Getting URL parameter in java and extract a specific text from that URL

Question

I have a URL and I need to get the value of v from this URL. Here is my URL: http://www.youtube.com/watch?v=_RCIP6OrQrE

How can I do that?

what is v ?? and please provide some more information and code if you can — Harmeet Singh
– Harmeet Singh, Commented Jul 31, 2012 at 5:15
what you are trying to do with youtube URL ? do you need to find the value v from a url-string using substring methods ? — sunil
– sunil, Commented Jul 31, 2012 at 5:20
@androidmaniac hey android, check out my answer btw.. I rewrote it. The function will support multiple sets of youtube urls. — classicjonesynz
– classicjonesynz, Commented Feb 14, 2013 at 9:46

Willi Mentzel · Accepted Answer · 2020-05-13 12:40:46Z

75

I think the one of the easiest ways out would be to parse the string returned by URL.getQuery() as

public static Map<String, String> getQueryMap(String query) {  
    String[] params = query.split("&");  
    Map<String, String> map = new HashMap<String, String>();

    for (String param : params) {  
        String name = param.split("=")[0];  
        String value = param.split("=")[1];  
        map.put(name, value);  
    }  
    return map;  
}

You can use the map returned by this function to retrieve the value keying in the parameter name.

edited May 13, 2020 at 12:40

Willi Mentzel

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answered Jul 31, 2012 at 5:31

verisimilitude

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5 Comments

DaBlick Over a year ago

This would not work for multivalued parameters. In the following example the EmployeeID parameter has three values: Northwest/Employee Sales Report&rs:Command=Render&EmployeeID=A&EmployeeID=B&EmployeeID=C You could make this work using a multi-valule map implementation instead of HashMap. I don't think the JRE has any, but multi-value map parameter implementations are easily found on the web.

Panthro Over a year ago

Also a ArrayIndexOutOfBoudndsException would be throw for urls like example.com.com?param or example.com?param=

Panthro Over a year ago

I recommend the solution here: stackoverflow.com/questions/5902090/…

Dr. Gadget Over a year ago

This will not work if the values contain "=" signs. I changed String value = param.split("=")[1]; with String value = param.split(name+"=")[1]; and now it seems to work.

Vishal_Kotecha Over a year ago

what if the param1=H&M, it doesn't parse if the parameter value itself has an ampersand(&)

Noah Andrews · Accepted Answer · 2017-09-10 03:41:20Z

31

If you're on Android, you can do this:

Uri uri = Uri.parse(url);
String v = uri.getQueryParameter("v");

edited Sep 10, 2017 at 3:41

Noah Andrews

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answered Nov 8, 2016 at 2:18

Deep P

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2 Comments

T J Over a year ago

Could you explain what package contains this?

Noah Andrews Over a year ago

@TJ It's in the Android SDK

Tomasz Szymulewski · Accepted Answer · 2017-07-11 14:26:22Z

9

I have something like this:

import org.apache.http.NameValuePair;
import org.apache.http.client.utils.URIBuilder;

private String getParamValue(String link, String paramName) throws URISyntaxException {
        List<NameValuePair> queryParams = new URIBuilder(link).getQueryParams();
        return queryParams.stream()
                .filter(param -> param.getName().equalsIgnoreCase(paramName))
                .map(NameValuePair::getValue)
                .findFirst()
                .orElse("");
    }

answered Jul 11, 2017 at 14:26

Tomasz Szymulewski

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Comments

classicjonesynz · Accepted Answer · 2015-03-19 04:53:49Z

3

I wrote this last month for Joomla Module when implementing youtube videos (with the Gdata API). I've since converted it to java.

