php - INSERT INTO SQL php 函数不起作用

Question

我已经设法编写了一个 php 脚本来检查用户名是否已存在于数据库中，并且仅在用户名不存在时才添加新用户。

这是我的 php 脚本:

<?php
require "init.php";

if(isset($_POST['username']) && isset($_POST['forename']) && isset($_POST['surname']) && isset($_POST['password'])){
    $username = $_POST['username'];
    $forename = $_POST['forename'];
    $username = $_POST['surname'];
    $password = $_POST['password'];

    $stmt = "SELECT username FROM users WHERE username = ?";
    $result = $dbcon -> prepare($stmt);
    $result->bind_param('s', $username);
    $result->execute();
    $result->bind_result($username);

    if($result->fetch()){
        echo "Can't add new user as it already exists!";
    }
    else{
        $stmt_two = "INSERT INTO users (username, forename, surname, password) 
                VALUES(?, ?, ?, ?)";
        $result_two = $dbcon -> prepare($stmt_two);
        $result_two->bind_param('ssss', $username, $forename, $surname, $password);
        $result_two->execute();
        $result_two->close();
        echo json_encode("Success");
    }
}
?>

我相信由于我有多个准备好的语句，记录没有被插入或被间歇性地插入。如果我只是用 SELECT FROM 语句单独执行 INSERT INTO 语句 - 记录几乎会立即添加。

这是为什么，我的代码有什么问题？

谢谢

Answer 1

正如我在评论中所说，不要过于复杂，只需检查找到的行数即可。不需要取任何东西。您只是在检查该用户是否存在。

$stmt = "SELECT username FROM users WHERE username = ?";
$result = $dbcon->prepare($stmt);
$result->bind_param('s', $username);
$result->execute();
$result->store_result();

if($result->num_rows() > 0) { // if it exists

} else {
    // make your insertions
}

还有一个注意事项:

isset 可以接受多个参数:

if(isset($_POST['username'], $_POST['forename'], $_POST['surname'], $_POST['password'])) {
    // and so on
}

编辑:另一种风格(使用 MySQL 的 COUNT()):

$stmt = "SELECT COUNT(username) FROM users WHERE username = ?";
$result = $dbcon->prepare($stmt);
$result->bind_param('s', $username);
$result->execute();
$result->bind_result($count);
$result->fetch();

if($count > 0) { // exists

} else {
    // do something else
}