Question

我正在尝试检查我的模拟Twitter项目上的riegistration是否存在句柄或电子邮件我称之为bleeter，但我收到此错误：

注意：类mysqli_result的对象无法在第20行的first.php中转换为int

这是我检查num_rows_handle == 0的行。我该怎么做？

$query_check_handle = "SELECT * FROM users WHERE handle = " . $handle;
$num_rows_handle = mysqli_query($dbc, $query_check_handle);
if ($num_rows_handle == 0) { // Line 20
    //check email
    $query_check_email = "SELECT * FROM users WHERE email = " . $email;
    $num_rows_email = mysqli_query($dbc, $query_check_email);
    if ($num_rows_email == 0) {
        $query_register = "INSERT INTO users (first_name,  last_name, handle, password, email, ACL) 
                        VALUES ('$fname', '$lname',  '$handle', '$pass', '$email', '$ACL')"
        or DIE ("error running the query.");
        mysqli_query($dbc, $query_register);        
        echo "<br>";
        echo "You have succesfully registered, please login";
    } else {
        echo "<br>";
        echo "Email already in use.  Please try again";
    }
} else {
    echo "<br>";
    echo "Handle already in use.  Please try again";
}

Answer 1

当查询为SELECT时，mysqli_query会返回类型为mysqli_result的对象，其对象的属性num_rows包含行数。因此

$result = mysqli_query($dbc, $query_check_email);
$num_rows_email = mysqli_num_rows($result);

检查PHP和mysqli中是否存在电子邮件和用户名

1 个答案: