错误使用PHP

Question

我想更新使用PHP这是在logcat中diplayed错误在数据库中的数据更新mysql中的数据是

Error: UPDATE usersSET Question1=null2null,Question3=nullnull,Question4=nullnullnullnullWHERE email=bb<br>You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '=null2null,Question3=nullnull,Question4=nullnullnullnullWHERE email=bb' at line 1{"error":false,"uid":"56bd5f88afb7b3.99372648","user":{"name":"Bb","email":"bb","created_at":"2016-02-12 09:58:56","updated_at":null,"Question1":"","Question3":"","Question4":""} 

的问题1，问题3和问题4列中的值不会得到更新，它应该分别为null2null，nullnull，nullnullnullnull。它被写入检查您的版本的语法，我在本地运行5.5.12 php版本和5.6.17 mysql版本。我检查了语法和不同的网站上显示的不同显示不同的syntasx如何得到正确的，下面是我的PHP代码

PHP

<?php 
$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "android_api"; 

// Create connection 
$conn = new mysqli($servername, $username, $password, $dbname); 
require_once 'include/DB_Functions.php'; 
$db = new DB_Functions(); 

// json response array 
$response = array("error" => FALSE); 

if (isset($_POST['Question1']) && isset($_POST['Question3']) && isset($_POST['Question4'])) { 

    // receiving the post params 
    $email = $_POST['email']; 
    $password = $_POST['password']; 

    $Question1 = $_POST['Question1']; 
    $Question3 = $_POST['Question3']; 
    $Question4 = $_POST['Question4']; 
/*$sql = "INSERT INTO users (Question1, Question2, Question4) 
    VALUES ('$Question1', '$Question3', '$Question4')"; */ 
    $user = $db->getUserByEmailAndPassword($email, $password); 
    // $result = mysql_query("UPDATE users"."SET Question1='$Question1',Question3='$Question3',Question4='$Question4'"."WHERE email=$email"; 
    $sql="UPDATE users"."SET Question1=$Question1,Question3=$Question3,Question4=$Question4"."WHERE email=$email"; 
    if ($conn->query($sql) === TRUE) { 
     echo "New record created successfully"; 
    } else { 
     echo "Error: " . $sql . "<br>" . $conn->error; 
    } 
    // get the user by email and password 
    $user = $db->getUserByEmailAndPassword($email, $password); 

    if ($user != false) { 
     // use is found 
     $response["error"] = FALSE; 
     $response["uid"] = $user["unique_id"]; 
     $response["user"]["name"] = $user["name"]; 
     $response["user"]["email"] = $user["email"]; 
     $response["user"]["created_at"] = $user["created_at"]; 
     $response["user"]["updated_at"] = $user["updated_at"]; 
     $response["user"]["Question1"] = $user["Question1"]; 
      $response["user"]["Question3"] = $user["Question3"]; 
      $response["user"]["Question4"] = $user["Question4"]; 
     echo json_encode($response); 
    } else { 
     // user is not found with the credentials 
     $response["error"] = TRUE; 
     $response["error_msg"] = "ABCD"; 
     echo json_encode($response); 
    } 
} else { 
    // required post params is missing 
    $response["error"] = TRUE; 
    $response["error_msg"] = "abcd"; 
    echo json_encode($response); 
} 
?> 

编辑

阅读答案后现在更改语法错误更改为

Error: UPDATE users SET Question1=null2null,Question3=nullnull,Question4=nullnullnullnull WHERE email=bb<br>Unknown column 'bb' in 'where clause'{"error":false,"uid":"56bd5f88afb7b3.99372648","user":{"name":"Bb","email":"bb","created_at":"2016-02-12 09:58:56","updated_at":null,"Question1":"","Question3":"","Question4":""}} 

Answer 1

Mi在这里和那里发现了几个白色空间和qutoes（'）。改变你的SQL语句来此

$sql ="UPDATE users" . " SET Question1 = '".$Question1."', Question3 = '".$Question3."', Question4= '".$Question4."' WHERE email = '".$email."'"; 

而且，在这些变量的值不正确的像，为什么你没有得到正确的价值观（或者是你逝去的是什么？）

Question1 = null2null 
Question3 = nullnull 
Question4 = nullnullnullnull 
email = bb 

Answer 2

在这里，你错过了3两件事：
1. usersSET
2.行情赋值（问题1 = null2null）之前WHERE关键字（问题4 = nullnullnullnullWHERE）
3.空间后，空间查询的正确版本如下：

UPDATE users SET Question1='null2null', Question3='nullnull' , 
    Question4='nullnullnullnull' WHERE email='b' 

所以取代下面一行在你的脚本：

$sql ="UPDATE users"."SET Question1=$Question1,Question3=$Question3,Question4=$Question4"."WHERE email=$email"; 

与

$sql =" UPDATE users SET Question1 = '".$Question1."', Question3 = '".$Question3."', Question4= '".$Question4."' WHERE email = '".$email."'"; 

Answer 3

嗨，你需要正确地保存细节问题归因于数据包含null作为字符串。

你需要如下准备查询：

Question1 = null2null

你要引用的所有参数保存到数据库中之前。这样

Question1 = 'null2null'

这里是下面的示例更新的代码，供大家参考，让习惯来清洁您的查询，同时具有更好的可读性，适当引用的字符串创建。

$email = $_POST['email']; 
$password = $_POST['password']; 

$Question1 = $_POST['Question1']; 
$Question3 = $_POST['Question3']; 
$Question4 = $_POST['Question4']; 

$user = $db->getUserByEmailAndPassword($email, $password); 

$sql="UPDATE users SET Question1 = '$Question1', Question3 = '$Question3', Question4 = '$Question4' WHERE email = '$email'"; 
if ($conn->query($sql) === TRUE) { 
    echo "New record created successfully"; 
} else { 
    echo "Error: " . $sql . "<br>" . $conn->error; 
}