Question

我是java腳本編碼的新手。我有一個需要用戶名，密碼和電子郵件的表單。它驗證其中三個，並不會進入下一個「logout.php」頁面，甚至不提醒，如果該ID已被採取。它只驗證文本框的條目。我想首先驗證這三個表單條目，然後如果它們是有效的，並且id是唯一的，那麼點擊提交「logout.php」應該被打開。 下面是我的代碼：

登錄up.php

<form method="post" name='registration' onSubmit="return formValidation();"> 

<input type="text" placeholder="Id*" required=" " name="uid"> 

<input type="password" placeholder="Password*" required=" " name="passid"> 

<input type="text" placeholder="Emailaddress*" required=" " name="uemail"> 

<input type="submit" value="Submit" style="font-face: 'Comic Sans MS'; font-size: larger; color: white; background-color: #FFA500; border: 3pt ridge lightgrey"> 
         </form> 

進行驗證的Java腳本代碼：

的JavaScript

<script type="text/javascript"> 
function formValidation() 
{ 
var uid = document.registration.uid; 
var passid = document.registration.passid; 
var uemail = document.registration.uemail; 


if(userid_validation(uid,5,12)) 
{ 

if(passid_validation(passid,7,12)) 
{ 

if(ValidateEmail(uemail)) 
{ 
} 
} 
} 
return false; 
} 


function userid_validation(uid,mx,my) 
{ 
var uid_len = uid.value.length; 
if (uid_len == 0 || uid_len >= my || uid_len < mx) 
{ 
alert("User Id should not be empty/length be between "+mx+" to "+my); 
uid.focus(); 
return false; 
} 
return true; 
} 

function passid_validation(passid,mx,my) 
{ 
var passid_len = passid.value.length; 
if (passid_len == 0 ||passid_len >= my || passid_len < mx) 
{ 
alert("Password should not be empty/length be between "+mx+" to "+my); 
passid.focus(); 
return false; 
} 
return true; 
} 


function ValidateEmail(uemail) 
{ 
var mailformat = /^\w+([\.-]?\w+)*@\w+([\.-]?\w+)*(\.\w{2,3})+$/; 
if(uemail.value.match(mailformat)) 
{ 
//alert("You have entered a valid email address!"); 
return true; 
} 
else 
{ 
alert("You have entered an invalid email address!"); 
uemail.focus(); 
return false; 
} 
} 

PHP的一部分將檢查身份證已被拿走或沒有。點擊提交按鈕後，這部分代碼不會被執行。這部分代碼中沒有顯示任何消息。

PHP

<?php 
session_start(); 
if(!empty($_POST)) { 
    class MyDB extends SQLite3 
    { 
     function __construct() 
     { 
      $this->open('trip.db'); 
     } 
    } 
    $db = new MyDB(); 
    if(!$db){ 
     echo $db->lastErrorMsg(); 
    } else { 

    } 
    $id=null; 
    $pass=null; 
    $email=null; 
    $id_exists=false; 

    if (isset($_POST['uid'])) { 
     $id = $_POST['uid']; 
    } 

    if (isset($_POST['passid'])) { 
     $pass = $_POST['passid']; 
    } 

     if (isset($_POST['uemail'])) { 
     $email = $_POST['uemail']; 
    } 



    $result= "SELECT COUNT(*) FROM Users WHERE ID = '".$id. "';" ; 
    $count= $db->querySingle($result); 

    if ($count > 0) 
    { 
     $id_exists = true; 
     echo "This id is not available. Please enter a valid id. "; 
    } 

    else 
    { 
     $sql= " INSERT INTO Users (ID, PASSWORD, EMAIL) 
      VALUES ('$id','$pass','$email'); " ; 
     $ret = $db->query($sql); 
     $_SESSION['Id'] = $id; 
     header("location:logout.php"); 
    } 


    $db->close(); 
} 
?> 

Answer 1

這是示例代碼。

function test_js() { 
 
if(document.test_form.id.value == "") { 
 
    alert('input name!'); 
 
    document.test_form.id.focus(); 
 
    return; 
 
} else { 
 
    document.test_form.action = "logout.php"; 
 
    document.test_form.submit(); 
 
} 
 
}

<form method="post" action="login.php" name="test_form"> 
 
     <input type="text" name="id"> 
 
     <input type="submit" value="Submit" onclick="javascript:test_js();"> 
 
</form>