如何在一个表中插入和检索数据到mysql数据库中的另一个表 [英] How to insert and retrive data from one table to another table in mysql database

问题描述

我有两个表页面和上传

PAGE：

  page_id，page_name，page_parent 
（4，test，0）
（5，test-sub，4）
（6，test-sub2，4）
  
 
 
 
 UPLOAD：
  upload_id， upload_name，upload_root 
（1，test，4）
（2，test-sub，4）
（3，test-sub2，4）
  
我非常成功地获得page_id到upload_root，但有一个问题，当页面有子菜单，并输入page_parent。 
 
 
 
我的插入命令如下：
 
 
 
  $ name = $ _POST ['name']; 
 $ root = $ _POST ['root']; 
 $ insert_sql =INSERT INTO tbl_upload VALUES（''，\$ name\，\$ root\）; 
 
 $ root = $ _POST ['page_id'];？> 
< form action =enctype =multipart / form-datamethod =post> 
< div>名称：< input name =titletype =textsize =50/>< / div> 
< input name =page_idtype =hiddenvalue =<？php echo$ page_id;？> 
< input type =submitvalue =上传id =上传class =上传/> 
< / form> 
  
 
 
 
我的fetch命令如下：
 
 
 
  $ fetch_sql =SELECT upload_id，upload_name，upload_root from tbl_upload WHERE upload_root = \$ page_id\; 
  
我无法获取正确的网页网址。我没有得到正确的话来描述我的问题。但我真正想要的是将page_id（这是自动递增）的确切值转换成upload_id，如果不可能，然后到upload_root。 
 
 
 
我想要做的是：
 
 
 
上传：
 
 $ b b 
  upload_id，upload_name，upload_root 
（4，test，4）
（5，test-sub，4）
 -sub2，4）
  
 
 
解决方案
 
 <？php 
 if（isset（$ _ GET ['submit']））
 {
 $ name = $ _POST ['title']; 
 $ pageid = $ _POST ['page_id']; 
 $ insert_sql =INSERT INTO tbl_upload（upload_id，upload_name，upload_root）VALUES（'。$ pageid。'，'。$ name。'，''）; 
 mysql_query（$ insert_sql）; 
} 
？> 
< form action =your_page.php？submit = yesenctype =multipart / form-datamethod =post> 
< div>名称：< input type =textname =titlesize =50>< / div> 
< input type =hiddenname =page_idvalue =<？php echo $ page_id;？>> 
< input type =submitvalue =上传id =上传class =上传> 
< / form> 
  
显然，请插入页面名称your_page.php
 
 
 
您的代码容易遭受SQL注入。请查看如何为未来的项目实施PDO。
 
I have two Tables Page and Upload

PAGE:  
page_id,   page_name,   page_parent 
( 4,   test,   0)
( 5,   test-sub,   4)
( 6,   test-sub2,   4)
UPLOAD:
upload_id,   upload_name,   upload_root
(1,    test,   4)
(2,    test-sub,   4)
(3,   test-sub2,   4)
I was quite successful to get the page_id into upload_root but got problem when the page has submenu and is entered into page_parent. 

My insert command goes like this:

$name =  $_POST['name'];
$root = $_POST['root'];
$insert_sql = "INSERT INTO tbl_upload VALUES ('' , \"$name\",  \"$root\")";

$root = $_POST['page_id'];?>
<form action="" enctype="multipart/form-data" method="post">
<div>Name:   <input name="title" type="text" size="50"/></div> 
<input name="page_id" type="hidden" value= "<?php echo"$page_id"; ?>"
<input type="submit" value="Upload" id="upload" class="upload"/>
</form>


My fetch command goes like this:

$fetch_sql = "SELECT upload_id, upload_name, upload_root from tbl_upload WHERE upload_root = \"$page_id\"";
I cannot get the correct url of the pages. I'm not getting the correct words to describe my problem. But what I exatly want is to retrive the exact value of page_id (which is auto increment) into upload_id if not possible then to upload_root. 

What I want to do is :

UPLOAD:
upload_id,   upload_name,   upload_root
(4,   test,   4)
(5,   test-sub,   4)
(6,   test-sub2,   4)

解决方案
<?php
if (isset($_GET['submit']))
{
$name =  $_POST['title'];
$pageid = $_POST['page_id'];
$insert_sql = "INSERT INTO tbl_upload (upload_id, upload_name, upload_root) VALUES ('".$pageid."','".$name."','')";
mysql_query($insert_sql);
}
?>
<form action="your_page.php?submit=yes" enctype="multipart/form-data" method="post">
<div>Name:   <input type="text" name="title" size="50"></div> 
<input type="hidden" name="page_id" value= "<?php echo $page_id; ?>">
<input type="submit" value="Upload" id="upload" class="upload">
</form>
Obviously, insert the name of page where it says 'your_page.php'

Your code is vulnerable to SQL injection. Please look up how to implement PDO for future projects.

                        
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