PHP的Android上传失败 [英] PHP Android upload failed
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问题描述
我目前使用hosting24.com为PHP文件上传的实施和将要上传的MP4文件。用于测试的HTML文件，它workds。当它由Android deivces来执行，它失败并显示$名称= $ _FILES [uploaded_file]为空。请你告诉我，有没有其他的因素$ P $从上传成功pventing我们吗？
以下是我的code
公共字符串uploadFIle（档案文件）{
字符串结果=; HttpURLConnection的康恩= NULL;
DataOutputStream类DOS = NULL;
字符串lineEnd =\\ r \\ n;
串twoHyphens = - ;
字符串边界=*****;
字符串文件名= file.getName（）;
INT读取动作，方bytesAvailable，缓冲区大小;
INT serverResponse code = 0;
字节[]缓冲区;
INT MAXBUFFERSIZE = 1 * 1024 * 1024; 的FileInputStream的FileInputStream;
尝试{
的FileInputStream =新的FileInputStream（文件）;
网址URL =新的URL（http://www.gallicalab.com/upload.php）; //打开HTTP连接的URL
康恩=（HttpURLConnection类）url.openConnection（）;
conn.setDoInput（真）; //允许输入
conn.setDoOutput（真）; //允许输出
conn.setUseCaches（假）; //不要使用缓存副本
conn.setRequestMethod（POST）;
conn.setRequestProperty（连接，保持活动）;
conn.setRequestProperty（ENCTYPE，多部分/表单数据）;
conn.setRequestProperty（内容类型，的multipart / form-data的;边界=+边界）;
conn.setRequestProperty（uploaded_file，文件名）; DOS =新的DataOutputStream类（conn.getOutputStream（））; dos.writeBytes（twoHyphens +边界+ lineEnd）;
dos.writeBytes（内容处置：表格数据;名称= \\uploaded_file \\;文件名=
+文件名+\\+ lineEnd）; dos.writeBytes（lineEnd）; //创建最大大小的缓冲区
参考bytesAvailable = fileInputStream.available（）; BUFFERSIZE = Math.min（方bytesAvailable，MAXBUFFERSIZE）;
缓冲区=新的字节[缓冲区大小] //读取文件，并将其写入形式...
读取动作= fileInputStream.read（缓冲液，0，缓冲区大小）; 而（读取动作大于0）{ dos.write（缓冲液，0，缓冲区大小）;
参考bytesAvailable = fileInputStream.available（）;
BUFFERSIZE = Math.min（方bytesAvailable，MAXBUFFERSIZE）;
读取动作= fileInputStream.read（缓冲液，0，缓冲区大小）; } //发送文件数据后necesssary多部分的表单数据...
dos.writeBytes（lineEnd）;
dos.writeBytes（twoHyphens +边界+ twoHyphens + lineEnd）; //服务器的响应（code和消息）
serverResponse code = conn.getResponse code（）;
InputStream为= conn.getInputStream（）; 如果（serverResponse code == 200）{
读者的BufferedReader =新的BufferedReader（新的InputStreamReader（是，UTF-8），8192）;
StringBuilder的SB =新的StringBuilder（）;
串线= NULL;
而（（行= reader.readLine（））！= NULL）{
sb.append（线）;
}
结果= sb.toString（）;
} //结果= serverResponseMessage;
}赶上（FileNotFoundException异常五）{
// TODO自动生成catch块
e.printStackTrace（）;
}赶上（MalformedURLException的E）{
// TODO自动生成catch块
e.printStackTrace（）;
}赶上（的ProtocolException E）{
// TODO自动生成catch块
e.printStackTrace（）;
}赶上（IOException异常五）{
// TODO自动生成catch块
e.printStackTrace（）;
}
返回结果;
}
服务器端
＆LT; PHP
使用error_reporting（E_ALL）; $ target_path =目标/;
// $ target_path = $ target_path。基本名（$ _FILES ['文件'] ['名']）;
$ NAME = $ _FILES [uploaded_file] [名称];
//回声$名称。＆LT; BR＆gt;中;
$ EXT =结束（爆炸（，$名）。）;
//回声$分机＆LT; BR＆gt;中。
$ randname = random_string（30）;
//回声$ randname＆LT; BR＆gt;中。
$ FNAME = $ randname。 。 。$ EXT;
//回声$ FNAME＆LT; BR＆gt;中。
$ target_path = $ target_path $ FNAME。
//回声$ target_path＆LT; BR＆gt;中。
如果（move_uploaded_file（$ _ FILES ['uploaded_file'] ['tmp_name的值']，$ target_path））{
//回声文件。 。基本名（$ _FILES ['文件'] ['名']）已上传;
//回声random_string（50）; // $ ARR =阵列（'数据'=＆GT;的http：//gallicalab.com/target/'.$_FILES ['文件'] ['名']）;
$ ARR =阵列（'数据'=＆GT;的http：//gallicalab.com/target/'.$fname）; 回声json_en code（$ ARR）; // {一：1，B：2，C：3，D：4，e的：5}
}
其他{
回声发生错误上传文件，请重试！
} 功能random_string（$长度）{
$关键='';
$键= array_merge（范围（0，9），范围（'一个'，'Z'））; 为（$ I = 0; $ I＆LT; $长度; $ I ++）{
$键= $键[array_rand（$键）。
} 返回$关键;
}？＆GT;
解决方案
你能登入你实际上是什么样接收在服务器端的文件？ Android应用程序可能无法正确地提出请求。
试试这个：
＆LT; PHP
使用error_reporting（E_ALL）;$ H =的fopen（'upload.log'，'W'）;
FWRITE（$ H，的print_r（$ _ POST，真）为\\ r \\ n --- \\ r \\ n）;
FWRITE（$ H，的print_r（$ _ FILES，真实））;
FCLOSE（$ H）;//你的code休息
然后检查 upload.log 文件看看从设备您的要求发生了什么。
P.S。试试这个替代上传code为Android：
私人无效uploadFile（字符串文件路径，字符串文件名）{
InputStream的InputStream的;
尝试{
的InputStream =新的FileInputStream（新文件（文件路径））;
字节[]数据;
尝试{
数据= IOUtils.toByteArray（InputStream的）; HttpClient的HttpClient的=新DefaultHttpClient（）;
HttpPost httpPost =新HttpPost（http://www.gallicalab.com/upload.php）; InputStreamBody inputStreamBody =新InputStreamBody（新ByteArrayInputStream的（数据），文件名）;
MultipartEntity multipartEntity =新MultipartEntity（）;
multipartEntity.addPart（文件，inputStreamBody）;
httpPost.setEntity（multipartEntity）; HTT presponse HTT presponse = httpClient.execute（httpPost）; //处理响应从脚本回来。
如果（HTT presponse！= NULL）{ }其他{//错误，没有任何反应。 }
}赶上（IOException异常五）{
e.printStackTrace（）;
}
}赶上（FileNotFoundException异常E1）{
e1.printStackTrace（）;
}
}
在服务器端：
$ OBJFILE =安培; $ _FILES [文件];
$ strPath的基本名称=（$ OBJFILE [名称]）;如果（move_uploaded_file（$ OBJFILE [tmp_name的值]，$ strPath中））{