Import These Libraries

    import java.net.URL;
    import java.util.regex.*;

Copy/Paste this function

    public String getVideoId( String videoId ) throws Exception {
        String pattern = "^(https?|ftp|file)://[-a-zA-Z0-9+&@#/%?=~_|!:,.;]*[-a-zA-Z0-9+&@#/%=~_|]";
        Pattern p = Pattern.compile(pattern);
        Matcher m = p.matcher(videoId);
        int youtu = videoId.indexOf("youtu");
        if(m.matches() && youtu != -1){
            int ytu = videoId.indexOf("http://youtu.be/");
            if(ytu != -1) { 
                String[] split = videoId.split(".be/");
                return split[1];
            }
            URL youtube = new URL(videoId);
            String[] split = youtube.getQuery().split("=");
            int query = split[1].indexOf("&");
            if(query != -1){
                String[] nSplit = split[1].split("&");
                return nSplit[0];
            } else return split[1];
        }
        return null; //throw something or return what you want
    }

URL's it will work with

http://www.youtube.com/watch?v=k0BWlvnBmIE (General URL)
http://youtu.be/k0BWlvnBmIE (Share URL)
http://www.youtube.com/watch?v=UWb5Qc-fBvk&list=FLzH5IF4Lwgv-DM3CupM3Zog&index=2 (Playlist URL)

edited Mar 19, 2015 at 4:53

answered Jul 31, 2012 at 5:17

classicjonesynz

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3 Comments

rtheunissen Over a year ago

You should use Arrays.toString(tokens).

classicjonesynz Over a year ago

@paranoid-android +1 rep yeah thanks I know, just forgot what the syntax was without Eclipse open.

Peter V. Mørch Over a year ago

Please don't advocate parsing URLs by hand. Use a library for that.

Jan Miller · Accepted Answer · 2015-04-23 07:04:29Z

Import these libraries

import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;

Similar to the verisimilitude, but with the capabilities of handling multivalue parameters. Note: I've seen HTTP GET requests without a value, in this case the value will be null.

public static List<NameValuePair> getQueryMap(String query)  
{  
    List<NameValuePair> queryMap = new ArrayList<NameValuePair>();
    String[] params = query.split(Pattern.quote("&"));  
    for (String param : params)
    {
        String[] chunks = param.split(Pattern.quote("="));
        String name = chunks[0], value = null;  
        if(chunks.length > 1) {
            value = chunks[1];
        }
        queryMap.add(new BasicNameValuePair(name, value));
    }
    return queryMap;
}

Example:

GET /bottom.gif?e235c08=1509896923&%49%6E%...

Roberto Rodriguez · Accepted Answer · 2020-07-21 15:29:15Z

3

Using pure Java 8

Assumming you want to extract param "v" from url:

             String paramV = Stream.of(url.split("?")[1].split("&"))
                        .map(kv -> kv.split("="))
                        .filter(kv -> "v".equalsIgnoreCase(kv[0]))
                        .map(kv -> kv[1])
                        .findFirst()
                        .orElse("");

answered Jul 21, 2020 at 15:29

Roberto Rodriguez

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Comments

Mohammad Naim Dahee · Accepted Answer · 2019-03-16 07:07:16Z

I believe we have a better approach to answer this question.

1: Define a function that returns Map values.

Here we go.

public Map<String, String> getUrlValues(String url) throws UnsupportedEncodingException {
    int i = url.indexOf("?");
    Map<String, String> paramsMap = new HashMap<>();
    if (i > -1) {
        String searchURL = url.substring(url.indexOf("?") + 1);
        String params[] = searchURL.split("&");

        for (String param : params) {
            String temp[] = param.split("=");
            paramsMap.put(temp[0], java.net.URLDecoder.decode(temp[1], "UTF-8"));
        }
    }

    return paramsMap;
}

2: Call your function surrounding with a try catch block

Here we go

try {
     Map<String, String> values = getUrlValues("https://example.com/index.php?form_id=9&page=1&view_id=78");
     String formId = values.get("form_id");
     String page = values.get("page");
     String viewId = values.get("view_id");
     Log.d("FormID", formId);
     Log.d("Page", page);
     Log.d("ViewID", viewId);
} catch (UnsupportedEncodingException e) {
     Log.e("Error", e.getMessage());
}

Umm · Accepted Answer · 2019-03-29 16:36:30Z

2

If you are using Jersey (which I was, my server component needs to make outbound HTTP requests) it contains the following public method:

var multiValueMap = UriComponent.decodeQuery(uri, true);

It is part of org.glassfish.jersey.uri.UriComponent, and the javadoc is here. Whilst you may not want all of Jersey, it is part of the Jersey common package which isn't too bad on dependencies...

