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Replace and modify String using regex in java

I have a part of HTML from a website in the below String format:
srcset=" /tesla_theme/assets/img/homepage/mobile/homepage-models--touch#200w.jpg?20170808 200w, /tesla_theme/assets/img/homepage/mobile/homepage-models--touch#338w.jpg?20170808 338w, /tesla_theme/assets/img/homepage/mobile/homepage-models--touch#445w.jpg?20170808 445w, tesla_theme/assets/img/homepage/mobile/homepage-models--touch#542w.jpg?20170808 542w, /tesla_theme/assets/img/homepage/mobile/homepage-models--touch#750w.jpg?20170808 750w"
I want to add http://tesla.com in front of all the urls in the srcset element like http://tesla_theme/assets/img/homepage/mobile/homepage-models--touch#750w.jpg?20170808 750w
I believe this could be done using regex, but I am not sure.
How do I do this using Java if I have multiple srcset elements in a html string variable, and I want to replace all of the srcset url.'s and add the server url in front?
Note: The /tesla_theme will not be consistent, so I cannot use replaceAll, instead, i will have to use regex.

You can simply use String Class replace method as below, It will replace all "/_tesla" in the given String. No special regex required unless you have a kind of pattern instead of "/tesla"
String srcset=" /tesla_theme/assets/img/homepage/mobile/homepage-models--touch#200w.jpg?20170808 200w, /tesla_theme/assets/img/homepage/mobile/homepage-models--touch#338w.jpg?20170808 338w, /tesla_theme/assets/img/homepage/mobile/homepage-models--touch#445w.jpg?20170808 445w, tesla_theme/assets/img/homepage/mobile/homepage-models--touch#542w.jpg?20170808 542w, /tesla_theme/assets/img/homepage/mobile/homepage-models--touch#750w.jpg?20170808 750w";
String requiredSrcSet = srcset.replace("/tesla_", "http://tesla_");

How do I split the rest of the URL from the last path of it

I have this file URL: http://xxx.xxx.xx.xx/resources/upload/2014/09/02/new sample.pdf which will be converted to http://xxx.xxx.xx.xx/resources/upload/2014/09/02/new%20sample.pdf later.
Now I can get the last path by:
public static String getLastPathFromUrl(String url) {
return url.replaceFirst(".*/([^/?]+).*", "$1");
}
which will give me new sample.pdf
but how do I get the remaining of the URL: http://xxx.xxx.xx.xx/resources/upload/2014/09/02/
?

Easier way to get last path from URL would be to use String.split function, like this:-
String url = "http://xxx.xxx.xx.xx/resources/upload/2014/09/02/new sample.pdf";
String[] urlArray = url.split("/");
String lastPath = urlArray[urlArray.length-1];
This converts your url into an Array which can then be used in many ways. There are various ways to get url-lastPath, one way could be to join the above generated Array using this answer. Or use lastIndexOf() and substring like this:-
String restOfUrl = url.substring(0,url.lastIndexOf("/"));
PS:- Although you can learn something by doing this but I think your best solution would be to replace space by %20 in the complete url String, that would be the fastest and make more sense.

I am not sure if I understood it correctly but when you say
I have this file URL: URL/new sample.pdf which will be converted to URL/new%20sample.pdf later.
It looks like you are trying to replace "space" with %20 in URL or said in simple words trying to take care of unwanted characters in URL. If that is what you need use pre-built
URLEncoder.encode(String url,String enc), You can us ÜTF-8 as encoding.
http://docs.oracle.com/javase/7/docs/api/java/net/URLEncoder.html
If you really need to split it, assuming that you interested in URL after http://, remove http:// and take store remaining URL in string variable called say remainingURL. then use
List myList = new ArrayList(Arrays.asList(remainingURL.split("/")));
You can iterate on myList to get rest of URL fragments.

I've found it:
File file=new File("http://xxx.xxx.xx.xx/resources/upload/2014/09/02/new sample.pdf");
System.out.println(file.getPath().replaceAll(file.getName(),""));
Output:
http://xxx.xxx.xx.xx/resources/upload/2014/09/02/

Spring solution:
List<String> pathSegments = UriComponentsBuilder.fromUriString(url).build().getPathSegments();
String lastPath = pathSegments.get(pathSegments.size()-1);

How to change the width and height of an html file using java

I wanted to change width="xyz" , where (xyz) can be any particular value to width="300". I researched on regular expressions and this was the one I am using a syntax with regular expression
String holder = "width=\"340\"";
String replacer="width=\"[0-9]*\"";
theWeb.replaceAll(replacer,holder);
where theWeb is the string
. But this was not getting replaced. Any help would be appreciated.

Your regex is correct. One thing you might be forgetting is that in Java all string methods do not affect the current string - they only return a new string with the appropriate transformation. Try this instead:
String replacement = 'width="340"';
String regex = 'width="[0-9]*"';
String newWeb = theWeb.replaceAll(regex, replacement); // newWeb holds new text

Better use JSoup for manipulating and extracting data, etc. from Html
See this link for more details:
http://jsoup.org/

Search and replace "/" at end of url's using regular expressions in java

Below is my regular expression :-
\\bhttps?://[-a-zA-Z0-9+&##/%?=~_|!:,.;]*[-a-zA-Z0-9+&##/%=~_|]\\b
when the request url is of type http://www.example.com/ , the last character is not replaced in my shortner url and / is appended at end.
The regex is not able to find the last /.
Please help with this.

I think that / would be a word boundary, so maybe it works better if you add a ? to the and, so it reads:
\\bhttps?://[-a-zA-Z0-9+&##/%?=~_|!:,.;]*[-a-zA-Z0-9+&##/%=~_|]\\b?

what about:
if(url.endsWith("/"))
url = url.substring(0,url.length()-1);
or if you need to use regular expressions you can do something like this:
url = url.replaceAll("(\\bhttps?://[-a-zA-Z0-9+&##/%?=~_|!:,.;]*)/(\\b?)","$1$2");

If all you want is to replace the trailing / (which is what your question directly asks), you can simply do:
url = url.substring(0, url.lastIndexOf('/'));
Remember to KISS often.

You could simply use:
url = url.replaceAll("\/+$","");

get the request parameter from url with & symbol

I am trying to get the request parameter which has '&' sybol in starting like:-
http://localhost:8080/simple.jsp?my=&234587
On other page I'm getting it like String value=request.getParameter("my");
value.substring(0,4);
I want to get &234, please suggest i am not getting any value.
Thanks,
Ars

In this example you have not one, but two parameters:
my
234
234 is not a value here. The & separates query parameters. If you need that ampersand to be part of the value of my, it needs to be escaped in the URL as %26.

thanks for resonse i found the answer, I used request.getQueryString(); to get the whole string i.e. &234587 and then parsed it accordingly.
:)
Thanks once again.

You will most likely have to encode the '&' in the url:
http://localhost:8080/simple.jsp?my=%26234587

If you will be handling only Get request. You can write something like this.
String requestURI=request.getRequestURI();
String pContent=requestURI.split("?")[1];
String[] pArray= pContent.split("&");
String value="";
for (int i=0 ;i<pArray.length;i++) {
if(pArray[i].equals("my"+"=")){
value="&" + pArray[i+1];
break;
}
}