Question

我有以下表单验证。

String[] strArr = {"p1", "p2", "p3"};
System.out.printf("%1$s %2$10s %2$10s", strArr );

验证工作正常，如果没有错误，它应该发出警报。但是，当我提交没有错误的表单时，我没有看到警告消息。我该如何解决这个问题？

Answer 1

试试这个

<?php 

$error_occured = 0;
$error_name = "";
$error_email = "";
$error_contact = "";
$error_comments = "";

if(isset($_POST["tx_name"])) {

        if(($tx_name == "") || (!preg_match("/^[a-zA-Z ]*$/",$tx_name))) {
                $error_occured = 1;
                $error_name = "Please enter a valid name";
            }

        if(($tx_email == "") || (!filter_var($tx_email, FILTER_VALIDATE_EMAIL))) {
                $error_occured = 1;
                $error_email = "Please enter a valid email";
            }

        if(($tx_contact == "") || !(is_numeric($tx_contact))) {
                $error_occured = 1;
                $error_contact = "Please enter a valid contact number";
            }

        if($tx_comments == "") {
                $error_occured = 1;
                $error_comments = "Please enter your message";
            }

     if($error_occured != 1) {
        echo "<script>alert('Hi!');</script>";
      }
 }



?>

如果没有错误，PHP表单验证会做警报

1 个答案: