ajax调用webservice失败

Question

跨域访问 ajax跨域访问

Answer 1

你这是跨域ajax吧！？跨域只能使用JSONP来实现，或者通过服务器端获取

Answer 2

楼主可以将webservice取数据放到后台去处理，然后数据返回页面就可以了。后台调用不涉及跨域的问题。

Answer 3

跨域了。不可以，你可以在后台调用webservice ,然后前台ajax' 获取就行了。

Answer 4

jsonp估计不好搞，callback这边控制不了。还是写个servlet吧。severlet调用webservice，然后前台调用这个servelet就行了。 <script type="text/javascript">
            $.ajax({
                url:'http://localhost:8080/test/severlet/weather',
                type:'GET',
                dataType: 'json',
                success:function(data){
                    alert(data.he);
                },
                error : function(e, text){
                    alert(e.status);
                    alert(text);
                }
            });</script>
public class WeatherServlet extends HttpServlet { private static final long serialVersionUID = 1L; @Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
doGet(req, resp);
} @Override
protected void doGet(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
request.setCharacterEncoding("utf-8");
response.setCharacterEncoding("utf-8");
PrintWriter out = response.getWriter();
response.setContentType("text/xml;charset=utf-8");
response.setHeader("Access-Control-Allow-Origin", "*"); String json = Test.getWeatherInfo();
json = new String(json.getBytes("gbk"), "utf-8");
JSONObject obj = new JSONObject();
out.write(obj.put("he", json.toString()).toString());
out.flush();
out.close();
}
}
public class Test { static String getWeatherInfo() throws MalformedURLException, IOException,
ProtocolException {
String s = "http://webservice.webxml.com.cn/WebServices/WeatherWS.asmx?op=getWeather";
URL url = new URL(s);
HttpURLConnection http = (HttpURLConnection) url.openConnection(); http.setDoOutput(true);
http.setDoInput(true);
http.setRequestMethod("POST");
http.setUseCaches(false);
http.setRequestProperty("Content-Type", "text/xml");
http.connect(); DataOutputStream out = new DataOutputStream(http.getOutputStream());
String city = "025";
String content = "<soap:Envelope xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:soap=\"http://schemas.xmlsoap.org/soap/envelope/\"><soap:Body><getWeather xmlns=\"http://WebXml.com.cn/\"><theCityCode>"
+ city
+ "</theCityCode><theUserID></theUserID></getWeather></soap:Body></soap:Envelope>";
out.writeBytes(content); out.flush();
out.close();
BufferedReader reader = new BufferedReader(new InputStreamReader(http
.getInputStream()));
String line;
StringBuffer buffer = new StringBuffer();
while ((line = reader.readLine()) != null) {
// System.out.println(line);
buffer.append(line);
}
reader.close();
http.disconnect();
System.out.println(new String(buffer.toString().getBytes("gbk"),
"utf-8"));
return buffer.toString();
}}web.xml
<servlet>
<servlet-name>weatherServlet</servlet-name>
<servlet-class>com.WeatherServlet
</servlet-class>
</servlet> <servlet-mapping>
<servlet-name>weatherServlet</servlet-name>
<url-pattern>/severlet/weather</url-pattern>
</servlet-mapping>这样就可以了，这是我以前做的一个，数据自己随便处理。