安卓HTTPGET问题 [英] android httpGet issue
本文介绍了安卓HTTPGET问题的处理方法,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
我想执行下面的链接使用下列code:
I am trying to execute the following link using following code:
class httpget{
HttpGet httpGet=null;
public void linkexecute(){
String url="http://<server>/<path>/action=send&msg=new message";
httpGet= new HttpGet(url); // line 1
....
}
at line 1 it is giving error "Illegal arguement exception"
java.lang.IllegalArgumentException: Illegal character in query at index 77: http://<server>/<path>/sms.json?action=send&msg=new message
at java.net.URI.create(URI.java:970)
at org.apache.http.client.methods.HttpGet.<init>(HttpGet.java:75)
at com.sms.login.LoginService.sendSms(LoginService.java:143)
虽然有有没有间隙的话没有错误的下面给出的URL味精=
Whereas there's no error for given below URL which has no gaps in words of "msg="
String url="http://<server>/<path>/action=send&msg=newmessage";
我怎样才能在URL解决的话差距问题?
How can I resolve the issue of gap in words in URL?
推荐答案
在这里,我给你一个功能,将删除所有无效字符的URL。请把您的网址在此功能,您将获得一个新的URL与EN codeD字符串。
Here, i give you one function that will remove all invalid characters from the url . Please pass your url in this function and you will get a new url with encoded strings.
public static String convertURL(String str) {
String url = null;
try{
url = new String(str.trim().replace(" ", "%20").replace("&", "%26")
.replace(",", "%2c").replace("(", "%28").replace(")", "%29")
.replace("!", "%21").replace("=", "%3D").replace("<", "%3C")
.replace(">", "%3E").replace("#", "%23").replace("$", "%24")
.replace("'", "%27").replace("*", "%2A").replace("-", "%2D")
.replace(".", "%2E").replace("/", "%2F").replace(":", "%3A")
.replace(";", "%3B").replace("?", "%3F").replace("@", "%40")
.replace("[", "%5B").replace("\\", "%5C").replace("]", "%5D")
.replace("_", "%5F").replace("`", "%60").replace("{", "%7B")
.replace("|", "%7C").replace("}", "%7D"));
}catch(Exception e){
e.printStackTrace();
}
return url;
}
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