PHP Form Get 和 Post Output 用 If 语句在同一页输出结果

Question

我正在学习 PHP 并正在处理表格。我有一个任务要创建一个表单，一个 GET 和一个 POST 在同一页面上，并将结果发布在每个表单下方的同一页面上。

我创建了一个 if/else 语句，但可能我的处理方式不正确。

<div>
<h1>POST Method Form</h1>
<form  method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
First Name:<br>
<input type="text" name="firstname"><br>
Last Name:<br>
<input type="text" name="lastname"><br>
E-mail: <input type="text" name="email"><br>
Database Utilized:  <br>
<input type="checkbox" name="dba" value="SQL Server">SQL Server<br>
<input type="checkbox" name="dba" value="Oracle">Oracle<br>
<input type="checkbox" name="dba" value="Access">Microsoft Access<br>
<br>
<input type="submit" name="postsubmit" value="Submit">
</form>
<br>
<h1>Database Consulting POST Form Results</h1>
<p>
<?php
if (isset($_POST['postsubmit'])) {

    $firstname = $_POST['firstname'];
    $lastname = $_POST['laststname'];
    $email = $_POST['email'];
    $dba = $_POST['dba'];
    echo $firstname; <br>
    echo $lastname; <br>
    echo $email; <br>
    echo $dba; <br>
}

else {
     echo "Please enter correct values in form and hit submit";
}

?>
</p>

<br>
<h1>GET Method Form</h1>
<form method="GET" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
First name:<br>
<input type="text" name="firstname_get"><br>
Last name:<br>
<input type="text" name="lastname_get"><br>
E-mail: <input type="text" name="email_get"><br>
Database Utilized<br>
<input type="checkbox" name="dba_get" value="SQL Server">SQL Server<br>
<input type="checkbox" name="dba_get" value="Oracle">Oracle<br>
<input type="checkbox" name="dba_get" value="Access">Microsoft Access<br>
<br>
<input type="submit" name="getsubmit" value="Submit">
</form>
<br>
<br>
<h1>Database Consulting GET Form Results</h1>

<p>

<?php 
if (isset($_GET['getsubmit'])) {
    $firstname_get = $_GET['firstname'];
    $lastname_get = $_GET['laststname'];
    $email_get = $_GET['email'];
    $dba_get = $_GET['dba'];
    echo $firstname_get; <br>
    echo $lastname_get; <br>
    echo $email_get; <br>
    echo $dba_get; <br>
}

else {
     echo "Please enter correct values in form and hit submit";
}

?>
</p>

</div>

Answer 1

首先，我建议有一个名为 action 的隐藏字段，例如：

<input type="hidden" name="action" value="submituserinfo" />

然后将您的 php 逻辑放在顶部而不是底部，在页面显示之前处理！你的 PHP_SELF 交易很好。但是你的逻辑不对。正确的逻辑应该是：

if(isset($_REQUEST['action']) && $_REQUEST['action']=='submituserinfo'){
    //analyze the data

    if($errors){
        //set a var to show errors below
    }else{
        //enter the data, whatever
    }
}

这应该会让您朝着正确的方向开始。如果将编码放在文档的头部，则可以更好地处理输出，甚至可以使用 header() 根据成功或错误进行重定向。

Answer 2

您将 $_GET['lastname'] 拼错为 $_GET['laststname'] ，将您的操作更改为 action="" 并将您的 <br> 包含在您的 echo 语句中，例如 echo $firstname .'<br>'; 而不是 echo $firstname; <br>

<h1>POST Method Form</h1>
 <form  method="POST" action="" >
  First Name:<br>
<input type="text" name="firstname"><br>
Last Name:<br>
 <input type="text" name="lastname"><br>
 E-mail: <input type="text" name="email"><br>
  Database Utilized:  <br>
 <input type="checkbox" name="dba[]" value="SQL Server">SQL Server<br>
 <input type="checkbox" name="dba[]" value="Oracle">Oracle<br>
 <input type="checkbox" name="dba[]" value="Access">Microsoft Access<br>
  <br>
 <input type="submit" name="postsubmit" value="Submit">
 </form>
  <br>
  <h1>Database Consulting POST Form Results</h1>
 <p>
  <?php
 if (isset($_POST['postsubmit'])) {

$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];
$dba = $_POST['dba'];
echo $firstname .'<br>';
echo $lastname .'<br>';
echo $email .'<br>';
foreach ($dba as $database) {
    echo $database .'<br>';
}
}

 else {
 echo "Please enter correct values in form and hit submit";
 }

 ?>
 </p>

PHP Form Get 和 Post Output 用 If 语句在同一页输出结果

PHP Form Get and Post Output with If statement to output results on same page

php

forms

post

if-statement

get