Develop Reference

Convert string to a valid URL link

I am parsing some values from a website with JSoup, some of them are also url links (href).
When I get the url link, which I set to a string. The string sometimes does not become a valid URL link as it has a special character like '!?()
Example: https://somelink.com/King's+Beak (The ' makes the link not valid).
Now I tackle this by replacing the characters with the default character set for UTF-8, which works as it should.
Example code:
String test = arTD.select("a.wiki_link").get(0).attr("href").replaceAll("'", "%27");
I also set JSoup to UTF-8 but that does not seem to work.
Document document = Jsoup.parse(response.body().string(), "UTF-8");
Now my question is, is there a more convenient way to tackle this?, as I need to escape more characters like '!?().
Thank you in advance.

One way to solve this issue is to use the URLEncoder.encode() method to encode the URL string. This method will replace special characters with their corresponding ASCII codes.
String test = arTD.select("a.wiki_link").get(0).attr("href");
String encodedUrl = URLEncoder.encode(test, StandardCharsets.UTF_8);
Another way to solve this issue is by using the Uri.Builder class in android to encode the URL.
Uri.Builder builder = new Uri.Builder()
.scheme("https")
.authority("link.com")
.appendPath("King's+Beak");
Uri uri = builder.build();
String encodedUrl = uri.toString();
The first method will encode the whole url while the second method will only encode the path of the url.
Choose the suitable one for you.

IllegalArgumentException IN ANDROID URL

Using the below url I got an error:
java.lang.IllegalArgumentException: Illegal character in path at index 47: http://safetracker-threetinker.rhcloud.com/api/{userid}/locations?lat={latitude}&lng={longitude}.
URL:
URL=http://safetracker-threetinker.rhcloud.com/api/{userid}/locations?lat={latitude}&lng={longitude}
how to solve the error. I don't have good knowledge in URL encoding. please help me to find the solution.

The problem is actually it is looking for long/integer value and you are passing a { just put a $ so that it will be replaced by the actual value pointed by the variable
http://safetracker-threetinker.rhcloud.com/api/1/locations?lat=5&lng=5
your Address is like this
URL=http://safetracker-threetinker.rhcloud.com/api/{userid}/locations?lat={latitude}&lng={longitude}
it should be like this
URL=http://safetracker-threetinker.rhcloud.com/api/${userid}/locations?lat=${latitude}&lng={longitude}

We can not use some special characters in URL, so we have to replace these special characters with its encoded form.
Replace your URL with following URL
URL=http://safetracker-threetinker.rhcloud.com/api/%7Buserid%7D/locations?lat=%7Blatitude%7D&lng=%7Blongitude%7D
May this help you.

URLEncoder should be the way to go. You only need to keep in mind to encode only the individual query string parameter name and/or value, not the entire URL, for sure not the query string parameter separator character & nor the parameter name-value separator character =.
String q = "replace_with user_id/locations?lat=replace with latitude&lng=replace with longitude";
String url = "http://safetracker-threetinker.rhcloud.com/api/=" + URLEncoder.encode(q, "UTF-8");

Equals sign in url not working

i got a problem which i tried hard to solve with searching with google because will be a simple solution.
I want to open the following url:
http://<<IP>>/query.html?sql="select * from ADAnreden"
To do this, i write this url in a string to open it with HttpGet...
String url = "http://"+ip+"/query.html?sql=\"select * from ADAnreden\"";
So i escaped the " infront of select and after ADAnreden. But the problem is that the following error is comming up:
Illegal character in query at index 36.
This is the equals sign. So how can i escape the = ? The backslash is not working.
Thanks for help

The issue is the escaping for the URL, not for Java. Spaces are not valid in URLs. See this answer for more about URL encoding in Android.

you must encode the query before using it as URL, see URLEncoder.encode(query);

Simple way to convert a URL into a filename

I'm writing an asynchronous image downloader for Android and was just wondering, given an arbitary URL such as:
http://www.android.com/images/brand/droid.gif
What would be the best way to convert the unique url to a filename. I thought about simply splitting the url and grabbing the last section, but I want the filename to be representative of the whole URL. The other alternatives I thought were replacing all the forward slashes with underscores or simply hashing the whole URL and storing this.
If anyone has any ideas I'd love to hear them!
Thanks

In case, usually uses MD5 hash. but I suggest to use 'aquery' library. In library you can simply download Image asynchronous and put it to view. It also support disk cache, memory cache simply.

This method will be fulfill your requirements. It will generate a name which will represent original URL. You can call generateNameFromUrl(String url) method like this.
String url = "http://www.android.com/images/brand/droid.gif";
String uniqueName = generateNameFromUrl(url));
Method is given below:
public static String generateNameFromUrl(String url){
// Replace useless chareacters with UNDERSCORE
String uniqueName = url.replace("://", "_").replace(".", "_").replace("/", "_");
// Replace last UNDERSCORE with a DOT
uniqueName = uniqueName.substring(0,uniqueName.lastIndexOf('_'))
+"."+uniqueName.substring(uniqueName.lastIndexOf('_')+1,uniqueName.length());
return uniqueName;
}
Input: "http://www.android.com/images/brand/droid.gif"
Output: "http_www_android_com_images_brand_droid.gif"

android java URLDecoder problem

i have a String displayed on a WebView as "Siwy & Para Wino"
i fetch it from url , i got a string "Siwy%2B%2526%2BPara%2BWino". // be corrected
now i'm trying to use URLDecoder to solve this problem :
String decoded_result = URLDecoder.decode(url); // the url is "Siwy+%26+Para+Wino"
then i print it out , i still saw "Siwy+%26+Para+Wino"
Could anyone tell me why?

From the documentation (of URLDecoder):
This class is used to decode a string which is encoded in the application/x-www-form-urlencoded MIME content type.
We can look at the specification to see what a form-urlencoded MIME type is:
The form field names and values are escaped: space characters are replaced by '+', and then reserved characters are escaped as per [URL]; that is, non-alphanumeric characters are replaced by '%HH', a percent sign and two hexadecimal digits representing the ASCII code of the character. Line breaks, as in multi-line text field values, are represented as CR LF pairs, i.e. '%0D%0A'.
Since the specification calls for a percent sign followed by two hexadecimal digits for the ASCII code, the first time you call the decode(String s) method, it converts those into single characters, leaving the two additional characters 26 intact. The value %25 translates to % so the result after the first decoding is %26. Running decode one more time simply translates %26 back into &.
String decoded_result = URLDecoder.decode(URLDecoder.decode(url));
You can also use the Uri class if you have UTF-8-encoded strings:
Decodes '%'-escaped octets in the given string using the UTF-8 scheme.
Then use:
String decoded_result = Uri.decode(Uri.decode(url));

thanks for all answers , i solved it finally......
solution:
after i used URLDecoder.decode twice (oh my god) , i got what i want.
String temp = URLDecoder.decode( url); // url = "Siwy%2B%2526%2BPara%2BWino"
String result = URLDecoder.decode( temp ); // temp = "Siwy+%26+Para+Wino"
// result = "Swy & Para Wino". !!! oh good job.
but i still don't know why.. could someone tell me?