Android的web视图POST [英] Android Webview POST

问题描述

我想实现的东西很简单，但我发现这个没有好的文档。我有一个web视图，我需要加载一个页面就需要POST数据。似乎是一个简单的过程，但我不能找到一种方法，结果显示在一个web视图。

这个过程应该是简单的：

查询（用POST数据） - > Web服务器 - > HTML响应 - > web视图

余可使用DefaultHttpClient提交数据，但是这不能被显示在一个web视图。

有什么建议？

许多感谢

解决方案

 私有静态最后弦乐URL_STRING =htt​​p://www.yoursite.com/postreceiver;

    公共无效POSTDATA（）抛出IOException异常，ClientProtocolException {

         名单＆LT;的NameValuePair＆GT; namevaluepairs中=新的ArrayList＆LT;的NameValuePair＆GT;（）;
         nameValuePairs.add（新BasicNameValuePair（富，12345））;
         nameValuePairs.add（新BasicNameValuePair（酒吧，23456））;

         HttpClient的HttpClient的=新DefaultHttpClient（）;
         HttpPost httppost =新HttpPost（URL_STRING）;
         httppost.setEntity（新UrlEn codedFormEntity（namevaluepairs中））;

         HTT presponse响应= httpclient.execute（httppost）;
         。字符串数据=新BasicResponseHandler（）用handleResponse（响应）;
         mWebView.loadData（数据，text / html的，UTF-8）;
    }
 

解决方案

试试这个：

 私有静态最后弦乐URL_STRING =htt​​p://www.yoursite.com/postreceiver;

公共无效POSTDATA（）抛出IOException异常，ClientProtocolException {

     名单＆LT;的NameValuePair＆GT; namevaluepairs中=新的ArrayList＆LT;的NameValuePair＆GT;（）;
     nameValuePairs.add（新BasicNameValuePair（富，12345））;
     nameValuePairs.add（新BasicNameValuePair（酒吧，23456））;

     HttpClient的HttpClient的=新DefaultHttpClient（）;
     HttpPost httppost =新HttpPost（URL_STRING）;
     httppost.setEntity（新UrlEn codedFormEntity（namevaluepairs中））;

     HTT presponse响应= httpclient.execute（httppost）;

}
 

我会建议这样做，为的AsyncTask的一部分，之后更新的WebView

I am trying to accomplish something quite simple, yet I have found no good documentation on this. I have a webView, and I need to load a page in it that requires POST data. Seems like a simple process, yet I cannot find a way to display the result in a webView.

The process should be simple:

query(with POST data) -> webserver -> HTML response -> WebView.

I can submit data using a DefaultHttpClient, but this cannot be displayed in a WebView.

Any suggestions?

Much Thanks

Solution

private static final String URL_STRING = "http://www.yoursite.com/postreceiver";

    public void postData() throws IOException, ClientProtocolException {  

         List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();  
         nameValuePairs.add(new BasicNameValuePair("foo", "12345"));  
         nameValuePairs.add(new BasicNameValuePair("bar", "23456"));

         HttpClient httpclient = new DefaultHttpClient();  
         HttpPost httppost = new HttpPost(URL_STRING);  
         httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));  

         HttpResponse response = httpclient.execute(httppost);  
         String data = new BasicResponseHandler().handleResponse(response);
         mWebView.loadData(data, "text/html", "utf-8");
    }

解决方案

Try this:

private static final String URL_STRING = "http://www.yoursite.com/postreceiver";

public void postData() throws IOException, ClientProtocolException {  

     List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();  
     nameValuePairs.add(new BasicNameValuePair("foo", "12345"));  
     nameValuePairs.add(new BasicNameValuePair("bar", "23456"));

     HttpClient httpclient = new DefaultHttpClient();  
     HttpPost httppost = new HttpPost(URL_STRING);  
     httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));  

     HttpResponse response = httpclient.execute(httppost);  

}

I would recommend doing this as part of an AsyncTask and updating the WebView afterwards

这篇关于Android的web视图POST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！