php - Remote Upload Amazon S3 -


i have made script upload remote files link amazon s3 not work. have no idea what's wrong. below code done , i'm using amazon sdk in latest version:

config.php

<?php  return [     's3' => [         'key' => 'mykey',         'secret' => 'mykey',         'bucket' => 'mybucket',         'region' => 'us-west-2',         'version' => 'latest'     ] ]  ?>

start.php

<?php use aws\s3\s3client; require 'aws/aws-autoloader.php';  $config = require('config.php');  //s3  $s3 = s3client::factory([     'key' => $config['s3']['key'],     'secret' => $config['s3']['secret'],     'region' => $config['s3']['region'],     'version' => $config['s3']['version']  ]); ?>

upload.php

<?php  use aws\s3\exception\s3exception; require 'start.php';  error_reporting(0); $get_url = $_post["url"]; $url = trim($get_url); if($url) {     $file = fopen($url,"rb");     $directory = "animes";     $valid_exts = array("php","jpeg","gif","png","doc","docx","jpg","html","asp","xml","jpeg","bmp");      $ext = end(explode(".",strtolower(basename($url))));     if(in_array($ext,$valid_exts))     {         $rand = rand(1000,9999);         $filename = "$rand.$ext";         $newfile = fopen($directory . $filename, "wb");         try{             $s3->putobject([                 'bucket' => $config['s3']['bucket'],                 'key' => "{$directory}/{$filename}",                 'acl' => 'public-read'         ]);       } catch(s3exception $e){         die("error.");     }         if($newfile)         {             while(!feof($file))             {                 fwrite($newfile,fread($file,1024 * 8),1024 * 8);             }             echo 'file uploaded successfully';             echo '**$$**'.$filename;         }         else         {             echo 'file not exists';         }     }     else     {         echo 'invalid url';     } } else {     echo 'please enter url'; } ?>

and index.php

<html>     <head>         <title>php file upload url</title>         <script type="text/javascript" src="jquery.js"></script>         <script type="text/javascript">         $(document).ready(function(){             $("#1").hide();         });         function uploadfile(){             $("#1").show();             $("#disp").html("");             var url = encodeuricomponent($("#url").val());             $.ajax({                 url: "upload.php",                 data: "url=" +url,                 type: 'post',                 success: function(data)                 {                     var findsucc = data.indexof("successfully");                     out=data.split('**$$**');                     if(findsucc!=-1)                     {                         $("#disp").css({"color": "green"});                         $("#disp").html(out[0]);                         $("#link").html("<a href='./upload/"+out[1]+"'>click here</a> view");                         $("#1").hide();                     }                     else                     {                         $("#1").hide();                         $("#disp").css({"color": "red"});                         $("#disp").html(data);                         $("#url").val("");                     }                 }             });          }         </script>     </head> <body> <div align='center' style='padding-top: 40px;color: #4e4e4e;'><h1>php file upload url</h1></div> <div align='center' style='padding-top: 30px;'> enter remote url: <input type="text" name="url" id='url' size="35"> <input type="submit" value="upload" name="submit" style='cursor: pointer;' onclick='uploadfile()'>&nbsp;&nbsp;<img src='ajax-loader.gif' id='1'>&nbsp;&nbsp;<br /><br /><div align='center'><span id='disp'></span></div><br> <div id='link'></div><br /> <div style=" padding-left: 20px;font-size: 10px;color: #dadada;" id="dumdiv"> </div> </body> </html>

the best way debug php run via command line version of php, show errors in terminal instead of digging through webserver logs may or may not give details.


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