从我的Android应用程序到Google App Engine,我正在发布一个zip文件;我也在发布一个字符串.当Google App Engine PHP脚本运行时,我将$_FILES的值转储到Log by …
ob_start();
var_dump($_FILES);
$result = ob_get_clean();
syslog(LOG_DEBUG, "VAR DUMP\n" . $result . "\n");
我还记录了字符串的值.日志具有字符串输出,但没有$_FILES转储输出.我读了THIS LINK,但这似乎与Web有关.我的POST文件的Android代码如下所示…
private void postExamplesZipToDrive(File file_zip){
RequestParams request_params = new RequestParams();
request_params.put("test", "Came thru");
//
try{ request_params.put("file_zip", file_zip); }
catch(FileNotFoundException e){
Log.e("postExamplesZipToDrive", "File not found");
Toast.makeText(this, "Unable to upload.", Toast.LENGTH_SHORT).show();
return;
}
AsyncHttpClient async_http_client = new AsyncHttpClient();
async_http_client.post(UPLOAD_URL, request_params, new JsonHttpResponseHandler(){
@Override
public void onSuccess(int code_status, Header[] headers, JSONArray success) {
Toast.makeText(getApplicationContext(), "" + code_status, Toast.LENGTH_SHORT).show();
super.onSuccess(code_status, headers, success);
}
@Override
public void onFailure(int code_status, Header[] headers, String string_response, Throwable throwable){
Toast.makeText(getApplicationContext(), "" + code_status, Toast.LENGTH_SHORT).show();
Log.e("onFailure", code_status + "\n" + string_response);
}
});
}
我的问题是,我是否正确发布.zip文件?如何捕获通常存储在$_FILES中的上传文件?
解决方法:
您可以在PHP55中使用直接文件上传功能(请参阅https://gae-php-tips.appspot.com/2015/03/09/direct-file-uploads-for-php-5-5/).