Question

我正在使用包含此代码的index.php页面

<form action="post.php" method="POST">
<textarea placeholder="write something.........." name="desc" "></textarea>
<input type='submit'  name='post' value='POST' />
<form>

然后当我点击index.php中的提交按钮时，会将我链接到pos.php

if(isset($_POST['post'])){
$description=$_POST['desc']; 
$sql = "INSERT INTO post(description) VALUES('$description')";
if(mysqli_query($con,$sql)){
    //i need to get another third php file which automaticaly load when you click the submit butto in the idex.php
<a href="third.php">third Page</a>
  }
  else{
    echo 'failure'.mysqli_error($con);
  }

当我尝试这个它使我链接比第三个PHP，但我不想要那个我想同时将数据插入到sql中，将我链接到第三页，而不是点击任何其他链接

任何人都可以帮助我

谢谢！

Answer 1

会包括做你想要的吗？ http://php.net/manual/en/function.include.php

include('third.php');

这会将third.php中的所有代码添加到你的post.php中，并加载它的内容。

Answer 2

如果我理解您的要求，则需要在将数据保存到数据库后进行重定向。

html表格：

<form action="do_something.php" method="post">
  <textarea placeholder="write something.........." name="desc" "></textarea>
  <input type='submit'  name='post' value='POST' />
<form>

do_something.php：

<?php
/*
* Do your sql; 
* use prepared statements; see http://php.net/manual/en/mysqli.quickstart.prepared-statements.php
* you'll have to look this up to implement it
*
*  $sql = "INSERT INTO post(description) VALUES(?)";
*
* bind value and execute.  
*/


/* 
* Note; there must be NO output before the header call!
*/

$newUrl = "http://yourDomain/third.php";
header('Location: '.$newURL);
die;

third.php

// page 3 loads immediately after do_something.php executes.

当您按照此模式操作时，后退按钮不会重新提交您的POST数据。

我想在使用第三页的同时链接两个php页面

2 个答案: