如何访问从Android的一个Web API方法的返回值？ [英] How do I access the returned value from a Web API Method in Android?
问题描述
混淆有关如何做到这一点（可以看出<经过href="http://stackoverflow.com/questions/22797194/how-to-call-a-restful-method-from-android">here和<一href="http://stackoverflow.com/questions/22798949/why-is-my-web-api-server-refusing-to-accept-calls-from-android">here,我现在成功地连接到我的服务器应用程序及相应的REST风格的方法与此code：
After being confused about how to do so (as can be seen here and here, I am now successfully connecting to my server app and the appropriate RESTful method with this code:
public void onFetchBtnClicked(View v){
if(v.getId() == R.id.FetchBtn){
Toast.makeText(getApplicationContext(), "You mashed the button, dude.", Toast.LENGTH_SHORT).show();
new CallAPI().execute("http://10.0.2.2:28642/api/Departments/GetCount?serialNum=4242");
}
}
public static class CallAPI extends AsyncTask<String, String, String> {
@Override
protected String doInBackground(String... params) {
String urlString=params[0]; // URL to call
String resultToDisplay = "";
InputStream in = null;
// HTTP Get
try {
URL url = new URL(urlString);
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
in = new BufferedInputStream(urlConnection.getInputStream());
} catch (Exception e ) {
System.out.println(e.getMessage());
return e.getMessage();
}
return resultToDisplay;
}
protected void onPostExecute(String result) {
Log.i("FromOnPostExecute", result);
}
} // end CallAPI
我意识到我需要分配的东西（而不是在初始化一个空字符串），以resultToDisplay，但什么？ 在我需要访问/隐蔽的字符串什么的一部分？
I realize I need to assign something (other than an empty string at initialization) to resultToDisplay, but what? What part of "in" do I need to access/covert to a string?
手工的方式为我工作，但fancypants apache的IO utils的没有那么多（当然，它编译...）。这是我的code：
The "manual" way is working for me, but the fancypants apache io utils "not so much" (well, it compiles...). This is my code:
try {
URL url = new URL(urlString);
HttpURLConnection urlConnection = (HttpURLConnection)url.openConnection();
in = new BufferedInputStream(urlConnection.getInputStream());
resultToDisplay = getStringFromInputStream(in);
total = IOUtils.toString(in);
resultToDisplay的分配工作（我得到的，18）。总的分配不会（我得到的，）。
resultToDisplay's assignment works (I get, "18"). total's assignment does not (I get, "").
注意：在getStringFromInputStream（）方法是Raghunandan的链接
Note: The "getStringFromInputStream()" method is from Raghunandan's link.
这工作只是花花公子（使用WIllJBD的主意，使用Apache的百科全书'IOUtils）：
This works just dandy (using WIllJBD's idea to use apache commons' IOUtils):
new CallWebAPI().execute("http://10.0.2.2:28642/api/Departments/GetCount?serialNum=4242");
. . .
private class CallWebAPI extends AsyncTask<String, String, String> {
@Override
protected String doInBackground(String... params) {
String urlString=params[0]; // URL to call
String result = "";
// HTTP Get
try {
URL url = new URL(urlString);
HttpURLConnection urlConnection =
(HttpURLConnection)url.openConnection();
InputStream inputStream = urlConnection.getInputStream();
if (null != inputStream)
result= IOUtils.toString(inputStream);
} catch (Exception e ) {
System.out.println(e.getMessage());
return e.getMessage();
}
return result;
}
@Override
protected void onPostExecute(String result) {
Log.i("RenameTheWashingtonFootballTeamTheRedskinPeanuts", result);
}
}
...等等显然它不必添加任何东西，如编译的文件（库/公地-IO-2.4.jar'）到build.gradle的依赖关系部分，因为似乎是至少必要在有一段时间，根据的这个。如果任何人都可以验证这样的[M，PP] endments到build.gradle不再需要，我会gradleful。
...and so apparently it is not necessary to add anything like "compile files('libs/commons-io-2.4.jar')" to the dependencies section of build.gradle, as seemingly was at least necessary at one time, according to this. If anybody can verify such a[m,pp]endments to build.gradle are no longer needed, I'd be gradleful.
我只注意到我无意中删除了@覆盖从onPostExecute（）方法，但它并没有区别 - 它工作得很好，没有它，它工作正常，一旦我恢复了它。那么，有什么的[不]有它的优势 - 它只是多余的绒毛
I just noticed that I inadvertently removed the "@Override" from the onPostExecute() method, but it made no difference - it worked fine without it, and it works fine once I restored it. So what's the advantage of [not] having it - is it just superfluous fluff?
推荐答案
为什么不使用类似IOUtils？
why not use something like IOUtils?
InputStream inputStream = urlConnection.getInputStream();
if (inputStream != null)
String content = IOUtils.toString(inputStream);
的http://commons.apache.org/proper/commons-io/apidocs/org/apache/commons/io/IOUtils.html
现在你可以解析字符串转换成JSON或XML，采用了许多图书馆之一。
now you can parse the string into Json or XML, using one of many libraries.
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