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我是 mysqli 的新手，我想用 php 和 mysqli 检查数据库中是否已经存在电子邮件

这是我到目前为止所做的：

ini.php

<?php
session_start();
require 'connect.php';
include 'user.func.php'; 
?>

连接.php

<?php
$con = new mysqli('host', 'user', 'password', 'db');
if($con->connect_errno > 0){
die('Sorry, We\'re experiencing some connection problems.'); 
}

?>

注册.php

<?php
include 'ini.php';
?>

<form action="" method="post">
<p>Username:<br/><input type="text" name="reg_name" maxlength="50"  ></p>
<p>Email:<br/><input type="text" name="reg_email" size="35" maxlength="50"  ></p>
<p>Password:<br/><input type="password" name="reg_password" maxlength="50" ></p>
<p>Re-type Password:<br/><input type="password" name="reg_re_password" maxlength="20"></p>
<p><input type="submit" value="Register" ></p>
</form>

<?php
include 'validate.php';
?>

验证.php

<?php
if(isset ( $_POST['reg_name'],$_POST['reg_email'], $_POST['reg_password'] )) {

$errors = array();
$name = $_POST['reg_name'];
$email = $_POST['reg_email'];
$password = $_POST['reg_password'];
$re_password = $_POST['reg_re_password'];
if(empty($name) || empty($email) || empty($password) || empty($re_password)){
$errors[] = 'All fields are required';
 } else {

if(strlen($name) > 50 || strlen($email) > 50 || strlen($password) > 50) {
 $errors[] = 'One or more field has too long Characters';
}

if(filter_var($email, FILTER_VALIDATE_EMAIL) === FALSE && $password !== $re_password){
$errors[] = 'Enter valid email address and password do not match';
}else{

if(filter_var($email, FILTER_VALIDATE_EMAIL) === FALSE) {
$errors[] = 'Please enter a valid email address';
}

if($password !== $re_password){
$errors[] = 'Both passwords do not match';
}
if(user_exists($email) > 0){
$errors[] = 'That email is already been registered';
}

}

}

if (!empty($errors)) {
foreach ($errors as $errors) {
    echo '<strong>',$errors,'</srtong><br />';
}
}else{
echo 'Registered';
}
}

?> user.func.php

<?php    
function user_exists($email) {

$con = new mysqli('host', 'user', 'password', 'db');
$query = "SELECT * FROM users WHERE email = ?";
$stmt = $con->prepare($query);
$stmt->bind_param("s", $email);

if ($stmt->execute()) {
return $stmt->num_rows;
}
return false;
}
?>

此代码显示所有其他错误，但不检查电子邮件是否存在。相反，即使我输入已经在数据库中的电子邮件，它也会显示已注册。

我有名为“测试”的数据库。和表“用户”与列 user_id、用户名、电子邮件和密码。

谁能告诉如何为此编写mysqli查询？

我在mysql中使用了这段代码

function user_exists($email) {
$email = mysql_real_escape_string($email);
$query = mysql_query("SELECT COUNT('user_id') FROM 'users' WHERE 'email'='$email'");
return (mysql_result($query, 0) == 1) ? true :false;
}
4

2 回答 2

2

使用此条件，因为您要从查询中返回行数。

if(user_exits($register_email) > 0){

编辑

function user_exists($email) {
    $mysqli = new mysqli("host", "my_user", "my_password", "your_db");
    $query = "SELECT COUNT(*) AS num_rows FROM users WHERE email = ?";
    $stmt = $mysqli->prepare($query);
    $stmt->bind_param("s", $email);

    if ($stmt->execute()) {
        return $stmt->num_rows;
    }
    return false;
}
于 2013-04-26T07:14:17.583 回答
1

我可以告诉如何用PDO做到这一点

<?php
function user_exists($email) {
    $query = "SELECT 1 FROM " . USER_TABLE . " WHERE email = ?";
    $stmt = $this->_db->prepare($query);
    $stmt->execute(array($email));
    return (bool)$stmt->fetchColumn();
}

虽然我认为严格检查没有意义

if(user_exits($register_email))

绰绰有余

另外，我不明白你从哪里得到这个$register_email变量。你有错误报告吗？

顺便说一句，您需要为此建立一个 PDO 连接。此代码不适用于 mysqli

于 2013-04-26T07:14:22.113 回答