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在mySQL数据库中检查现有用户的最佳方法?

mysql php

Saladin Akara · 技术社区 · 14 年前

function createUser($uname,$pword) {
        $server->connect(DB_HOST,DB_USER,DB_PASS,DB_NAME);
        $this->users = $server->query("SELECT * FROM user_list");
        while ($check = mysql_fetch_array($this->users) {
            if ($check['uname'] == $uname) {

我不确定的是这样做的最佳逻辑。我正在考虑添加一个布尔变量来执行以下操作(在if语句之后):

$boolean = true;
}
if ($boolean) {
    echo "User already exists!";
    }
else {
    $server->query("INSERT USER INTO TABLE");
    echo "User added Successfully";
    }

但这似乎有点低效-有没有更有效的方法来做到这一点?抱歉,如果这有一个基本的解决方案-我是一个相对较新的PHP程序员。

2 回复 | 直到 11 年前

Gumbo 14 年前

使用 WHERE 子句仅获取具有给定用户名的行:

"SELECT * FROM user_list WHERE uname='".$server->real_escape_string($uname)."'"

然后检查查询是否导致选择具有 MySQLi_Result::num_rows :

function createUser($uname,$pword) {
    $server->connect(DB_HOST,DB_USER,DB_PASS,DB_NAME);
    $result = $server->query("SELECT * FROM user_list WHERE uname='".$server->real_escape_string($uname)."'");
    if ($result->num_rows() === 0) {
        if ($server->query("INSERT INTO user_list (uname) VALUES ('".$server->real_escape_string($uname)."'")) {
            echo "User added Successfully";
        } else {
            echo "Error while adding user!";
        }
    } else {
        echo "User already exists!";
    }
}

Martin Bean 14 年前

<?php
$errors = array();
$alerts = array();

if (isset($_POST['register'])) {

    $pdo = new PDO('[dsn]', '[user]', '[pass]');

    // first, check user name has not already been taken
    $sql = "SELECT COUNT(*) AS count FROM user_list WHERE uname = ?";
    $smt = $pdo->prepare($sql);
    $smt->execute(array($_POST['uname']));
    $row = $smt->fetch(PDO::FETCH_ASSOC);
    if (intval($row['count']) > 0) {
        $errors[] = "User name " . htmlspecialchars($_POST['uname']) . " has already been taken.";
    }

    // continue if there are no errors
    if (count($errors)==0) {
        $sql = "INSERT INTO user_list ([fields]) VALUES ([values])";
        $res = $pdo->exec($sql);
        if ($res==1) {
            $alerts[] = "Member successfully added.";
        } else {
            $errors[] = "There was an error adding the member.";
        }
    }
}