

1 个答案: