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

问题

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

Any useful and fruitful help is highly appreciated..

回答1:

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.

回答2:

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

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

回答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)

回答4:

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%...

回答5:

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);

回答6:

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());
}

回答7:

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("");
    }

回答8:

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...

回答9:

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

回答10:

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

回答11:

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;

}

来源：https://stackoverflow.com/questions/11733500/getting-url-parameter-in-java-and-extract-a-specific-text-from-that-url

标签

java

url

text-extraction