From the string I need received result like below (using regex).
string: [Raxy] GTS_R-1/GTS_CNF-1/NA-15/SDA-1/MGTS-8002/PQM-1/APM-1/RF-3/NTL-15
result: [Raxy] MGTS-8002/PQM-1/APM-1
I know how to get:
"[Raxy] " -> \[.+\][\s?]
"MGTS-8002/PQM-1/APM-1" -> MGTS-\d+/PQM-\d+/PQM-\d+/APM-\d+
but I have a problem how to combine them.
Use
/^(\[[^\]\[]+]).*\/(MGTS-\d+\/PQM-\d+\/APM-\d+)/
See proof. Extract captures from the match, and combine into a single string.
Here is a brief demo:
const string = "[Raxy] GTS_R-1/GTS_CNF-1/NA-15/SDA-1/MGTS-8002/PQM-1/APM-1/RF-3/NTL-15";
const match = /^(\[[^\]\[]+]).*\/(MGTS-\d+\/PQM-\d+\/APM-\d+)/.exec(string);
console.log(match[1] + " " + match[2]);
Related
I have strings in this format:
object[i].base.base_x[i] and I get lists like List(0,1).
I want to use regular expressions in scala to find the match [i] in the given string and replace the first occurance with 0 and the second with 1. Hence getting something like object[0].base.base_x[1].
I have the following code:
val stringWithoutIndex = "object[i].base.base_x[i]" // basically this string is generated dynamically
val indexReplacePattern = raw"\[i\]".r
val indexValues = List(0,1) // list generated dynamically
if(indexValues.nonEmpty){
indexValues.map(row => {
indexReplacePattern.replaceFirstIn(stringWithoutIndex , "[" + row + "]")
})
else stringWithoutIndex
Since String is immutable, I cannot update stringWithoutIndex resulting into an output like List("object[0].base.base_x[i]", "object[1].base.base_x[i]").
I tried looking into StringBuilder but I am not sure how to update it. Also, is there a better way to do this? Suggestions other than regex are also welcome.
You couldloop through the integers in indexValues using foldLeft and pass the string stringWithoutIndex as the start value.
Then use replaceFirst to replace the first match with the current value of indexValues.
If you want to use a regex, you might use a positive lookahead (?=]) and a positive lookbehind (?<=\[) to assert the i is between opening and square brackets.
(?<=\[)i(?=])
For example:
val strRegex = """(?<=\[)i(?=])"""
val res = indexValues.foldLeft(stringWithoutIndex) { (s, row) =>
s.replaceFirst(strRegex, row.toString)
}
See the regex demo | Scala demo
How about this:
scala> val str = "object[i].base.base_x[i]"
str: String = object[i].base.base_x[i]
scala> str.replace('i', '0').replace("base_x[0]", "base_x[1]")
res0: String = object[0].base.base_x[1]
This sounds like a job for foldLeft. No need for the if (indexValues.nonEmpty) check.
indexValues.foldLeft(stringWithoutIndex) { (s, row) =>
indexReplacePattern.replaceFirstIn(s, "[" + row + "]")
}
I want to split the following string around +, but I couldn't succeed in getting the correct regex for this.
String input = "SOP3a'+bEOP3'+SOP3b'+aEOP3'";
I want to have a result like this
[SOP3a'+bEOP3', SOP3b'+aEOP3']
In some cases I may have the following string
c+SOP2SOP3a'+bEOP3'+SOP3b'+aEOP3'EOP2
which should be split as
[c, SOP2SOP3a'+bEOP3'+SOP3b'+aEOP3'EOP2]
I have tried the following regex but it doesn't work.
input.split("(SOP[0-9](.*)EOP[0-9])*\\+((SOP)[0-9](.*)(EOP)[0-9])*");
Any help or suggestions are appreciated.
Thanks
You can use the following regex to match the string and by replacing it using captured group you can get the expected result :
(?m)(.*?)\+(SOP.*?$)
see demo / explanation
Following is the code in Java that would work for you:
public static void main(String[] args) {
String input = "SOP3a'+bEOP3'+SOP3b'+aEOP3'";
String pattern = "(?m)(.*?)\\+(SOP.*?$)";
Pattern regex = Pattern.compile(pattern);
Matcher m = regex.matcher(input);
if (m.find()) {
System.out.println("Found value: " + m.group(0));
System.out.println("Found value: " + m.group(1));
System.out.println("Found value: " + m.group(2));
} else {
System.out.println("NO MATCH");
}
}
The m.group(1) and m.group(2) are the values that you are looking for.
