请向我解释这是为什么第二个警告没有显示

Question

看来下面的代码中的第二个警报不会出现; alert("The user is succesfully been added!");不会显示。我有一个用于用户名，名字，姓氏和密码的每个文本框的脚本。

首先，我试着做保存btn和文本框单独警报;方案是当我点击带有空字段的保存按钮时，它将显示用户添加成功的警报。那么它会显示另一个提醒，我需要填写文本框。所以我在想，如果我检查空的领域成功警报不会显示，它只会显示当我输入正确的东西。但发生的事情是它不会显示我期待展示的警报。

代码为全添加用户表单：

<div id="add_user" class="modal hide fade" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" 
    aria-hidden="true"> 
    <div class="modal-body"> 
     <div class="alert alert-info"><strong>Add User</strong></div> 
     <form class="form-horizontal" method="post"> 
      <div class="control-group"> 
       <label class="control-label" for="inputEmail">Username</label> 
       <div class="controls"> 
        <input type="text" id="inputEmail" name="username" placeholder="Username" required> 
       </div> 
      </div> 
      <div class="control-group"> 
       <label class="control-label" for="inputPassword">Password</label> 
       <div class="controls"> 
        <input type="password" name="password" id="inputPassword" placeholder="Password" required> 
       </div> 
      </div> 
      <div class="control-group"> 
       <label class="control-label" for="inputEmail">Firstname</label> 
       <div class="controls"> 
        <input type="text" id="inputEmail" name="firstname" placeholder="Firstname" pattern="[A-Za-z].{1,}" 
          required> 
       </div> 
      </div> 
      <div class="control-group"> 
       <label class="control-label" for="inputEmail">Lastname</label> 
       <div class="controls"> 
        <input type="text" id="inputEmail" name="lastname" placeholder="Lastname" pattern="[A-Za-z].{1,}" 
          required> 
       </div> 
      </div> 
      <div class="control-group"> 
       <div class="controls"> 
        <button name="submit" onClick="return saveuservalidation()" class="btn btn-success"><i 
          class="icon-save icon-large"></i>&nbsp;Save 
        </button> 
       </div> 
      </div> 
     </form> 
    </div> 
    <div class="modal-footer"> 
     <button class="btn" data-dismiss="modal" aria-hidden="true"><i class="icon-remove icon-large"></i>&nbsp;Close 
     </button> 
    </div> 
</div> 
<script> 
    function saveuservalidation() { 
     if (firstname == "" || lastname == "" || username == "" || password == "") { 
      return false; 
     } else { 
      return true; 
      alert("The user is succesfully been added!"); 
     } 
    } 
</script> 
<?php 
if (isset($_POST['submit'])){ 
$username=$_POST['username']; 
$password=$_POST['password']; 
$firstname=$_POST['firstname']; 
$lastname=$_POST['lastname']; 

mysql_query("insert into users (username,password,firstname,lastname) values('$username','$password','$firstname','$lastname')")or die(mysql_error()); 
} 
?> 

的JavaScript：（？这是正确的方式或不）

<script> 
    function saveuservalidation() { 
     if (firstname == "" || lastname == "" || username == "" || password == "") { 
      return false; 
     } else { 
      return true; 
      alert("The user is succesfully been added!"); 
     } 
    } 
</script> 

OR

<script> 
    function saveuservalidation() { 
     if (firstname == "" || lastname == "" || username == "" || password == "") { 
      return false; 
     } else { 
      alert("The user is succesfully been added!"); 
      return true; 
     } 
    } 
</script> 

它不会显示任何警报。

Answer 1

只要您return，该函数返回，并且之后的所有内容都不会执行（无法访问）。将其更改为

else{ 
    alert("The user is succesfully been added!"); 
    return true; 
} 

并且将显示警报。