I have a string as : "This is a URL http://www.google.com/MyDoc.pdf which should be used"
I just need to extract the URL that is starting from http and ending at pdf :
http://www.google.com/MyDoc.pdf
String sLeftDelimiter = "http://";
String[] tempURL = sValueFromAddAtt.split(sLeftDelimiter );
String sRequiredURL = sLeftDelimiter + tempURL[1];
This gives me the output as "http://www.google.com/MyDoc.pdf which should be used"
Need help on this.
This kind of problem is what regular expressions were made for:
Pattern findUrl = Pattern.compile("\\bhttp.*?\\.pdf\\b");
Matcher matcher = findUrl.matcher("This is a URL http://www.google.com/MyDoc.pdf which should be used");
while (matcher.find()) {
System.out.println(matcher.group());
}
The regular expression explained:
\b before the "http" there is a word boundary (i.e. xhttp does not match)
http the string "http" (be aware that this also matches "https" and "httpsomething")
.*? any character (.) any number of times (*), but try to use the least amount of characters (?)
\.pdf the literal string ".pdf"
\b after the ".pdf" there is a word boundary (i.e. .pdfoo does not match)
If you would like to match only http and https, try to use this instead of http in your string:
https?\: - this matches the string http, then an optional "s" (indicated by the ? after the s) and then a colon.
why don't you use startsWith("http://") and endsWith(".pdf") mthods of String class.
Both the method returns boolean value, if both returns true, then your condition succeed else your condition is failed.
Try this
String StringName="This is a URL http://www.google.com/MyDoc.pdf which should be used";
StringName=StringName.substring(StringName.indexOf("http:"),StringName.indexOf("which"));
You can use Regular Expression power for here.
First you have to find Url in original string then remove other part.
Following code shows my suggestion:
String regex = "\\b(http|ftp|file)://[-a-zA-Z0-9+&##/%?=~_|!:,.;]*[-a-zA-Z0-9+&##/%=~_|]";
String str = "This is a URL http://www.google.com/MyDoc.pdf which should be used";
String[] splited = str.split(regex);
for(String current_part : splited)
{
str = str.replace(current_part, "");
}
System.out.println(str);
This snippet code cans retrieve any url in any string with any pattern.
You cant add customize protocol such as https to protocol part in above regular expression.
I hope my answer help you ;)
public static String getStringBetweenStrings(String aString, String aPattern1, String aPattern2) {
String ret = null;
int pos1,pos2;
pos1 = aString.indexOf(aPattern1) + aPattern1.length();
pos2 = aString.indexOf(aPattern2);
if ((pos1>0) && (pos2>0) && (pos2 > pos1)) {
return aString.substring(pos1, pos2);
}
return ret;
}
You can use String.replaceAll with a capturing group and back reference for a very concise solution:
String input = "This is a URL http://www.google.com/MyDoc.pdf which should be used";
System.out.println(input.replaceAll(".*(http.*?\\.pdf).*", "$1"));
Here's a breakdown for the regex: https://regexr.com/3qmus
Related
I am trying to split a string having multi-delimiters in it but want to first check if the string satisfies the regex and then split based on it.
Example:-
The testString will contain ://,:,#,:,/ these characters in specific order and I need to first check if the given string satisfies the pattern or not and if satisfy then split it. The string other characters can also contain these in it but I need to split based on order of these ://,:,#,:,/
String testString = "aman://jaspreet:raman!#127.0.0.1:5031/test";
String[]tokens = testString.split("://|\\:|#|\\:|\\/");
for(String s:tokens) {
System.out.println(s);
}
Here above I have tried the regex to split but it doesn't split by checking in order. It just checks if any given regex character exists in string or not and then splits based on it.
If you first validate the pattern, then you shouldn't do split() after. Use capturing groups to gather the data you already validated.
E.g. in a simple case, foo#bar, with separator #, you would validate with ^([^#]+)#(.+)$, i.e. match and capture text up to #, match but don't capture the #, then match and capture the rest:
Pattern p = Pattern.compile("^([^#]+)#(.+)$");
Matcher m = p.matcher("foo#bar");
if (! m.matches()) {
// invalid data
} else {
String a = m.group(1); // a = "foo"
String b = m.group(2); // b = "bar"
// use a and b here
}
For the matching in the question, a lenient pattern could be:
^(.*?)://(.*?):(.*?)#(.*?):(.*?)/(.*)$
You would then use code above, but with:
String scheme = m.group(1); // "aman"
String user = m.group(2); // "jaspreet"
String password = m.group(3); // "raman!"
String host = m.group(4); // "127.0.0.1"
String port = m.group(5); // "5031"
String path = m.group(6); // "test"
For a stricter matching, replace .*? with a pattern that only matches allowed characters, e.g. [^:]+ if it cannot be empty and cannot contain colons.