Do you really need to use split method?
And what are the rules? They are unclear to me.
Anyway, considering the regex you provided, I've only removed some unnecessary groups and I've found what you are looking for, however, instead of split, I just joined the matches as splitting it would generate some empty elements.
const str = "SOP1a+bEOP1+SOP2SOP3a'+bEOP3'+SOP3b'+aEOP3'EOP2";
const regex = RegExp(/(SOP[0-9].*EOP[0-9])*\+(SOP[0-9].*EOP[0-9])*/)
const matches = str.match(regex);
console.log('Matches ', matches);
console.log([matches[1],matches[2]]);
I want to replace all the consecutive underscores with a single space. This is the code that I have written. But it is not replacing anything. Below is the code that I have written. What am I doing wrong?
import scala.util.matching.Regex
val regex: Regex = new Regex("/[\\W_]+/g")
val name: String = "cust_id"
val newName: String = regex.replaceAllIn(name, " ")
println(newName)
Answer: "cust_id"
You could use replaceAll to do the job without regex :
val name: String = "cust_id"
val newName: String = name.replaceAll("_"," ")
println(newName)
The slashes in your regular expression don't belong there.
new Regex("[\\W_]+", "g").replaceAllIn("cust_id", " ")
// "cust id"
A string in Scala may be treated as a collection, hence we can map over it and in this case apply pattern matching to substitute characters, like this
"cust_id".map {
case '_' => " "
case c => c
}.mkString
Method mkString glues up the vector of characters back onto a string.
I'm trying to replace a bunch of function calls using regular expressions but can't seem to be getting it right. This is a simplified example of what I'm trying to do:
GetPetDog();
GetPetCat();
GetPetBird();
I want to change to:
GetPet<Animal_Dog>();
GetPet<Animal_Cat>();
GetPet<Animal_Bird>();
Use below regex:
(GetPet)([^(]*) with subsitution \1<Animal_\2>
Demo
You can use the following regex and code for that:
std::string ss ("GetPetDog();");
static const std::regex ee ("GetPet([^()]*)");
std::string result;
result = regex_replace(ss, ee, "GetPet<Animal_$1>");
std::cout << result << endl;
Regex:
GetPet - Matches GetPet literally (we need no capturing group here)
([^()]*) - A capturing group to match any characters other than ( or ) 0 or more times (*)
Output:
I have a string like '[1]-[2]-[3],[4]-[5],[6,7,8],[9]' or '[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]', I'd like the Pattern to get the list result, but don't know how to figure out the pattern. Basically the comma is the split, but [6,7,8] itself contains the comma as well.
the string: [1]-[2]-[3],[4]-[5],[6,7,8],[9]
the result:
[1]-[2]-[3]
[4]-[5]
[6,7,8]
[9]
or
the string: [Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]
the result:
[Computers]-[Apple]-[Laptop]
[Cables]-[Cables,Connectors]
[Adapters]
,(?=\[)
This pattern splits on any comma that is followed by a bracket, but keeps the bracket within the result text.
The (?=*stuff*) is known as a "lookahead assertion". It acts as a condition for the match but is not itself part of the match.
In C# code:
String inputstring = "[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]";
foreach(String s in Regex.Split(inputstring, #",(?=\[)"))
System.Console.Out.WriteLine(s);
In Java code:
String inputstring = "[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]";
Pattern p = Pattern.compile(",(?=\\[)"));
for(String s : p.split(inputstring))
System.out.println(s);
Either produces:
[Computers]-[Apple]-[Laptop]
[Cables]-[Cables,Connectors]
[Adapters]
Although I believe the best approach here is to use split (as presented by #j__m's answer), here's an approach that uses matching rather than splitting.
Regex:
(\[.*?\](?!-))
Example usage:
String input = "[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]";
Pattern p = Pattern.compile("(\\[.*?\\](?!-))");
Matcher m = p.matcher(input);
while (m.find()) {
System.out.println(m.group(1));
}
Resulting output:
[Computers]-[Apple]-[Laptop]
[Cables]-[Cables,Connectors]
[Adapters]
An answer that doesn't use regular expressions (if that's worth something in ease of understanding what's going on) is:
substitute "]#[" for "],["
split on "#"