一尘不染

如何使用PHP检查MySQL中是否已经存在用户

mysql

我正在使用以下对我不起作用的代码。

$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/>";
    }
}

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



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2020-05-17

共1个答案

一尘不染

该代码对您来说很好…

$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/>";
    }
}
2020-05-17