如何用PHP检查MySQL中是否已经存在用户

Pro*_*jee 0 php mysql

我正在使用以下代码,这对我不起作用.

$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/>";
    }
}
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mysqli_result类的对象无法在C:\ xampp\htdocs\Exp\welcome.php中转换为int

Vij*_*jay 6

这段代码适合你......

$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,$newUser))
    {
        echo "You are now registered<br/>";
    }
    else
    {
        echo "Error adding user in database<br/>";
    }
}
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  • 这可能会导致通知.如果未找到记录,则数据不是数组.可以通过将分配替换为数据并使用_IF检查数据[0]来修复($ data = mysqli_fetch_array($ rs,MYSQLI_NUM)){_ (2认同)