在另一個系統上部署PHP項目顯示爲權限被拒絕

當我試圖在另一個系統我得到一個錯誤這樣在另一個系統上部署PHP項目顯示爲權限被拒絕

Warning: imagejpeg() [function.imagejpeg]: Unable to open 'image.jpg' for writing: Permission denied in /Applications/XAMPP/xamppfiles/htdocs/WISMAN/pmhome.php on line 61

我在這裏調用後得到一個JSON部署PHP項目的錯誤網絡服務來獲取一個JSON作爲其中包含字符串形式，對此我其編碼爲圖像並將其保存爲一個image.jpeg圖像的響應，但表示這一

<?php 
$userid=$_SESSION[user_id]; 
//echo $userid; 
$useridofuser=array(
'user_id'=> "$userid"); 
    //echo json_encode($useridofuser);//coverting the vlaues collected from form into json 
     //calling the web service 
     $url='webservice url'; 
    $data=$useridofuser; 
//echo("Input to Server : ".$data."\n"); 
    $ch=curl_init(); 
    curl_setopt($ch, CURLOPT_URL, $url); 
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
    curl_setopt($ch, CURLOPT_POST, true); 
    curl_setopt($ch, CURLOPT_POSTFIELDS, json_encode($useridofuser)); 
    curl_setopt($ch, CURLOPT_HTTPHEADER, array("Content-Type: application/json")); 
    $response= curl_exec($ch); 
    echo('\n'."Server response : \n \n".$response); 
    curl_close($ch); 
    //parsing the json response from server 
    $jsonde="$response"; 
    $json_o=json_decode($jsonde); 
    $json_a=json_decode($jsonde,true); 
    //echo("$json_a"); 
    $company_name=$json_a[user_data][company_name]; 
    $name=$json_a[user_data][name]; 
    $registration_date=$json_a[user_data][registration_date]; 
    $user_id=$json_a[user_data][user_id]; 
    $product_id=$json_a[product_data][0][product_id]; 
    $product_registration_id=$json_a[product_data][0][product_registration_id]; 
    $product_name=$json_a[product_data][0][product_name]; 
    $registration_date=$json_a[product_data][0][registration_date]; 
    $image=$json_a[product_data][0][image]; 
    $expiry_date=$json_a[product_data][0][expiry_date]; 
    $product_evaluation_level=$json_a[product_data][0][product_evaluation_level]; 
$_SESSION[regid]=$product_registration_id; 
if($product_evaluation_level=="1") 
{ 
    $level=trial; 
} 
$image_base64="$image"; 
$img = imagecreatefromstring(base64_decode($image_base64)); 
if($img != false) 
{ 
imagejpeg($img, 'image.jpg'); 
} ?>

來源

2013-04-17 Sree