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java – how to insert %20 in place of space in android

Posted by: admin April 23, 2020 Leave a comment


I have a xml URL file in which there are white spaces i want to replace white spaces with %20.. how to do this????

SAXParserFactory spf = SAXParserFactory.newInstance();
SAXParser sp = spf.newSAXParser();
XMLReader xr = sp.getXMLReader();

/** Send URL to parse XML Tags */
URL sourceUrl = new URL(

XMLHandlerartistspace myXMLHandler = new XMLHandlerartistspace();
xr.parse(new InputSource(sourceUrl.openStream()));
How to&Answers:

Try this:

String temp = http://www.arteonline.mobi/iphone/output.php?gallery=MALBA%20-%20MUSEO%20DE%20ARTE%20LATINOAMERICANO%20DE%20BUENOS%20AIRES

temp = temp.replaceAll(" ", "%20");
URL sourceUrl = new URL(temp);


When you build your URL you should use URLEncoder to encode the parameters.

StringBuilder query = new StringBuilder();
query.append(URLEncoder.encode(value, "UTF-8"));

If you already have the whole URL in a String or a java.net.URL, you could grab the query part and rebuild while URLEncoding each parameter value.


Just one addition to sudocode’s response:

Use android.net.Uri.encode instead of URLEncoder.encode to avoid the “spaces getting converted into +” problem. Then you get rid of the String.replaceAll() and it’s more elegant 🙂

StringBuilder query = new StringBuilder();


I guess you want to replace all spaces, not only white.

the simplest way is to use

"url_with_spaces".replaceAll(" ", "%20);

However you should consider also other characters in the url. See Recommended method for escaping HTML in Java


String s = "my string";
s=s.replaceAll(" ", "%20");


Try using URIUtil.encodePath method from the api org.apache.commons.httpclient.util.URIUtil.

This should do the trick for you.


For anyone that needs space characters to be encoded as a %20 value instead of a + value, use:

String encodedString = URLEncoder.encode(originalString,"UTF-8").replaceAll("\+", "%20")