2015-05-19 53 views
-5

嗨,我已經建立了一個上傳器,應該上傳一些數據和圖像。我的文件和信息不會上傳(mysqli和PHP)

表單顯示在網頁上,沒有錯誤但不會上傳。

這裏是一個截圖也代碼附加。

http://simplicity-designs.tk/images/Screenshot%20%28104%29.png

<?php 
    include("includes/connect.php"); 

    if(isset($_POST['submit'])){ 

    $title = $_POST['title']; 
    $date = date('d-m-y'); 
    $author = $_POST['author'];  
    $image = $_FILES['image']['name']; 
    $keywords = $_POST['keywords'];  
    $content = $_POST['content']; 
    $image_tmp = $_FILES['image']['tmp_name']; 

    if($title == '' or $author == '' or $keywords == '' or $content == ''){ 
     echo " 
      <script> 
       alert('PLEASE CHECK YOUR FORM IS COMPLETE') 
      </script>"; 
     exit(); 

     move_uploaded_file($image_tmp, "../images/$image"); 

    }else{ 


     $sqli = "INSERT INTO posts (title, date, author, image, keywords, content) VALUES ('$title', ''$date , '$author', '$image', '$keywords', '$content')"; 
    }}; 
?> 





<!doctype html> 
<html> 
    <head> 
     <title>Inserting Data</title> 
     <meta charset="utf-8"> 

    </head> 

    <body> 

     <form action="insert_data.php" id="InsertDataForm" method="post" enctype="multipart/form-data"> 
      <table width="600" align="center" border="10"> 
       <tr> 
        <td align="center" bgcolor="#CC6600" colspan="6"><h1>Insert New Data Here</h1></td> 
       </tr> 


       <tr> 
        <td align="right">Data Title</td> 
        <td><input style="width:300px;" type="text" name="title"/></td> 
       </tr> 


       <tr> 
        <td align="right">Data Author</td> 
        <td><input style="width:300px;" type="text" name="author"/></td> 
       </tr> 


       <tr> 
        <td align="right">Data Keywords</td> 
        <td><input style="width:300px;" type="text" name="keywords"/></td> 
       </tr> 


       <tr> 
        <td align="right">Data Image</td> 
        <td><input style="width:300px; height:50px;" type="file" name="image"/></td> 
       </tr> 


       <tr> 
        <td align="right">Data Content</td> 
        <td><textarea style="max-width:300px; max-height:300px;" name="content" cols="48" rows="20"></textarea></td> 
       </tr> 

       <tr> 
        <td align="center" colspan="6"><input style="width:200px; height:50px; 
        margin:5px 0px;" type="submit" name="submit" value="Insert Data Now"/></td> 
       </tr> 

      </table> 
     </form> 


    </body> 
</html> 

連接腳本

<?php 
    $connect = mysqli_connect("localhost","########","####","#######"); 
?> 

無不良帖子只是幫助PLZ我不想拼寫糾正或消極我需要幫助。

的問候和感謝事先

+1

縮進你的代碼,你必須突出顯示它並按下'CTRL + k'。請這樣做。 – ZekeDroid

+1

(要在棧溢出代碼中按壓空格4次而不是使用製表符) – monxas

+1

*「沒有不良的帖子只是幫助PLZ我不想拼寫糾正或消極的我想要幫助。」* - 反諷不羈。 –

回答

1

首先確保你的php.ini是file_uploads是註釋。

file_uploads=On  instead of   ;file_uploads=On 

另外,您在INSERT中有錯誤。在$日期值。

你有

''$date  instead of  '$date' 

更改爲...

$sqli = "INSERT INTO posts (title, date, author, image, keywords, content) VALUES ('$title', '$date' , '$author', '$image', '$keywords', '$content')"; 
+0

可能會感謝我將在大約一小時內嘗試並更新如果工作:) –