Alternatively, you could just use the URI class to parse the URL string.
String testString = "aman://jaspreet:raman!#127.0.0.1:5031/test";
URI uri = URI.create(testString);
String scheme = uri.getScheme(); // "aman"
String userInfo = uri.getUserInfo(); // "jaspreet:raman!"
String host = uri.getHost(); // "127.0.0.1"
String port = uri.getPort(); // "5031"
String path = uri.getPath(); // "test"
I'm trying to extract CANseIqFMnf from the URL https://www.instagram.com/p/CANseIqFMnf/ using regex in Android studio. Please help me to get a regex expression eligible for Android Studio.
Here is the code for my method:
String url = "https://www.instagram.com/p/CANseIqFMnf/";
String REGEX = "/p\//";
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(url);
boolean match = matcher.matches();
if (match){
Log.e("success", "start = " + matcher.start() + " end = " + matcher.end() );
}else{
Log.e("failed", "failed");
}
But it gives me failed in return!
Method 1
You just need to use replaceAll method in String, no need to compile a pattern and complicate things:
String input = "https://www.instagram.com/p/CANseIqFMnf/";
String output = input.replaceAll("https://www.instagram.com/p/", "").replaceAll("/", "");
Log.v(TAG, output);
Note that the first replaceAll is to remove the url and the second replaceAll is to remove any slashes /
Method 2
Pattern pattern = Pattern.compile("https://www.instagram.com/p/(.*?)/");
Matcher matcher = pattern.matcher("https://www.instagram.com/p/CANseIqFMnf/");
while(matcher.find()) {
System.out.println(matcher.group(1));
}
Note that if matcher.find() returns true then if you used modifiers like this in your REGEX (.*?) then the part found there will be in group(1), and group(0) will hold the entire regex match which is in your case the entire url.
Alternate option w/o regex can be implemented in a simpler manner as below using java.nio.file.Paths APIs
public class Url {
public static void main(String[] args) {
String url = "https://www.instagram.com/p/CANseIqFMnf/";
String name = java.nio.file.Paths.get(url).getFileName().toString();
System.out.println(name);
}
}
I have the follow urls.
https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258
https://docs.google.com/a/example.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY/edit#gid=1842172258
https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY
Foreach url, I need to extract the sheet id: 1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY into a java String.
I am thinking of using split but it can't work with all test cases:
String string = "https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258";
String[] parts = string.split("/");
String res = parts[parts.length-2];
Log.d("hello res",res );
How can I that be possible?
You can use regex \/d\/(.*?)(\/|$) (regex demo) to solve your problem, if you look closer you can see that the ID exist between d/ and / or end of line for that you can get every thing between this, check this code demo :
String[] urls = new String[]{
"https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258",
"https://docs.google.com/a/example.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY/edit#gid=1842172258",
"https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY"
};
String regex = "\\/d\\/(.*?)(\\/|$)";
Pattern pattern = Pattern.compile(regex);
for (String url : urls) {
Matcher matcher = pattern.matcher(url);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
}
Outputs
1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY
1mrsetjgfZI2BIypz7SGHMOfHGv6PTKTzY0xOM5c6TXY
1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY
it looks like the id you are looking for always follow "/spreadsheets/d/" if it is the case you can update your code to that
String string = "https://docs.google.com/spreadsheets/d/1mrsetjgfZI2BIypz7SGHMOfHGv6kTKTzY0xOM5c6TXY/edit#gid=1842172258";
String[] parts = string.split("spreadsheets/d/");
String result;
if(parts[1].contains("/")){
String[] parts2 = parts[1].split("/");
result = parts2[0];
}
else{
result=parts[1];
}
System.out.println("hello "+ result);
Using regex
Pattern pattern = Pattern.compile("(?<=\\/d\\/)[^\\/]*");
Matcher matcher = pattern.matcher(url);
System.out.println(matcher.group(1));
Using Java
String result = url.substring(url.indexOf("/d/") + 3);
int slash = result.indexOf("/");
result = slash == -1 ? result
: result.substring(0, slash);
System.out.println(result);
Google use fixed lenght characters for its IDs, in your case they are 44 characters and these are the characters google use: alphanumeric, -, and _ so you can use this regex:
regex = "([\w-]){44}"
match = re.search(regex,url)
I am getting a piece of JSON text from a url connection and saving it to a string currently as such:
...//setting up url and connection
BufferedReader in = new BufferedReader(new InputStreamReader(connection.getInputStream()));
String str = in.readLine();
When I print str, I correctly find the data {"build":{"version_component":"1.0.111"}}
Now I want to extract the 111 from str, but I am having some trouble.
