如何插入包含外键的表

Question

我是新手，我得到问题， 我无法更新表上载，因为我从表请求中插入外键 任何人都可以帮助我吗？

table request: 
id_request; Primary Key 
subjek; 
email; 
reportto; 
pelaksana; 
isi; 

table upload: 
id_upload; Primary Key 
id_request; Foreign Key 
filename; 
filetype; 
filesize; 
filedata; 

，如果我想创造新的请求时，它会创建表请求记录，插入数据表的要求，而且在表上载插入id_request，我填与LAST_INSERT_ID

在同一时间，我被插入到表上传id_request后，我用查询通过插入或更新查询 插入附件但老是报错

的错误一直在寻找这样的：

警告：mysql_qu注意：mysql_query（）[function.mysql-query]：error（）[function.mysql-query]：MySQL服务器已在C：\ wamp \ www \ beta \ insert.php上消失32行

警告：读取结果集的头在C：上线32

这个\ WAMP \ WWW \测试\ insert.php是线32 mysql_query($up)or die('Error upload file');

这是我的房源码

<?php 
include('config.php'); 
session_start(); 

$jenis = $_POST['jenis']; 
$subjek = $_POST['subject']; 
$username = $_SESSION['username']; 
$email = $_SESSION['email']; 
$reportto = $_SESSION['reportto']; 
$pelaksana = $_POST['pelaksana']; 
$ket = $_POST['isi']; 

$uploaddir = 'attach/'; 
$filedata = addslashes(fread(fopen($_FILES['uploadfile']['tmp_name'], 'r'), 
      $_FILES['uploadfile']['size'])); 
$filetype = $_FILES['uploadfile']['type']; 
$filesize = $_FILES['uploadfile']['size']; 
$filename = $_FILES['uploadfile']['name']; 

$query = "INSERT INTO request (waktu, jenis_request, subject, customer, isi, pelaksana) 
    VALUES (NOW(), '".$jenis."', '".$subjek."', '".$username."', '".$ket."', '".$pelaksana."')"; 
mysql_query($query)or die('Error, insert query failed'); 
$ff = mysql_query("insert into upload (id_request) select id_request from request where id_request = LAST_INSERT_ID()"); 

$up = "update upload set deskripsi = '".$subjek."' , filetype = '".$filetype."', filename = '".$filename."', filedata = '".$filedata."', filesize = '".$filesize."' where id_request =  last_insert_id()"; 
mysql_query($up)or die('Error upload file'); 

$uploadfile = $uploaddir . $filename; 
if (move_uploaded_file($_FILES['uploadfile']['tmp_name'], $uploadfile)) 
{ 
echo "File telah diupload\n"; 
echo '$filename\n'; 
// header("location: home.php"); 

} 
else 
{ 
echo "File gagal diupload"; 
} 
?> 

Answer 1

首先，记得清理你的数据！您的代码易受SQL注入攻击。要获取插入的ID，你这样做

$sql = "INSERT INTO ..."; 
$result = mysql_query($sql); 

$new_row_id = mysql_insert_id(); 

在你的代码，你不打开一个MySQL连接，选择一个数据库，等等。如果你不这样做，在包括config.php文件，然后这是导致错误。

编辑：

我无法真正了解你的脚本，因为它似乎是在另一种语言，但是这是我想出了

<?php 
include('config.php'); 
session_start(); 

//filter the data 
$jenis  = filter_var($_POST['jenis'], FILTER_SANITIZE_STRING); 
$subjek  = filter_var($_POST['subject'], FILTER_SANITIZE_STRING); 
$pelaksana   = filter_var($_POST['pelaksana'], FILTER_SANITIZE_STRING); 
$ket  = filter_var($_POST['isi'], FILTER_SANITIZE_STRING); 


$username = $_SESSION['username']; 
$email  = $_SESSION['email']; 
$reportto = $_SESSION['reportto']; 


$uploaddir = 'attach/'; 
$filedata = addslashes(fread(fopen($_FILES['uploadfile']['tmp_name'], 'r'), 
      $_FILES['uploadfile']['size'])); 
$filetype = $_FILES['uploadfile']['type']; 
$filesize = $_FILES['uploadfile']['size']; 
$filename = $_FILES['uploadfile']['name']; 

$sql = "INSERT INTO request (waktu, jenis_request, subject, customer, isi, pelaksana) VALUES (NOW(), '{$jenis}', '{$subjek}', '{$username}', '{$ket}', '{$pelaksana}')"; 

$result = mysql_query($sql); 
$id_inserted_request = mysql_insert_id(); 

$sql = mysql_query("INSERT INTO upload (deskripsi, filetype, filename, filedata, filesize, id_request) VALUES('{$subjek}', '{$filetype}', '{$filename}', '{$filedata}', '{$filedata}', '{$filesize}', '{$id_inserted_request}') "); 

mysql_query($sql) 

$uploadfile = $uploaddir . $filename; 
if (move_uploaded_file($_FILES['uploadfile']['tmp_name'], $uploadfile)) 
{ 
echo "File telah diupload\n"; 
echo "{$filename}\n"; 
// header("location: home.php"); 

} 
else 
{ 
echo "File gagal diupload"; 
} 
?>