Question

我是一名android/java開發人員，我正在努力使用php。 我已經做到了在數據庫中插入新用戶，現在我想獲得他們的ID。

我該如何處理結果？我想將它分配給一個變量。 PS：我使用的是mysql，而不是mysqli。我不知道如何使用mysql，而不是mysqli。

編輯：這是我做的：

$query = "SELECT user_id FROM users WHERE user_email = '" . $user_email ."';"; 
$store_info = mysql_fetch_array(mysql_query($query)); 
$user_id = $store_info['user_id']; 
$response["message"] = "User created with id: " . $user_id; 
echo json_encode($response); 

和插入（成功）後的錯誤信息在數據庫中的用戶：

null{"success":3,"message":"User created with id: "} 

Answer 1

我假設你正在使用的MySQLi API

$query = "SELECT user_id FROM users WHERE user_email = '$user_email'"; //Your Query 

$store_info = mysqli_fetch_array(mysqli_query($connection, $query)); 
//Execute the query, fetch the result, it's just one result so no need for a while loop 

echo $store_info['user_id']; //echo id 

---

按照與c omments，你請求mysql_()版本，所以在這裏你去...

$query = "SELECT user_id FROM users WHERE user_email = '$user_email'"; //Your Query 

$store_info = mysql_fetch_array(mysql_query($query)); 
//Execute the query, fetch the result, it's just one result so no need for a while loop 

echo $store_info['user_id']; //echo id 

不過考慮使用mysqli_()或PDO代替。爲什麼？因爲mysql_()現在已經廢棄，閱讀文檔頁面它說紅色框...

參考this答案PDO教程

Answer 2

這裏是PDO變種：

<?php 
//credentials 
$host = 'localhost'; 
$user = "user"; 
$password = ''; 
$db_name = 'test'; 
$port = 3306; 

//connection to the database 
try 
{ 
    $connection = new PDO("mysql:host=$host;port=$port;dbname=$db_name", $user, $password); 
    $connection->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
} 
catch (PDOException $e) 
{ 
    echo 'Connection failed: ' . $e->getMessage(); 
} 

//prepare and execute SELECT statement 
$sth = $connection->prepare("SELECT user_id FROM users WHERE user_email = :email"); 
$sth->execute(array(':email' => $user_email)); 

$record = $sth->fetch(PDO::FETCH_ASSOC); 
print $record["user_id"]; 

Answer 3

如果你使用mysql（但你不應該這樣做，它已被廢棄）：

$result = mysql_query("SELECT user_id FROM users WHERE user_email = '$user_email'"); 
$row = mysql_fetch_row($result); 

echo $row[0]; // you result (id) 

Answer 4

連接：

define("HOST","localhost"); 
define("USER","mysql_username"); 
define("PASS","password"); 

$conn = mysql_connect(HOST,USER,PASS) or die("<h3>Sorry, could not connect to MySQL. Please Try Again</h3>"); 
$db = mysql_select_db(DBNAME,$conn) or die("<h3>Sorry, could not connect to Database. Please Try Again</h3>") 

查詢：

$query = "SELECT user_id FROM users WHERE user_email = = '" . $user_email . "'"; 
$result = mysql_query($query); 
$row=mysql_fetch_assoc($result); 

Answer 5

你的錯誤出現，因爲在SQL查詢中的錯誤：你用=操作兩次：

$query = "SELECT user_id FROM users WHERE user_email = = '" . $user_email . "'"; 

必須是：

$query = "SELECT user_id FROM users WHERE user_email = '" . $user_email . "'";