如何使用PHP检查MySQL中是否已经存在用户 [英] How to check if user already exists in MySQL with PHP

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本文介绍了如何使用PHP检查MySQL中是否已经存在用户的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在使用以下对我不起作用的代码.

I am using following code which is not working for me.

$con=mysqli_connect("localhost","root","","my_db");
$check="SELECT COUNT(*) FROM persons WHERE Email = '$_POST[eMailTxt]'";
if (mysqli_query($con,$check)>=1)
{
    echo "User Already in Exists<br/>";
}
else
{
    $newUser="INSERT INTO persons(Email,FirstName,LastName,PassWord) values('$_POST[eMailTxt]','$_POST[NameTxt]','$_POST[LnameTxt]','$_POST[passWordTxt]')";
    if (mysqli_query($con,$newUser))
    {
        echo "You are now registered<br/>";
    }
    else
    {
        echo "Error adding user in database<br/>";
    }
}

无法将mysqli_result类的对象转换为int C:\ xampp \ htdocs \ Exp \ welcome.php

Object of class mysqli_result could not be converted to int in C:\xampp\htdocs\Exp\welcome.php

推荐答案

此代码适合您...

$con=mysqli_connect("localhost","root","","my_db");
$check="SELECT * FROM persons WHERE Email = '$_POST[eMailTxt]'";
$rs = mysqli_query($con,$check);
$data = mysqli_fetch_array($rs, MYSQLI_NUM);
if($data[0] > 1) {
    echo "User Already in Exists<br/>";
}

else
{
    $newUser="INSERT INTO persons(Email,FirstName,LastName,PassWord) values('$_POST[eMailTxt]','$_POST[NameTxt]','$_POST[LnameTxt]','$_POST[passWordTxt]')";
    if (mysqli_query($con,$newUser))
    {
        echo "You are now registered<br/>";
    }
    else
    {
        echo "Error adding user in database<br/>";
    }
}

这篇关于如何使用PHP检查MySQL中是否已经存在用户的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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