answered Mar 29, 2019 at 16:36

Umm

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Comments

Oleksandr H · Accepted Answer · 2019-10-27 09:09:42Z

2

I solved the problem like this

public static String getUrlParameterValue(String url, String paramName) {
String value = "";
List<NameValuePair> result = null;

try {
    result = URLEncodedUtils.parse(new URI(url), UTF_8);
    value = result.stream().filter(pair -> pair.getName().equals(paramName)).findFirst().get().getValue();
    System.out.println("-------------->  \n" + paramName + " : " + value + "\n");
} catch (URISyntaxException e) {
    e.printStackTrace();
} 
return value;

}

answered Oct 27, 2019 at 9:09

Oleksandr H

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2 Comments

tfga Feb 14 at 0:19

Where does URLEncodedUtils come from?

Oleksandr H Apr 13 at 19:14

Standart lib from org.apache.http.client.utils.URLEncodedUtils

rtheunissen · Accepted Answer · 2012-07-31 05:27:31Z

0

Assuming the URL syntax will always be http://www.youtube.com/watch?v= ...

String v = "http://www.youtube.com/watch?v=_RCIP6OrQrE".substring(31);

or disregarding the prefix syntax:

String url = "http://www.youtube.com/watch?v=_RCIP6OrQrE";
String v = url.substring(url.indexOf("v=") + 2);

answered Jul 31, 2012 at 5:27

rtheunissen

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2 Comments

rtheunissen Over a year ago

What do you mean, @smonff?

smonff Over a year ago

I mean that I appreciated the simplicity of your answer :-D

Harmeet Singh · Accepted Answer · 2012-07-31 05:40:28Z

0

this will work for all sort of youtube url :
if url could be

youtube.com/?v=_RCIP6OrQrE
youtube.com/v/_RCIP6OrQrE
youtube.com/watch?v=_RCIP6OrQrE
youtube.com/watch?v=_RCIP6OrQrE&feature=whatever&this=that

Pattern p = Pattern.compile("http.*\\?v=([a-zA-Z0-9_\\-]+)(?:&.)*");
String url = "http://www.youtube.com/watch?v=_RCIP6OrQrE";
Matcher m = p.matcher(url.trim()); //trim to remove leading and trailing space if any

if (m.matches()) {
    url = m.group(1);        
}
System.out.println(url);

this will extract video id from your url

further reference

edited Jul 31, 2012 at 5:40

answered Jul 31, 2012 at 5:33

Harmeet Singh

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Comments

T J · Accepted Answer · 2018-10-31 15:18:58Z

0

My solution mayble not good

        String url = "https://www.youtube.com/watch?param=test&v=XcHJMiSy_1c&lis=test";
        int start = url.indexOf("v=")+2;
        // int start = url.indexOf("list=")+5; **5 is length of ("list=")**
        int end = url.indexOf("&", start);

        end = (end == -1 ? url.length() : end); 

        System.out.println(url.substring(start, end));
        // result: XcHJMiSy_1c

work fine with:

https://www.youtube.com/watch?param=test&v=XcHJMiSy_1c&lis=test
https://www.youtube.com/watch?v=XcHJMiSy_1c

edited Oct 31, 2018 at 15:18

T J

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answered May 13, 2014 at 5:23

vuhung3990

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Comments

Hank · Accepted Answer · 2020-02-14 05:39:14Z

0

public static String getQueryMap(String query) {        
    String[] params = query.split("&");     
    for (String param : params) {           
       String name = param.split("=")[0];
       if ("YourParam".equals(name)) {
           return param.split("=")[1]; 
       }
    }
    return null;
}

edited Feb 14, 2020 at 5:39

Hank

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answered Feb 14, 2020 at 4:49

Shubham Goswami

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Comments

tfga · Accepted Answer · 2025-02-18 19:06:24Z

0

Re: Oleksandr's answer

URLEncodedUtils is deprecated in the latest version of Apache HttpClient (5.4.2). It's been replaced by URIBuilder. Here's my solution:

static String getYouTubeId(String url) {

    try
    {
        NameValuePair nv = new URIBuilder(url).getFirstQueryParam("v");

        return nv != null ? nv.getValue()
                          : null
                          ;
    }
    catch (URISyntaxException e)
    {
        return null;
    }
}

answered Feb 18 at 19:06

tfga

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