php - 如何使用 PHP 检查用户是否已存在于 MySQL 中

Question

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/>";
    }
}

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

Answer 1

这段代码对你来说很好......

$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/>";
    }
}

Answer 2

mysqli_query函数返回结果集句柄。然后，您需要从中读取行：

$rs = mysqli_query($con,$check);
$data = mysqli_fetch_array($rs, MYSQLI_NUM);
if($data[0] > 1) {
    //user exists;
}

另请注意，这SELECT count(1) FROM ...将比SELECT count(*) FROM ... 您在小表中看不到太大差异要快，但是对于数十万行的大表，差异可能很大。

Answer 3

mysqli_query返回一个mysqli_result对象。与其将其与整数进行比较，不如尝试使用num_rows：

$res = mysqli_query($con,$check);
if($res->num_rows){
    //User exists
}

编辑：上面假设查询正在使用SELECT *，这不适用于SELECT COUNT(*). 查看 Aleks G 的答案。

Answer 4

您可以尝试以下代码：

$query_code = "SELECT COUNT(itemCode) FROM masterData WHERE itemCode='{$itemCode}'";
$result_login = mysqli_query($conn,$query_code);
$anything_found = mysqli_num_rows($result_login);

if($anything_found > 0) {
    $formOk = false;
    echo "ITEM CODE ALREADY EXISTS! Please try again.";  
}