String str = "internet address : http://test.com Click this!";
I want to get "http://test.com", so I wrote like this.
String[] split = str.split(" ");
for ( int i = 0 ; i < split.length ; i++ ) {
if ( split[i].contains("http://") ) {
return split[i];
}
}
but I think this is ineffective. how to get that more easily?
Assuming you always have the same format (some text : URL more text) this can work:
public static void main(String[] args) throws IOException {
String str = "internet address : http://test.com Click this!";
String first = str.substring(str.indexOf("http://"));
String second = first.substring(0, first.indexOf(" "));
System.out.println(second);
}
But better is regex as suggested in different answer
substring twice: use the indexOf(String, int) overload to get the end. But remember that you (may) need to handle the cast of there being no following space.
Usually, this is either done with a regular expression or with indexOf and substring.
With a regular expression, this can be done like that:
// This is using a VERY simplified regular expression
String str = "internet address : http://test.com Click this!";
Pattern pattern = Pattern.compile("[http:|https:]+\\/\\/[\\w.]*");
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
System.out.println(matcher.group(0));
}
You can read here why it's simplified: https://mathiasbynens.be/demo/url-regex - tl;dr: the problem with URLs is they can have so many different patterns which are valid.
With split, there would be a way utilizing the URL class of Java:
String[] split = str.split(" ");
for (String value : split) {
try {
URL uri = new URL(value);
System.out.println(value);
} catch (MalformedURLException e) {
// no valid url
}
}
You can check their validation in the OpenJDK source here.
My try with regex
String regex = "http?:\\/\\/(www\\.)?[-a-zA-Z0-9@:%._\\+~#=]{2,256}\\.[a-z]{2,6}\\b([-a-zA-Z0-9@:%_\\+.~#?&//=]*)";
String str = "internet address : http://test.com Click this!";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
System.out.println(matcher.group(0));
}
result:
http://test.com
source: here
but must clarify the pattern carefully, that's why I put the link in my answer for further investigation/understanding all the risk to choose which pattern suit you the most.Find the http:// in the string, then look forwards and backwards for the space:
int pos = str.indexOf("http://");
if (pos >= 0) {
// Look backwards for space.
int start = Math.max(0, str.lastIndexOf(' ', pos));
// Look forwards for space.
int end = str.indexOf(' ', pos + "http://".length());
if (end < 0) end = str.length();
return str.substring(start, end);
}
if (start < 0) start = 0; if you want (and that would be consistent with the looking forwards); but really, that is nano-optimization.
pos?http://. Had the original code been split[i].startsWith("http://"), you wouldn't need to search backwards.It is not clear if the structure of the input string is constant, however, I would do something like this:
String str = "internet address : http://test.com Click this!";
// get the index of the first letter of an url
int urlStart = str.indexOf("http://");
System.out.println(urlStart);
// get the first space after the url
int urlEnd = str.substring(urlStart).indexOf(" ");
System.out.println(urlEnd);
// get the substring of the url
String urlString = str.substring(urlStart, urlStart + urlEnd);
System.out.println(urlString);
I just made a quick solution for the same. It should work for you perfectly.
package Main.Kunal;
import java.util.ArrayList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class URLOutOfString {
public static void main(String[] args) {
String str = "internet address : http://test.com Click this!, internet address : http://tes1t.com Click this!";
List<String> result= new ArrayList<>();
int counter = 0;
final Pattern urlPattern = Pattern.compile(
"(?:^|[\\W])((ht|f)tp(s?):\\/\\/|www\\.)"
+ "(([\\w\\-]+\\.){1,}?([\\w\\-.~]+\\/?)*"
+ "[\\p{Alnum}.,%_=?&#\\-+()\\[\\]\\*$~@!:/{};']*)",
Pattern.CASE_INSENSITIVE | Pattern.MULTILINE | Pattern.DOTALL);
Matcher matcher = urlPattern.matcher(str);
while (matcher.find()) {
result.add(str.substring(matcher.start(1), matcher.end()));
counter++;
}
System.out.println(result);
}
}
This will find all URLs in your string and add it to arraylist. You can use it as per your business need.
You could use regex for it
String str = "internet address : http://test.com Click this!";
Pattern pattern = Pattern.compile("((http|https)\\S*)");
Matcher matcher = pattern.matcher(str);
if (matcher.find())
{
System.out.println(matcher.group(1));
}