Question

有人可以告诉我下面filter_var函数中对validateEmail的调用有什么问题吗？如果我发表评论，可以在FireBug或Chrome控制台中看到该行hohum，另一方面Chrome报告的内容如下：This request has no response。来自java背景，我想知道我是否需要include一些库用于filter_var函数？

<?php

function validateInput($data, $con) {
  $data = trim($data);
  $data = stripslashes($data);
  $data = mysqli_real_escape_string($con, $data);
  return $data;
}

function validateEmail($email, $con) {
   $email = validateInput($email, $con);
   if (filter_var($email, FILTER_VALIDATE_EMAIL) {
      return $email;
   } else {
      return "";
   }
}

function validateText($text, $con) {
    $text = validateInput($text, $con);
    if(preg_match('/[^a-zA-Z]/', $text) == 0) {
        return $text;
    } else {
        return "";
    }
 }

 $hostname="localhost";
 $username="my_user";
 $password="my_psswd";
 $dbname="my_db";
 $con = mysqli_connect($hostname, $username, $password, $dbname);


 if (!$con) {
    header('HTTP', true, 500);
    echo('Could not connect: ' . mysql_error());
 } else {
    echo ("Hohum: ");
 }
?>

Answer 1

您错过了)，

if (filter_var($email, FILTER_VALIDATE_EMAIL)) {

FILTER_VALIDATE_EMAIL的filter_var导致未知错误

1 个答案: