小编use*_*932的帖子

我得到Exception时如何设置消息

public class XMLParser {

    // constructor
    public XMLParser() {

    }


    public String getXmlFromUrl(String url) {
        String responseBody = null;

        getset d1 = new getset();
        String d = d1.getData(); // text
        String y = d1.getYear(); // year
        String c = d1.getCircular();
        String p = d1.getPage();

        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
        nameValuePairs.add(new BasicNameValuePair("YearID", y));

        nameValuePairs.add(new BasicNameValuePair("CircularNo", c));

        nameValuePairs.add(new BasicNameValuePair("SearchText", d));
        nameValuePairs.add(new BasicNameValuePair("pagenumber", p));
        try {

            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost(url);
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);

            HttpEntity …
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java android

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