打印文件。 $ strPath的。 已上传。
}其他{
打印有错误上传文件，请重试！
}
来源：的http://www.$c$cpuppet.com/2013/03/26/android-uploading-a-file-to-a-php-server/
I am currently using hosting24.com for php file upload implementation and going to upload mp4 files. The html file for testing , it workds. When it comes to the execution by Android deivces , it fails and shows the $name = $_FILES["uploaded_file"] is null. Would you please tell me is there any other factors preventing us from successful upload ?
The below is my code
public String uploadFIle(File file){
String result = "";
HttpURLConnection conn = null;
DataOutputStream dos = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
String fileName = file.getName();
int bytesRead, bytesAvailable, bufferSize;
int serverResponseCode = 0;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
FileInputStream fileInputStream;
try {
fileInputStream = new FileInputStream(file);
URL url = new URL("http://www.gallicalab.com/upload.php");
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("uploaded_file", fileName);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename="
+ fileName + "\"" + lineEnd);
dos.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
InputStream is = conn.getInputStream();
if(serverResponseCode==200){
BufferedReader reader = new BufferedReader(new InputStreamReader( is, "utf-8"), 8192);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line);
}
result = sb.toString();
}
// result = serverResponseMessage;
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (MalformedURLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return result;
}
Server side
<?php
error_reporting(E_ALL);
$target_path = "target/";
// $target_path = $target_path . basename( $_FILES['file']['name']);
$name = $_FILES["uploaded_file"]["name"];
// echo $name."<br>";
$ext = end(explode(".", $name));
// echo $ext."<br>";
$randname = random_string(30);
// echo $randname."<br>";
$fname = $randname . "." .$ext;
//echo $fname."<br>";
$target_path = $target_path .$fname;
// echo $target_path."<br>";
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $target_path)) {
//echo "The file ". basename( $_FILES['file']['name'])." has been uploaded";
//echo random_string(50);
// $arr = array ('data'=>'http://gallicalab.com/target/'.$_FILES['file']['name']);
$arr = array ('data'=>'http://gallicalab.com/target/'.$fname);
echo json_encode($arr); // {"a":1,"b":2,"c":3,"d":4,"e":5}
}
else {
echo "There was an error uploading the file, please try again!";
}
function random_string($length) {
$key = '';
$keys = array_merge(range(0, 9), range('a', 'z'));
for ($i = 0; $i < $length; $i++) {
$key .= $keys[array_rand($keys)];
}
return $key;
}
?>
解决方案
Could you log what you are actually receiving at the server side to a file? Android app might not be making the request correctly.
Try this:
<?php
error_reporting(E_ALL);
$h = fopen('upload.log', 'w');
fwrite($h, print_r($_POST, true) ."\r\n---\r\n");
fwrite($h, print_r($_FILES, true));
fclose($h);
// rest of your code
Then inspect the upload.log file and see what happened with your request from device.
P.S. Try this alternate upload code for android:
private void uploadFile(String filePath, String fileName) {
InputStream inputStream;
try {
inputStream = new FileInputStream(new File(filePath));
byte[] data;
try {
data = IOUtils.toByteArray(inputStream);
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://www.gallicalab.com/upload.php");
InputStreamBody inputStreamBody = new InputStreamBody(new ByteArrayInputStream(data), fileName);
MultipartEntity multipartEntity = new MultipartEntity();
multipartEntity.addPart("file", inputStreamBody);
httpPost.setEntity(multipartEntity);
HttpResponse httpResponse = httpClient.execute(httpPost);
// Handle response back from script.
if(httpResponse != null) {
} else { // Error, no response.
}
} catch (IOException e) {
e.printStackTrace();
}
} catch (FileNotFoundException e1) {
e1.printStackTrace();
}
}
At server side:
$objFile = & $_FILES["file"];
$strPath = basename( $objFile["name"] );
if( move_uploaded_file( $objFile["tmp_name"], $strPath ) ) {
print "The file " . $strPath . " has been uploaded.";
} else {
print "There was an error uploading the file, please try again!";
}
Source: http://www.codepuppet.com/2013/03/26/android-uploading-a-file-to-a-php-server/
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