Question

当表单操作是另一个页面（php代码）时，在php中进行表单验证。如何在视图页面中验证表单并显示错误消息？

查看页面

<html>
<body>
<form action="redirect.php" method="POST" enctype="multipart/form-data">  
    <label for="name"><b>Name: </b></label>
    <input type="text"  name="name" id="name" >
<br>
    <label for="email"><b>Email: </b></label>
    <input type="text"  name="email" id="email" >
<br>
    <label for="email"><b>Project Type: </b></label>
    <input type="text"  name="projecttype" id="projecttype">
<br>
    <label for="email"><b>Project Description: </b></label>
    <input type="text"  name="pdescription" id="pdescription" >
<br>
    <label for="file"><b>File Upload: </b></label>
    <input type="file"  name="file"  id="file">
<br>
    <div class="clearfix">
      <input type="submit" name="submit1" id="submit1" value="submit1">
    </div>  
</form>
</body>
</html>

redirect.php

<?php
if(isset($_POST['submit1'])) 
{
    $name=$_POST['name'];
    $email=$_POST['email'];
    $projecttype=$_POST['projecttype'];
    $projectdetails=$_POST['pdescription'];
    $attachment=$_FILES['file']['name'];
    $tmp_name=$_FILES['file']['tmp_name'];
    move_uploaded_file($tmp_name,"upload/".$attachment);
    $query=mysql_query("insert into projectdetails(Customer_name,Customer_email,Customer_attachment,Customer_project_type,Customer_project_details) values('$name','$email','$attachment','$projecttype','$projectdetails')");
    if($query)
    {   
        header('Location: test2.php');      
    }
    ?>

Answer 1

为name field required in php using session

创建简单的演示

你可以这样试试

<强> form.php

<?php
    session_start();
    if(isset($_SESSION['msg'])){
        $name=$_SESSION['msg']['name'];
    }
?>
<html>
<body>
<form action="redirect.php" method="POST" enctype="multipart/form-data">  
    <label for="name"><b>Name: </b></label>
    <input type="text"  name="name" id="name" > <span style="color:red"><?php echo $name ?></span>
<br>
    <label for="email"><b>Email: </b></label>
    <input type="text"  name="email" id="email" >
<br>
    <label for="email"><b>Project Type: </b></label>
    <input type="text"  name="projecttype" id="projecttype">
<br>
    <label for="email"><b>Project Description: </b></label>
    <input type="text"  name="pdescription" id="pdescription" >
<br>
    <label for="file"><b>File Upload: </b></label>
    <input type="file"  name="file"  id="file">
<br>
    <div class="clearfix">
      <input type="submit" name="submit1" id="submit1" value="submit1">
    </div>  
</form>
</body>
</html>

<强> redirect.php

<?php
session_start();
if(isset($_POST['submit1'])) 
{

    $name=$_POST['name'];

    if(isset($name) && empty($name)){
        $_SESSION['msg']['name']="Name must be required!";
        header('location: form.php');  
        exit;
    }
    $email=$_POST['email'];
    $projecttype=$_POST['projecttype'];
    $projectdetails=$_POST['pdescription'];
    $attachment=$_FILES['file']['name'];
    $tmp_name=$_FILES['file']['tmp_name'];
    move_uploaded_file($tmp_name,"upload/".$attachment);
    $query=mysql_query("insert into projectdetails(Customer_name,Customer_email,Customer_attachment,Customer_project_type,Customer_project_details) values('$name','$email','$attachment','$projecttype','$projectdetails')");
    if($query)
    {   
        header('Location: test2.php');      
    }
}
?>

Answer 2

我建议使用HTML5的简单方法有很多种方法

 <script src="http://code.jquery.com/jquery-1.9.1.js"></script>

$(function () {
        $('form').bind('click', function (event) {
         event.preventDefault();
         $.ajax({
            type: 'POST',
            url: 'post.php',
            data: $('form').serialize(),
            success: function () {
              alert('form was submitted');
            }
          });

        });
      });

    <html>
    <body>
    <form action="redirect.php" method="POST" enctype="multipart/form-data">  
        <label for="name"><b>Name: </b></label>
        <input type="text"  name="name" id="name" required>
    <br>
        <label for="email"><b>Email: </b></label>
        <input type="email"  name="email" id="email" required>
    <br>
        <label for="email"><b>Project Type: </b></label>
        <input type="text"  name="projecttype" id="projecttype" required>
    <br>
        <label for="email"><b>Project Description: </b></label>
        <input type="text"  name="pdescription" id="pdescription" required>
    <br>
        <label for="file"><b>File Upload: </b></label>
        <input type="file"  name="file"  id="file" required>
    <br>
        <div class="clearfix">
          <input type="submit" name="submit1" id="submit1" value="submit1">
        </div>  
    </form>
    </body>
    </html>

在PHP PAge中：

if(empty($_POST['name'])){
 echo "Name is required";
}

Answer 3

在重定向之前在会话中设置错误，在视图页面上回显它们然后在redirect.php顶部取消设置错误，通过在顶部取消设置它们，您将确保始终拥有最新的错误消息。

Answer 4

您的代码中存在严重问题

Do not use mysql_* functions in PHP

可以在您的代码中完成微调

您不需要将您的输入发送到另一个页面进行验证

您的一些示例

Client Side form Validation

PHP MySQLi Basic Examples

如果您需要继续执行当前的实施

声明一个名为$formavalid的变量，并将其设为true和false

<html>
<body>
<?php 
$formvalid = true;
?>
<form action="redirect.php" method="POST" enctype="multipart/form-data">  
    <label for="name"><b>Name: </b></label>
    <input type="text"  name="name" id="name" >
<br>
    <label for="email"><b>Email: </b></label>
    <input type="text"  name="email" id="email" >
<?php
//Do like this for all fields
if(isset($_POST['submit1'])) 
{
if(!isset($_POST['email'])) 
{
echo 'Email field is missing';
}else{
$formvalid = false;// Make the $formvalid as false if your input is dirty
}
}
<br>
    <label for="email"><b>Project Type: </b></label>
    <input type="text"  name="projecttype" id="projecttype">
<br>
    <label for="email"><b>Project Description: </b></label>
    <input type="text"  name="pdescription" id="pdescription" >
<br>
    <label for="file"><b>File Upload: </b></label>
    <input type="file"  name="file"  id="file">
<br>
    <div class="clearfix">
      <input type="submit" name="submit1" id="submit1" value="submit1">
    </div>  
</form>
<?php
if($formvalid==true){
//Form is valid so do the action for it
}else{
//form is not valid, so let the user input to make it valid
}
?>
</body>
</html>

希望，这有助于你

表单操作是另一个页面（php代码）如何验证表单并在视图页面中显示错误消息

4 个答案: