Get Specific URL from the Text/String Using Java

Question

I'm newbie to Java, I want to get all of the URL in the text below

WEBSITE1 https://localhost:8080/admin/index.php?page=home
WEBSITE2 https://192.168.0.3:8084/index.php
WEBSITE3 https://192.168.0.5:9090/controller/index.php?page=home
WEBSITE4 https://192.168.0.1:8080/home/index.php?page=forum

the result that I want is:

https://localhost:8080
https://192.168.0.3:8084
https://192.168.0.5
https://192.168.0.1:8080

I want to store it into the Linked List or Array too. Can somebody teach me? Thank You

Have you tried anything?

fge
– fge

2013-06-20 13:23:04 +00:00
Commented Jun 20, 2013 at 13:23 — fge
– fge, Commented Jun 20, 2013 at 13:23
stackoverflow.com/questions/163360/…

juniperi
– juniperi

2013-06-20 13:26:02 +00:00
Commented Jun 20, 2013 at 13:26 — juniperi
– juniperi, Commented Jun 20, 2013 at 13:26
@juniperi VERY bad idea, when you have URI

fge
– fge

2013-06-20 13:29:07 +00:00
Commented Jun 20, 2013 at 13:29 — fge
– fge, Commented Jun 20, 2013 at 13:29
@fge Second answer uses URI

juniperi
– juniperi

2013-06-20 13:30:15 +00:00
Commented Jun 20, 2013 at 13:30 — juniperi
– juniperi, Commented Jun 20, 2013 at 13:30

grepit · Accepted Answer · 2013-06-20 13:30:37Z

1

This is how you can do this. I did one for you and you do the rest :)

try {
            ArrayList<String> urls = new ArrayList<String>();
            URL aURL = new URL("https://localhost:8080/admin/index.php?page=home");
             System.out.println("protocol = " + aURL.getProtocol()+aURL.getHost()+aURL.getPort());
             urls.add(aURL.getProtocol()+aURL.getHost()+aURL.getPort());
        } catch (MalformedURLException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

answered Jun 20, 2013 at 13:30

grepit

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Comments

Alex · Accepted Answer · 2013-06-22 11:19:34Z

0

Use a simple regexp to locate what's starting with https?:// and then just extract this until the first /

Matcher m = Pattern.compile("(https?://[^/]+)").matcher(//
        "WEBSITE1 https://localhost:8080/admin/index.php?page=home\r\n" + //
        "WEBSITE2 https://192.168.0.3:8084/index.php\r\n" + //
        "WEBSITE3 https://192.168.0.5:9090/controller/index.php?page=home\r\n" + //
        "WEBSITE4 https://192.168.0.1:8080/home/index.php?page=forum");
List<String> urls = new ArrayList<String>();
while (m.find()) {
    urls.add(m.group(1));
}
System.out.println(urls);

Now if you do want to get only the WEBSITE. part you will only have to change the regular expression "(https?://[^/]+)" with the following one: "(.*?)\\s+https?". The rest of the code stays untouched.

edited Jun 22, 2013 at 11:19

answered Jun 20, 2013 at 13:31

Alex

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1 Comment

user1973423 Over a year ago

thank you, and what if I want to get the list of website name? for example, the result that I want: website1 website2 website3 website4 thank you agan :-)

darijan · Accepted Answer · 2013-06-20 13:28:39Z

0

Let's say the line represents a single line (probably in a loop):

//get the index of "https" in the string
int indexOfHTTPS= line.indexOf("https://");
//get the index of the first "/" after the "https"
int indexOfFirstSlashAfterHTTPS= line.indexOf("/", indexOfHTTPS + "https://".length());

//take a string between "https" and the first "/"
String url = line.substring(indexOfHTTPS, indexOfFirstSlashAfterHTTPS);

Later on, add this url to an ArrayList<String>:

ArrayList<String> urlList= new ArrayList<String>();
urlList.add(url);

answered Jun 20, 2013 at 13:28

darijan

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Comments

Suresh Atta · Accepted Answer · 2013-06-20 13:30:05Z

0

You can do it with the help of URL class.

 public static void main(String[] args) throws MalformedURLException { 

        String string ="https://192.168.0.5:9090/controller/index.php?page=home";
        URL url= new URL(string);
        String result ="https://"+url.getHost()+":"+url.getPort();
        System.out.println(result);
    }

Output :https://192.168.0.5:9090

answered Jun 20, 2013 at 13:30

Suresh Atta

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1 Comment

fge Over a year ago

That won't grab the port -- hence the interest of using URI instead. What is more, URI will never attempt to resolve hostnames

Mena · Accepted Answer · 2013-06-20 13:30:59Z

You could either try to find the index of the protocol substring ("http[s]") in the Strings, or use a simple Pattern (only for matching the "website[0-9]" head, not to apply to the URLs).

Here's a solution with the Pattern.

String webSite1 = "WEBSITE1 https://localhost:8080/admin/index.php?page=home";
String webSite2 = "WEBSITE2 https://192.168.0.3:8084/index.php";
String webSite3 = "WEBSITE3 https://192.168.0.5:9090/controller/index.php?page=home";
String webSite4 = "WEBSITE4 https://192.168.0.1:8080/home/index.php?page=forum";
ArrayList<URI> uris = new ArrayList<URI>();
Pattern pattern = Pattern.compile("^website\\d+\\s+?(.+)", Pattern.CASE_INSENSITIVE);
Matcher matcher;
matcher = pattern.matcher(webSite1);
if (matcher.find()) {
    try {
        uris.add(new URI(matcher.group(1)));
    }
    catch (URISyntaxException use) {
        use.printStackTrace();
    }
}
matcher = pattern.matcher(webSite2);
if (matcher.find()) {
    try {
        uris.add(new URI(matcher.group(1)));
    }
    catch (URISyntaxException use) {
        use.printStackTrace();
    }
}
matcher = pattern.matcher(webSite3);
if (matcher.find()) {
    try {
        uris.add(new URI(matcher.group(1)));
    }
    catch (URISyntaxException use) {
        use.printStackTrace();
    }
}
matcher = pattern.matcher(webSite4);
if (matcher.find()) {
    try {
        uris.add(new URI(matcher.group(1)));
    }
    catch (URISyntaxException use) {
        use.printStackTrace();
    }
}
System.out.println(uris);

Output:

[https://localhost:8080/admin/index.php?page=home, https://192.168.0.3:8084/index.php, https://192.168.0.5:9090/controller/index.php?page=home, https://192.168.0.1:8080/home/index.php?page=forum]

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