如何使用过程php从mysql获取user_id？

Question

我是一名android / java开发人员，我正在努力学习php。 我已经做到了在数据库中插入新用户，现在我想获得他们的ID。

我该如何处理结果？我想将它分配给变量。

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

PS：我使用的是mysql，而不是mysqli。

编辑：这就是我所做的：

$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);

将用户插入（成功）到db：

后的错误消息

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

Answer 1

我假设您使用的是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

---

根据评论，您要求mysql_()版本，所以在这里...

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

仍然考虑使用mysqli_()或PDO。为什么？由于mysql_()现已弃用，请阅读文档页面上的红色框，上面写着......

请参阅PDO教程的this答案

Answer 2

以下是PDO变体：

<?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（但你不应该使用它，它已被弃用）：

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

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

Answer 4

连接：

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>")

查询：

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

Answer 5

您的错误是由于SQL查询中的错误引起的：您使用了=运算符两次：

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

必须：

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