通过GET方法发送的值 [英] Send values by GET method
本文介绍了通过GET方法发送的值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我一直在试图通过GET方法的数据发送到服务器，但我无法找到一个方法来做到这一点。我曾尝试一些codeS的异步任务，但一无所获。 Web服务是由在CakePHP中，格式是这样的：
Base_URI /用户/ add.json JSON = {电子邮件：xxx@x.com，密码：XXXXXXXXX，FIRST_NAME：XYZ，姓氏：XYZ}
Android的专家被要求想法子出这个问题。谢谢
下面是code：
名单，其中，的NameValuePair＆GT; namevaluepairs中=新的ArrayList＆LT;的NameValuePair＆GT;（）;
nameValuePairs.add（新BasicNameValuePair（电子邮件，UserDummy.email））;
nameValuePairs.add（新BasicNameValuePair（密码，UserDummy.password））;
nameValuePairs.add（新BasicNameValuePair（FIRST_NAME，UserDummy.fname））;
nameValuePairs.add（新BasicNameValuePair（姓氏，UserDummy.lname））;
//使HTTP请求
尝试 {
//检查请求的方法
如果（方法==POST）{
//请求方法是POST
// defaultHttpClient
DefaultHttpClient的HttpClient =新DefaultHttpClient（）;
HttpPost httpPost =新HttpPost（URL）;
httpPost.setEntity（新UrlEn codedFormEntity（PARAMS））;
HTT presponse HTT presponse = httpClient.execute（httpPost）;
HttpEntity httpEntity = HTT presponse.getEntity（）;
是= httpEntity.getContent（）;
}否则，如果（方法==GET）{
//请求方法是GET
DefaultHttpClient的HttpClient =新DefaultHttpClient（）;
字符串中的paramString = URLEn codedUtils.format（参数，可以HTTP.UTF_8）;
？JSON = URL + = {+ +的paramString}; ;
HTTPGET HTTPGET =新HTTPGET（URL）;
HTT presponse HTT presponse = httpClient.execute（HTTPGET）;
HttpEntity httpEntity = HTT presponse.getEntity（）;
是= httpEntity.getContent（）;
}
}赶上（UnsupportedEncodingException E）{
e.printStackTrace（）;
Log.v（XXX，e.getMessage（））;
}赶上（ClientProtocolException E）{
e.printStackTrace（）;
Log.v（XXX，e.getMessage（））;
}赶上（IOException异常E）{
e.printStackTrace（）;
Log.v（XXX，e.getMessage（））;
}
尝试 {
的BufferedReader读卡器=新的BufferedReader（新的InputStreamReader（
是，ISO-8859-1），8）;
StringBuilder的SB =新的StringBuilder（）;
串线= NULL;
而（（行= reader.readLine（））！= NULL）{
sb.append（行+\ N）;
}
is.close（）;
JSON = sb.toString（）;
}赶上（例外五）{
Log.e（缓冲区错误，转换的结果错误+ e.toString（））;
}
//尝试解析字符串到一个JSON对象
尝试 {
jObj =新的JSONObject（JSON）;
}赶上（JSONException E）{
Log.e（JSON解析器，错误分析数据+ e.toString（））;
}
//返回JSON字符串
返回jObj;
HTTPGET不接受URL这种格式，并提供了错误，但是当我尝试在浏览器中正常工作。错误是以下内容：
在指数56非法字符查询
解决方案
最后调试，并尝试不同的解决方案了整整一天后，我解决了这个问题：）
我需要连接code中的部分参数，而不是整个URL是这样的：
字符串URL =Base_URI /用户/ add.json JSON =？;
URL = URL + URLEn coder.en code（{\电子邮件\：\+电子邮件+\，\密码\：\+密码+\}， UTF-8）;
谢谢大家的支持！
I have been trying to send data to server through GET method but I am unable to find a way to do it. I have tried few codes in asynchronous task but nothing. The web services are made in cakePhp and the format is like this:
Base_URI/users/add.json?json={"email": xxx@x.com, "password": "xxxxxxxxx", "first_name": "Xyz", "last_name": "Xyz"}
Android experts are requested to figure a way out of this problem. Thanks
Here is the code:
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("email", UserDummy.email));
nameValuePairs.add(new BasicNameValuePair("password", UserDummy.password));
nameValuePairs.add(new BasicNameValuePair("first_name", UserDummy.fname));
nameValuePairs.add(new BasicNameValuePair("last_name", UserDummy.lname));
// Making HTTP request
try {
// check for request method
if (method == "POST") {
// request method is POST
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} else if (method == "GET") {
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, HTTP.UTF_8);
url += "?json={" + paramString+"}"; ;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
Log.v("XXX", e.getMessage());
} catch (ClientProtocolException e) {
e.printStackTrace();
Log.v("XXX", e.getMessage());
} catch (IOException e) {
e.printStackTrace();
Log.v("XXX", e.getMessage());
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
HttpGet is not accepting this format of url and is giving the error but when I try it on browser it works fine. Error is following:
Illegal character in query at index 56
解决方案
Finally after debugging and trying different solutions the whole day, I solved the problem :)
I needed to encode the parameters part and not the whole URL like this:
String url = "Base_URI/users/add.json?json=";
url =url + URLEncoder.encode("{\"email\":\""+email+"\",\"password\":\""+password+"\"}", "UTF-8");
Thanks everyone for your support !
这篇关于通过GET方法发送的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
查看全文
