【发布时间】:2014-12-10 13:00:44
【问题描述】:
我正在尝试检查数据库中是否已存在用户 ID。但由于某种原因,它不起作用。我尝试了不同的解决方案,但都失败了。
db_connect.php 工作正常,我仔细检查了!
create_user.php
<?php
$response = array();
if (isset($_POST['user_id'])){
$id = $_POST['user_id'];
$fullName = $_POST['user_fullName'];
$firstName = $_POST['user_firstName'];
$lastName = $_POST['user_lastName'];
$birthday = $_POST['user_birthday'];
$friends = $_POST['user_friends'];
$totalFriends = $_POST['user_totalFriends'];
$email = $_POST['user_email'];
$gender = $_POST['user_gender'];
$likes = $_POST['user_likes'];
$events = $_POST['user_events'];
$hometown = $_POST['user_hometown'];
require_once __DIR__ . '/db_connect.php';
$db = new DB_CONNECT();
$query = mysql_query("SELECT * FROM 'userData' WHERE 'id' = '$id'");
$user_data = mysql_num_rows($query);
if(empty($user_data)) {
$result = mysql_query("INSERT INTO userData(id, fullName, firstName, lastName, birthday, friends, totalFriends, email, gender, likes, events, hometown) VALUES ('$id', '$fullName', '$firstName', '$lastName', '$birthday', '$friends', '$totalFriends', '$email', '$gender', '$likes', '$events', '$hometown')");
if($result){
$response["success"] = 1;
$response["message"] = "User successfully created";
echo json_encode($response);
}
else{
$response["success"] = 0;
$response["message"] = "Error occurred";
echo json_encode($response);
}
}
else {
$response["success"] = 1;
$response["message"] = "User already in database";
echo json_encode($response);
}
}
else{
$response["success"] = 0;
$response["message"] = "Required userdata is missing";
echo json_encode($response);
}
?>
嗯。如果我在浏览器中运行此脚本,它会完美运行。当我在我的 android 应用程序中运行它时,它总是会创建一个新用户。
谢谢! T
【问题讨论】:
-
SELECT * FROM 'userData' WHERE 'id'去掉表名和列名上的单引号。 -
什么不起作用?是否有可以显示的错误或可以描述的行为?
-
请阅读 SQL 注入。