2015-11-29 81 views
0

我試圖插入用戶輸入到我的數據庫。我想將標題和描述添加到廣告表中,並將圖像添加到圖片表中。 這是我到目前爲止所提出的,但它根本不起作用。這是一個posthandler.php頁面,當用戶按下提交時會被調用。頂部的變量從html頁面分配輸入名稱。目前我的照片數據庫中都有路徑名和blob,以查看哪個選項最適合。MYSQL php數據庫插入包括圖片上傳

請幫助

<?php 

session_start(); 
echo "You are signed in as ". $_SESSION['username']; 

include 'mysql.php'; 

$title='title'; 
$des='des'; 
$image='image'; 

if(isset($_POST['submit'])) { 
    $error = ""; 
    if (!empty($_POST['title'])) { 
      $title= $_POST['title']; 
     } else { 
      $error = "Please enter a title for your Ad. <br />"; 
     } 
     if (!empty($_POST['des'])) { 
      $des = $_POST['des']; 
     } else { 
      $error = "Please describe your item <br />"; 
     } 
     if (!empty($_POST['image'])) { 

      $image=addslashes($_FILES['image']['tmp_name']); 
      $name=addslashes($_FILES['image']['name']); 
      $image=file_get_contents($image); 
      $image=base64_encode($image); 
      $filepath = "images/".$filename; 
      move_uploaded_file($filetmp,$filepath); 
      $image = $_POST['image']; 

     } else { 

     } 

     if (empty($error)) { 

     $conn= mysql_connect("localhost","km","data"); 
     if (!$conn){ 
      die ("Failed to connect to MySQL: " . mysql_error()); 

     mysql_select_db("mdb_km283",$conn); 

     $sql = "INSERT INTO Advert (title, description) VALUES ('$_POST[title]','$_POST[des]')"; 
     $sql2 = "INSERT INTO Pictures(image, Path) VALUES ('$_POST[image]' $filepath)"; 

     mysql_query($sql,$sql2, $conn); 
     //mysql_query($sql2,$conn); 
     mysql_close($conn); 


      echo 'Upload successfull'; 
     } 

     } 
} 


    ?> 

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" 
    "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> 
<html version="-//W3C//DTD XHTML 1.1//EN" 
     xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" 
     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
     xsi:schemaLocation="http://www.w3.org/1999/xhtml 
          http://www.w3.org/MarkUp/SCHEMA/xhtml11.xsd" 
> 

<head> 
<link rel="stylesheet" type=text/css" href="stylesheet.css"> 
<a href="signout.php">Logout</a> 

<h1>ERROR - Please go back and fix the below:</h1> 
</head> 
<body> 


<?php 
    if (!empty($error)) { 
     echo'<p class="error"><strong>Upload failed. Please go back and fix the below <br/> The following error(s) returned:</strong><br/>' . $error . '</p>'; 
    } else { 

    echo '<meta http-equiv="refresh" content="0; URL=user.php">'; 


    } 


?> 


</body> 
</html> 

回答

2
$filename = $_FILES['image']['name']; 
$filepath = 'image/' . $filename; 
$filetmp = $_FILES['image']['temp_name']; 
$result = move_uploaded_file($filetmp,$filepath); 
if(!$result) 
     echo "Photo field was blank"; 
else 
     echo "uploaded"; 

那麼這裏寫烏爾MySQL代碼....