PHP和JavaScript表單驗證腳本的問題

Question

我在使用此腳本進行表單驗證時遇到了一些麻煩，最初使用JavaScript，然后再次進行驗證，然后使用PHP提交。 有人可以在這里看到問題嗎？ 打開文件時我什么也沒看到。 我對PHP相當陌生。 忽略表格格式。 這是一本書的實驗室，因此重點是PHP和JavaScript。 我知道CSS布局等。

任何幫助將不勝感激。 提前致謝

<?php

error_reporting(~0); ini_set('display_errors', 1);

//PHP adduser.php

//Start with the PHP code

$forename=$surname=$password=$age=$email="";

if(isset($_POST['forename']))
$forename = fix_string($_POST['forename']);
if(isset($_POST['surname']))
$surname = fix_string($_POST['surname']);
if(isset($_POST['password']))
$password = fix_string($_POST['password']);
if(isset($_POST['age']))
$age = fix_string($_POST['age']);
if(isset($_POST['email']))
$email = fix_string($_POST['email']);

$fail= validate_forename($forename);
$fail.= validate_surname($surname);
$fail.= validate_password($password);
$fail.= validate_age($age);
$fail.= validate_email($email);

echo "<html><head><title>An example form</title></head><body>";

if($fail==""){
  echo "Form data successfully validated: $forename,
  $surname, $password, $age, $email";        
}



//This is where you'd enter post fields to the DB
exit;

//Now the HTML & JavaScript goes here

回聲<<< _ SOQ

<style type="text/css">.signup{
  border:1px solid #999999;
  font:normal 14px helvetica;
  color:#444444;
  }
</style>


 <script type="text/javascript">

   function validate(form){

    fail = validateForename(form.forename.value);
    fail += validateSurname(form.surname.value);
    fail += validatePassword(form.password.value);
    fail += validateAge(form.age.value);
    fail += validateEmail(form.email.value);

      if(fail=="") return true;
      else alert(fail); return false;

    }

</script>

</head>
    <body>
      <table class="signup" border="0" cellpadding="2" cellspacing="5" 
       bgcolor="#eeeeee">
        <th colspan="2" align="center">Sign up form</th>

  <tr>
    <td colspan="2">Sorry, the following errors were found<br/>
     in your form: <i>$fail</i>
    </td>
  </tr>


<form method="post" action="adduser.php" onsubmit="validate(this.form)">

<tr><td>Forename:</td><td><input type="text" maxlength="32"
  name="forename" value="$forename"/></td>
</tr>

<tr><td>Surname:</td><td><input type="text" maxlength="32"
  name="surname" value="$surname"/></td>
</tr>

<tr><td>Password:</td><td><input type="text" maxlength="32"
  name="password" value="$password"/></td>
</tr>

<tr><td>Age:</td><td><input type="text" maxlength="32"
  name="age" value="$age"/></td>
</tr>

<tr><td>Email:</td><td><input type="text" maxlength="32"
  name="email" value="$email"/></td>
</tr>

<tr><td colspan="2" align="center">
   <input type="submit" value="Sign-up"/></td>
</tr>

  </form>
</table>

<script type="text/javascript">

function validateForename(field){
  if(field=="") return "No surname was entered";
  return "";
}

function validateSurname(field){
  if(field=="") return "No surname was entered";
  return "";
}

function validatePassword(field){
   if(field=="") return "No surname was entered";
   else if(field.length<6) return "Passwords, must be at least 6 characters";
   else if([^/a-zA-Z0-9_-/]) return "Only a-zA-Z0-9_- characters are allowed ";
   return "";
 }

function validateAge(field){
   if((field=="") || (isNaN(field)) return "No age was entered";
   else if((field<18) || (field>101))  return "Age must be between 18 and 101 years";
   else if(/[^a-zA-Z0-9_-]/.test(field)) 
   return "Only a-zA-Z0-9_- characters are allowed ";
   return "";
 }

 function validateEmail(field){
   if(field=="") return "No surname was entered";
   else if(!((field.indexOf('@')>0) && (field.indexOf('.')>0)) ||
   /[^a-zA-Z0-9-]/.test(field)) return "E-mail address invalid";
   return "";
 }

</script>

  </body>
</html>

_SOQ;

//最后是PHP函數

function validate_forename($field){
  if($field=="") return "No forename was entered";
  return "";
}

function validate_surname($field){
  if($field=="") return "No surname was entered";
  return "";
}

function validate_password($field){
  if($field=="") return "No password was entered";
  elseif(strlen($field) <6) return "Passwords, must be at least 6 characters";
  elseif(!preg_match("/[^a-zA-Z0-9_-]/", $field)) 
  return "Only a-zA-Z0-9_- characters are allowed ";
  return "";
}

function validate_age($field){
  if($field<18 || field>101) return "Age must be between 18 and 101 years";
  return "";
}

function validate_email($field){
  if($field=="") return "No surname was entered";
  elseif(!(strpos($field, ".")>0) && 
         (strpos($field, "@")>0) ||
          preg_match("/[^a-zA-Z0-9_]/",$field)) 
        return "E-mail address invalid";
        return "";
}

//sanitise the PHP input

function fix_string($string){
  if(get_magic_quotes_gpc($string)) stripslashes($string);
  return htmlentities($string);

 }

？>

Answer 1

如果您什么都沒有看到，則說明您的PHP錯誤

轉到php.ini並將錯誤報告設置為E_ALL並將錯誤顯示為on或

<?php
     error_reporting(E_ALL);
     ini_set("display_errors", 1);
?>

Answer 2

if($fail=""){

您分配而不是檢查是否相等。 然后，由於分配的值為false，因此將永遠不會執行相應的成功代碼。

Answer 3

當您實際上想要$fail==""時，您似乎正在使用$fail="" $fail=="" 。 第一個是賦值表達式，第二個帶有==的表達式是比較表達式。 不管之前將$fail設置為什么，您的用法將始終解析為false （通過您剛剛設置的空字符串），因為您只是將其重置：

$fail = "hello";
var_dump($fail); //string(5) "hello"
var_dump($fail=""); //string(0) ""
var_dump(""); //string(0) ""

所以你的if語句就變成

if("") {
    //never get here
} else {
    //always get here, but you have nothing defined
}

PHP和JavaScript表單驗證腳本的問題

問題描述

3 個解決方案

解決方案1
1 2012-09-01 12:05:44

解決方案2
1 2012-09-01 13:31:44

解決方案3
1 2012-09-01 13:32:18

PHP和JavaScript表單驗證腳本的問題

問題描述

3 個解決方案

解決方案1 1 2012-09-01 12:05:44

解決方案2 1 2012-09-01 13:31:44

解決方案3 1 2012-09-01 13:32:18

解決方案1
1 2012-09-01 12:05:44

解決方案2
1 2012-09-01 13:31:44

解決方案3
1 2012-09-01 13:32:18