I tried
String afterLastDot = inputLine.substring(inputLine.lastIndexOf(".") + 1);
but I end up with 111"}}
I need a solution that is generic so that if I have String str = {"build":{"version_component":"1.0.111111111"}}; the solution still works and extracts 111111111 (ie, I don't want to hard code extract the last three digits after the decimal point)
If you cannot use a JSON parser then you can this regex based extraction:
String lastNum = str.replaceAll("^.*\\.(\\d+).*", "$1");
RegEx Demo
^.* is greedy match that matches everything until last DOT and 1 or more digits that we put in group #1 to be used in replacement.
Find the start and the end indexes of the String you need and substring(start, end) :
// String str = "{"build":{"version_component":"1.0.111"}};" cannot compile without escaping
String str = "{\"build\":{\"version_component\":\"1.0.111\"}}";
int start = str.lastIndexOf(".")+1;
int end = str.lastIndexOf("\"");
String substring = str.substring(start,end);
just use JSON api
JSONObject obj = new JSONObject(str);
String versionComponent= obj.getJSONObject("build").getString("version_component");
Then just split and take the last element
versionComponent.split("\\.")[2];
Please, your can try the following code :
...
int index = inputLine.lastIndexOf(".")+1 ;
String afterLastDot = inputLine.substring(index, index+3);
With Regular Expressions (Rexp),
You can solve your problem like this ;
Pattern pattern = Pattern.compile("111") ;
Matcher matcher = pattern.matcher(str) ;
while(matcher.find()){
System.out.println(matcher.start()+" "+matcher.end());
System.out.println(str.substring(matcher.start(), matcher.end()));
}
I have a string line
String user_name = "id=123 user=aron name=aron app=application";
and I have a list that contains: {user,cuser,suser}
And i have to get the user part from string. So i have code like this
List<String> userName = Config.getConfig().getList(Configuration.ATT_CEF_USER_NAME);
String result = null;
for (String param: user_name .split("\\s", 0)){
for(String user: userName ){
String userParam = user.concat("=.*");
if (param.matches(userParam )) {
result = param.split("=")[1];
}
}
}
But the problem is that if the String contains spaces in the user_name, It do not work.
For ex:
String user_name = "id=123 user=aron nicols name=aron app=application";
Here user has a value aron nicols which contain spaces. How can I write a code that can get me exact user value i.e. aron nicols
If you want to split only on spaces that are right before tokens which have = righ after it such as user=... then maybe add look ahead condition like
split("\\s(?=\\S*=)")
This regex will split on
\\s space
(?=\\S*=) which has zero or more * non-space \\S characters which ends with = after it. Also look-ahead (?=...) is zero-length match which means part matched by it will not be included in in result so split will not split on it.
Demo:
String user_name = "id=123 user=aron nicols name=aron app=application";
for (String s : user_name.split("\\s(?=\\S*=)"))
System.out.println(s);
output:
id=123
user=aron nicols
name=aron
app=application
From your comment in other answer it seems that = which are escaped with \ shouldn't be treated as separator between key=value but as part of value. In that case you can just add negative-look-behind mechanism to see if before = is no \, so (?<!\\\\) right before will require = to not have \ before it.
BTW to create regex which will match \ we need to write it as \\ but in Java we also need to escape each of \ to create \ literal in String that is why we ended up with \\\\.
So you can use
split("\\s(?=\\S*(?<!\\\\)=)")
Demo:
String user_name = "user=Dist\\=Name1, xyz src=activedirectorydomain ip=10.1.77.24";
for (String s : user_name.split("\\s(?=\\S*(?<!\\\\)=)"))
System.out.println(s);
output:
user=Dist\=Name1, xyz
src=activedirectorydomain
ip=10.1.77.24
Do it like this:
First split input string using this regex:
" +(?=\\w+(?<!\\\\)=)"
This will give you 4 name=value tokens like this:
id=123
user=aron nicols
name=aron
app=application
Now you can just split on = to get your name and value parts.
Regex Demo
Regex Demo with escaped =
CODE FISH, this simple regex captures the user in Group 1: user=\\s*(.*?)\s+name=
It will capture "Aron", "Aron Nichols", "Aron Nichols The Benevolent", and so on.
It relies on the knowledge that name= always follows user=
However, if you're not sure that the token following user is name, you can use this:
user=\s*(.*?)(?=$|\s+\w+=)
Here is how to use the second expression (for the first, just change the string in Pattern.compile:
String ResultString = null;
try {
Pattern regex = Pattern.compile("user=\\s*(.*?)(?=$|\\s+\\w+=)", Pattern.CASE_INSENSITIVE | Pattern.UNICODE_CASE);
Matcher regexMatcher = regex.matcher(subjectString);
if (regexMatcher.find()) {
ResultString = regexMatcher.group(1);
}
} catch (PatternSyntaxException ex) {
// Syntax error in the regular expression
}