One of my URL is like the following: "h--p://www.test.com///rss.xml"
When I run the following code:
private String RSSFEEDURL = Uri.encode("h--p://www.test.com/path/*/*/rss.xml");
URL url = null;
try {
url = new URL(xml);
} catch (MalformedURLException e1) {
e1.printStackTrace();
}
I am getting "java.net.MalformedURLException: Protocol not found: http%3A%2F%2Ftest.com%2Fpath%2F*%2F*%2Frss.xml"
I have already done Uri encode as shown above. Any idea, what is causing this issue and how I could resolve it?
if you call new URL the thing you put in there should be a valid URL.
You're putting this in there: http%3A%2F%2Ftest.com%2Fpath%2F*%2F*%2Frss.xml, and that's not a valid URL, so the exception is expected.
You shouldn't encode your whole URL.
I am getting this error:
java.net.MalformedURLException: Protocol not found[java.lang.StringBuilder]
When the following line is getting executed:
url = new URL(urlString.toString());
urlString stores the following value:
http://maps.google.com/maps?f=d&hl=en&saddr=25.04202,121.534761&daddr=25.05202,121.554761&ie=UTF8&0&om=0&output=kml
What causes this Exception?
Chances are that you didn't clean after changing from
url = new URL( urlString );
to
url = new URL(urlString.toString());
You should log the value of the parameter passed to the constructor of URL.
It's not what you think it should be.
urlString would print a value in the form of java.lang.StringBuilder#
thus throwing the exception if you try to build a url out of that.
But using to String will print the value of the content string built by the stringbuilder.
Here is an example with exception handling ...
String urlString = "https://www.example.com/";
URL url = null;
// handle Exception
try {
url = new URL(urlString);
} catch (MalformedURLException e) {
System.out.println("The URL is not valid.");
System.out.println(e.getMessage());
}
// print
if (url != null) {
System.out.println(url.toString());
}
You can also get this exception if your network firewall blocking that URL.
replace url with php file address placed over server instedof html
i am working with JSoup and Android to get image urls from some site but some urls contains special characters like (é,è,à...) example :
http://www.mysite.com/détail du jour.jpg
the element.attr("abs:src") returns the same url as above
till now no problem to retrieve the url but when i submit this url in the code below it returns file not found (i grabbed this function from an example on the internet) :
public Object fetch(String address) throws MalformedURLException,IOException {
try {
URL url = new URL(address);
Object content = url.getContent();
return content;
} catch (Exception e) {
return null;
}
}
i think the problem is in the url format because when i get the real address of the image in google chrome :
http://www.mysite.com/d%C3%A9tail%20du%20jour.jpg
and submit it in the code like :
URL url = new URL("http://www.mysite.com/d%C3%A9tail%20du%20jour.jpg");
the image loads correctly so how to get this formatted url from JSoup?
thanks
You need to use URLEncoder for the extracted url from the JSoup.
Something like:
URL url = new URL(URLEncoder.encode(address));
The spaces between will be replaced with special character values %something
i would like to parse out some text from a page.
Is there an easy way to save the product info in to a string for example? Example url: http://upcdata.info/upc/7310870008741
Thanks
Jsoup is excellent at parsing simple HTML from Android applications:
http://jsoup.org/
To get the page, just do this:
URL url = new URL("http://upcdata.info/upc/7310870008741");
Document document = Jsoup.parse(url, 5000);
Then you can parse out whatever you need from the Document. Check out this link for a brief description of how to extract parts of the page:
http://jsoup.org/cookbook/extracting-data/dom-navigation
If you want to read from a URL into a String:
StringBuffer myString = new StringBuffer();
try {
String thisLine;
URL u = new URL("http://www.google.com");
DataInputStream theHTML = new DataInputStream(u.openStream());
while ((thisLine = theHTML.readLine()) != null) {
myString.append(thisLine);
}
} catch (MalformedURLException e) {
} catch (IOException e) {
}
// call toString() on myString to get the contents of the file your URL is
// pointing to.
This will give you a plain old string, HTML markup and all.
String tmpHtml = "<html>a whole bunch of html stuff</html>";
String htmlTextStr = Html.fromHtml(tmpHtml).toString();
I am using grid view for displaying image using xml parsing,i got some exception like
java.lang.IllegalArgumentException: Illegal character in path at
index 80:
http://www.theblacksheeponline.com/party_img/thumbspps/912big_361999096_Flicking
Off Douchebag.jpg
How to solve this problem? I want to display all kind of url,anybody knows please give sample code for me.
Thanks All
URL encoding is done in the same way on android as in Java SE;
try {
String url = "http://www.example.com/?id=123&art=abc";
String encodedurl = URLEncoder.encode(url,"UTF-8");
Log.d("TEST", encodedurl);
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
Also you can use this
private static final String ALLOWED_URI_CHARS = "##&=*+-_.,:!?()/~'%";
String urlEncoded = Uri.encode(path, ALLOWED_URI_CHARS);
it's the most simple method
As Ben says in his comment, you should not use URLEncoder.encode to full URLs because you will change the semantics of the URL per the following example from the W3C:
The URIs
http://www.w3.org/albert/bertram/marie-claude
and
http://www.w3.org/albert/bertram%2Fmarie-claude
are NOT identical, as in the second
case the encoded slash does not have
hierarchical significance.
Instead, you should encode component parts of a URL independently per the following from RFC 3986 Section 2.4
Under normal circumstances, the only
time when octets within a URI are
percent-encoded is during the process
of producing the URI from its
component parts. This is when an
implementation determines which of the
reserved characters are to be used as
subcomponent delimiters and which can
be safely used as data. Once
produced, a URI is always in its
percent-encoded form.
So, in short, for your case you should encode/escape your filename and then assemble the URL.
You don't encode the entire URL, only parts of it that come from "unreliable sources" like.
String query = URLEncoder.encode("Hare Krishna ", "utf-8");
String url = "http://stackoverflow.com/search?q=" + query;
URLEncoder should be used only to encode queries, use java.net.URI class instead:
URI uri = new URI(
"http",
"www.theblacksheeponline.com",
"/party_img/thumbspps/912big_361999096_Flicking Off Douchebag.jpg",
null);
String request = uri.toASCIIString();
you can use below method
public String parseURL(String url, Map<String, String> params)
{
Builder builder = Uri.parse(url).buildUpon();
for (String key : params.keySet())
{
builder.appendQueryParameter(key, params.get(key));
}
return builder.build().toString();
}
I tried with URLEncoder that added (+) sign in replace of (" "), but it was not working and getting 404 url not found error.
Then i googled for get better answer and found this and its working awesome.
String urlStr = "http://www.example.com/test/file name.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();
This way of encoding url its very useful because using of URL we can separate url into different part. So, there is no need to perform any string operation.
Then second URI class, this approach takes advantage of the URI class feature of properly escaping components when you construct a URI via components rather than from a single string.
I recently wrote a quick URI encoder for this purpose. It even handles unicode characters.
http://www.dmurph.com/2011/01/java-uri-encoder/