Question

如何使用PHP将上传的swf文件存储在MySQL数据库中？

Answer 1

最好的方法是将文件存储在磁盘上，然后将元数据（如文件名，以及可能的其他字段，如上载程序）存储在数据库中。通过这种方式，您将获得更快的数据库，因为您不需要在表中包含大型BLOB字段，并且您还可以更快地提供文件，因为它们可以直接从磁盘读取而不是通过网络通过MySQL连接发送

因此，在上传文件时，请为其指定唯一的文件名并将其存储在磁盘上的文件夹中。然后将元数据（包括唯一文件名）存储在数据库中。如果您在定义唯一文件名时遇到问题，只需先保存元数据，然后在数据库中使用AUTOINCREMENTed列作为文件名。

Answer 2

// here u can upload swf files also 

<?php
//define a maxim size for the uploaded images in Kb
 define ("MAX_SIZE","100"); 

//This function reads the extension of the file. It is used to determine if the file  is an image by checking the extension.
 function getExtension($str) {
         $i = strrpos($str,".");
         if (!$i) { return ""; }
         $l = strlen($str) - $i;
         $ext = substr($str,$i+1,$l);
         return $ext;
 }

//This variable is used as a flag. The value is initialized with 0 (meaning no error  found)  
//and it will be changed to 1 if an errro occures.  
//If the error occures the file will not be uploaded.
 $errors=0;
//checks if the form has been submitted
 if(isset($_POST['Submit'])) 
 {
    //reads the name of the file the user submitted for uploading
    $image=$_FILES['image']['name'];
    //if it is not empty
    if ($image) 
    {
    //get the original name of the file from the clients machine
        $filename = stripslashes($_FILES['image']['name']);
    //get the extension of the file in a lower case format
        $extension = getExtension($filename);
        $extension = strtolower($extension);
    //if it is not a known extension, we will suppose it is an error and will not  upload the file,  
    //otherwise we will do more tests
 if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif") && ($extension != "swf")) 
        {
        //print error message
            echo '<h1>Unknown extension!</h1>';
            $errors=1;
        }
        else
        {
//get the size of the image in bytes
 //$_FILES['image']['tmp_name'] is the temporary filename of the file
 //in which the uploaded file was stored on the server
 $size=filesize($_FILES['image']['tmp_name']);

//compare the size with the maxim size we defined and print error if bigger
if ($size > MAX_SIZE*1024)
{
    echo '<h1>You have exceeded the size limit!</h1>';
    $errors=1;
}

//we will give an unique name, for example the time in unix time format
$image_name=time().'.'.$extension;
//the new name will be containing the full path where will be stored (images folder)
$newname="images/".$image_name;
//we verify if the image has been uploaded, and print error instead
$copied = copy($_FILES['image']['tmp_name'], $newname);
if (!$copied) 
{
    echo '<h1>Copy unsuccessfull!</h1>';
    $errors=1;
}}}}


//If no errors registred, print the success message
 if(isset($_POST['Submit']) && !$errors) 
 {
    echo "<h1>File Uploaded Successfully! Try again!</h1>";


 }

 ?>

 <!--next comes the form, you must set the enctype to "multipart/frm-data" and use an input type "file" -->
 <form name="newad" method="post" enctype="multipart/form-data"  action="">
 <table>
    <tr><td><input type="file" name="image"></td></tr>
    <tr><td><input name="Submit" type="submit" value="Upload image"></td></tr>
 </table>   
 </form>

如何使用PHP将上传的swf文件存储在MySQL数据库中？

2 个答案: