Question

我正在使用android并尝试创建一个能够将多个图像上传到服务器的应用程序。我曾尝试使用xampp将图像上传到我的localhost，效果很好。但是，当我尝试上传到我的企业服务器时，我找不到我的文件，换句话说。该文件无法写入。我不知道是什么让它失败了？这是我的代码

上传tp XAMPP

连接字符串private static final String url_photo = "http://192.168.7.110/blabla/base.php";

路径static final String path = "C:\\xampp\\htdocs\\psn_asset_oracle\\Images\\";

上传到实际的企业服务器

连接字符串private static final String url_photo = "http://192.168.4.27/oracle/logam/am/data_images/android_image/base.php";

路径static final String path = "http://192.168.4.27/oracle/logam/am/data_images/android_image/";

我上传到服务器的代码

params_p.add(new BasicNameValuePair("image_name_1",
                        image_name_1));
                params_p.add(new BasicNameValuePair("image_name_2",
                        image_name_2));
                params_p.add(new BasicNameValuePair("image_name_3",
                        image_name_3));
                params_p.add(new BasicNameValuePair("image_name_4",
                        image_name_4));
json_photo = jsonParser.makeHttpRequest(url_photo, "POST", params_p);
ArrayList<NameValuePair> params_p = new ArrayList<NameValuePair>();

PHP代码

if(isset($_POST["image_name_1"]) && isset($_POST["image_name_2"]) && isset($_POST["image_name_3"]) && isset($_POST["image_name_4"]) 
&& isset($_POST["image_1"]) && isset($_POST["image_2"]) && isset($_POST["image_3"]) && isset($_POST["image_4"]))
{
$image_name_1 = $_POST["image_name_1"];
$image_name_2 = $_POST["image_name_2"];
$image_name_3 = $_POST["image_name_3"];
$image_name_4 = $_POST["image_name_4"];
$image_1 = $_POST["image_1"];
$image_2 = $_POST["image_2"];
$image_3 = $_POST["image_3"];
$image_4 = $_POST["image_4"];

/*---------base64 decoding utf-8 string-----------*/
$binary_1=base64_decode($image_1);
$binary_2=base64_decode($image_2);
$binary_3=base64_decode($image_3);
$binary_4=base64_decode($image_4);

/*-----------set binary, utf-8 bytes----------*/
header('Content-Type: bitmap; charset=utf-8');
/*---------------open specified directory and put image on it------------------*/
$file_1 = fopen($image_name_1, 'wb');
$file_2 = fopen($image_name_2, 'wb');
$file_3 = fopen($image_name_3, 'wb');
$file_4 = fopen($image_name_4, 'wb');

/*---------------------assign image to file system-----------------------------*/
fwrite($file_1, $binary_1);
fclose($file_1);
fwrite($file_2, $binary_2);
fclose($file_2);
fwrite($file_3, $binary_3);
fclose($file_3);
fwrite($file_4, $binary_4);
fclose($file_4);

$response["message"] = "Success";
echo json_encode($response);

}

我已经联系了我的DBA，并要求我允许编写该文件，但它仍然无效。错误是json没有将“Success”作为指示文件写入失败的消息。我将不胜感激任何帮助。谢谢。

Answer 1

文件服务器是否允许写访问权限？ chmod的示例，http://catcode.com/teachmod/

您的企业服务器是否在线？ IP地址对我来说是私密的，192.168.4.27。

Answer 2

不要使用json使用http post。

见下面的代码

    HttpClient client = new DefaultHttpClient();
    String postURL = "your url";
    HttpPost post = new HttpPost(postURL);

    try {
        MultipartEntity reqEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);

        ByteArrayBody bab = new ByteArrayBody(img, "image.jpg");
        reqEntity.addPart("image", bab); 
        post.setEntity(reqEntity);
        HttpResponse response = client.execute(post);

这里img是ByteArray格式的图像

Answer 3

在我的DBA给我另一个文件路径后问题解决了。

无法在实际服务器上写入图像文件

3 个答案: