What's the best way to replace a path url placeholder. I have the following that needs to be replaced
/user/:name/password/:password
as
/user/{name}/password/{password}
Is there a library that could do this for me in Java?
Since the format is so simple, and : isn't a valid URL character, I would just use a basic regex matching : followed by any word, capturing the word for reprint.
"/user/:name/password/:password".replaceAll(":(\\w+)","{$1}")
Just using String#replaceAll can achieve your way.
"/user/:name/password/:password".replaceAll(":(\\w+)","{$1}")
Did you try to use replaceAll like this :
String str = "/user/:name/password/:password";
String result = str.replaceAll(":(\\w+)", "{$1}");
Output
/user/{name}/password/{password}
Related
Now I know about FilenameUtils.getExtension() from apache.
But in my case I'm processing extensions from http(s) urls, so in case I have something like
https://your_url/logo.svg?position=5
this method is gonna return svg?position=5
Is there the best way to handle this situation? I mean without writing this logic by myself.
You can use the URL library from JAVA. It has a lot of utility in this cases. You should do something like this:
String url = "https://your_url/logo.svg?position=5";
URL fileIneed = new URL(url);
Then, you have a lot of getter methods for the "fileIneed" variable. In your case the "getPath()" will retrieve this:
fileIneed.getPath() ---> "/logo.svg"
And then use the Apache library that you are using, and you will have the "svg" String.
FilenameUtils.getExtension(fileIneed.getPath()) ---> "svg"
JAVA URL library docs >>>
https://docs.oracle.com/javase/7/docs/api/java/net/URL.html
If you want a brandname® solution, then consider using the Apache method after stripping off the query string, if it exists:
String url = "https://your_url/logo.svg?position=5";
url = url.replaceAll("\\?.*$", "");
String ext = FilenameUtils.getExtension(url);
System.out.println(ext);
If you want a one-liner which does not even require an external library, then consider this option using String#replaceAll:
String url = "https://your_url/logo.svg?position=5";
String ext = url.replaceAll(".*/[^.]+\\.([^?]+)\\??.*", "$1");
System.out.println(ext);
svg
Here is an explanation of the regex pattern used above:
.*/ match everything up to, and including, the LAST path separator
[^.]+ then match any number of non dots, i.e. match the filename
\. match a dot
([^?]+) match AND capture any non ? character, which is the extension
\??.* match an optional ? followed by the rest of the query string, if present
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_");
I have a 2 column tsv file with column one having
1-FN3Z1-206329557431
1-FN411-153115736976
Where I am trying to remove the first two parts of the value (i.e to extract 206329557431 and 153115736976). I've used online regex tool to generate the patterns
pattern
".*?\\d+.*?\\d+.*?(\\d+)" AND ".*?\\d+.*?\\d+.*?\\d+.*?(\\d+)"
Independently they work fine. I'm trying to look for a combined regex pattern. Any pointers as to how this can be done.
Why don't use split for example :
String spl = "1-FN411-153115736976".split("-")[2];
If you want a regex you can use (.*?-){2}(.*), which mean get everything after the second -
regex demo
Output
206329557431
153115736976
If the strings in your TSV file all have the same widths and patterns, then you can just use substring here:
String tsv = "1-FN3Z1-206329557431";
System.out.println(tsv.substring(8));
Demo
How about this regexp : .-.{5}- looks like it can matches all statements but it depends on your format.
Here is Java code example :
#Test
public void test() {
String test = "1-FN3Z1-206329557431 1-FN411-153115736976";
String result = test.replaceAll(".-.{5}-", "");
assertEquals("206329557431 153115736976", result);
}
what is the regular expression to find the path param from the url?
http://localhost:8080/domain/v1/809pA8
https://localhost:8080/domain/v1/809pA8
Want to retrieve the value(809pA8) from the above URL using regular expression, java is preferable.
I would suggest you do something like
url.substring(url.lastIndexOf('/') + 1);
If you really prefer regexps, you could do
Matcher m = Pattern.compile("/([^/]+)$").matcher(url);
if (m.find())
value = m.group(1);
I would try:
String url = "http://localhost:8080/domain/v1/809pA8";
String value = String.valueOf(url.subSequence(url.lastIndexOf('/'), url.length()-1));
No need for regex here, I think.
EDIT: I'm sorry I made a mistake:
String url = "http://localhost:8080/domain/v1/809pA8";
String value = String.valueOf(url.subSequence(url.lastIndexOf('/')+1, url.length()));
See this code working here: https://ideone.com/E30ddC
For your simple case, regex is an overkill, as others noted. But, if you have more cases and this is why you prefer regex, give Spring's AntPathMatcher#extractUriTemplateVariables a look, if you're using Spring. It's actually better equipped for extracting path variables than regex directly. Here are some good examples.
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$2");
If all you want is to replace the trailing / (which is what your question directly asks), you can simply do:
url = url.substring(0, url.lastIndexOf('/'));
Remember to KISS often.
You could simply use:
url = url.replaceAll("\/+$","");