帮我修改此代码以写入数据库

| 所以我必须写一个表格将图像上传到网站...我找到了下面的代码,将其添加到我的网站,并对其进行了测试。似乎工作很棒。问题是写入数据库的代码中没有任何地方。。。它只是将图像上传到文件夹,并且没有数据库可以对其进行跟踪……我知道SQL足够好,可以完成编写工作代码...。但是我不太了解实际的sql代码在哪里.....如果我单击GO按钮上传图像,则用户将从此页面移走...因此sql也会运行?...我如何修改它以添加sql代码 有任何想法吗? 继承人的代码...任何帮助将是巨大的!
<?php
//define a maxim size for the uploaded images in Kb
define (\"MAX_SIZE\",\"500\"); 

//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\")) 
        {
        //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>
    
已邀请:
将您的代码放在以下代码块中:
//If no errors registred, print the success message
if(isset($_POST[\'Submit\']) && !$errors) {
   // do database stuff here...
   echo \"<h1>File Uploaded Successfully! Try again!</h1>\";
}
至此,您知道上传已成功。您没有提供表架构或SQL风格。我将假设MySQL。在这种情况下,请检查mysql_query()。 PHP文档为连接和运行查询提供了一个很好的示例。     

要回复问题请先登录注册