This question already has answers here:
Java replace all square brackets in a string
(8 answers)
Closed 4 years ago.
I am having string as below and want prefix to be removed [*TESTABC*]
String a = "[*TESTABC*]test#test.com";
Expected result: test#test.com
I tried below code but it is removing the bracklets only. I need to remove the inner content also.
The string with come in this pattern only. Please help.
String s = "[*TESTABC*]test#test.com";
String regex = "\\[|\\]";
s = s.replaceAll(regex, "");
System.out.println(s); //*TESTABC*test#test.com
int index;
String s = "[*TESTABC*]test#test.com";
index=s.indexOf(']');
s=s.substring(index+1);
System.out.println(s);
**i solve your question according to your problem and output **
Related
This question already has answers here:
How to find and replace all occurrences of a substring in a string?
(9 answers)
Replace substring with another substring C++
(18 answers)
How to find and replace string?
(11 answers)
How do I replace all instances of a string with another string?
(6 answers)
Closed 6 months ago.
I'm using C++ and I have a problem. Instead of creating a new line it prints \n. My Code:
std::string text;
std::cout << text;
It prints:Hello\nWorld
It was supposed to read \n as a new line and print something like this:
"Hello
World"
So i've tried to use replace(text.begin(), text.end(), '\n', 'a') for testing purposes and nothing happened. It contiuned to print Hello\nWorld
std::replace() won't work in this situation. When called on a std::string, it can replace only single characters, but in your case your string actually contains 2 distinct characters '\' and 'n'. So you need to use std::string::find() and std::string::replace() instead, eg:
string::size_type index = 0;
while ((index = str.find("\\n", index)) != string::npos) {
str.replace(index, 2, "\n");
++index;
}
This question already has answers here:
RegExp in TypeScript
(5 answers)
Closed 2 years ago.
I am new to Typescript and trying to make a webhook in my Google Cloud Functions.
I have a string: C1234567890A460450P10TS1596575969702
I want to use regex to extract the number 1234567890 from that string.
The first character C is fixed and does not change, the character A after the number is variable and can be any other alphabet.
The regex that matches the number is (?<=C)(\d{10})(?=\w).
I want to know how to execute this regex in Typescript so that I can get the number into a variable(eg: const number = [the number extracted from the string] //value 1234567890)
Edit 1:
Based on the provided suggestions (which I had tried already before posting this question), here is the code I could make out of it:
const string = request.body.string;
let regxp = new RegExp('(?<=C)(\d{10})(?=\w)');
const number = regxp.exec(string);
response.send(number);
This gives a blank response.
There is two problems, you never parsed the returned string to a number with parseInt and (?<=C) (positive lookbehind) is not always supported.
Second, your regular expression can be simplified into ^C\d{10} and a .splice(1) to remove the C.
const string: string = request.body.string;
const matches = s.match(/^C\d{10}/);
let number: number;
if(matches !== null) {
number = parseInt(matches[0].slice(1));
} else {
res.status(400).end(); // Assuming this is express
return;
}
res.send(number); // 1234567890
Playground
This question already has answers here:
Match text between two strings with regular expression
(3 answers)
Closed 4 years ago.
there are some Japanese sentences like following:
{keyword: 部屋}いいね!
{keyword: 公園}は綺麗です.
私は{keyword: 部屋捜査}です。
..........
.........
I want to replace the substring like :{keyword: 部屋},{keyword: 公園}..... with 'keyword'.
For example:
input: 私は{keyword: 部屋捜査}です
output: 私はkeywordです
My trying code is following and but it is wrong, the result is same:
import re
s = '{keyword: 賃貸}'
t = re.sub(r"\{keyword:[あ-んア-ン一-]+\}", 'keyword', s)
print(t)
Thanks!
Use the following:-
inputString = "私は{keyword: 部屋捜査}です"
t = re.sub(r"\{keyword:[^}]*}", 'keyword', inputString)
print(t)
This question already has answers here:
Replace part of a string with another string
(17 answers)
Closed 7 years ago.
I have a string such as:
AAAbbbbbAAA
I'd like to remove all the occurances of a pattern AAA to get:
bbbbb
The pattern can occur anywhere in the string.
Given:
string yourstring("AAAbbbAAA");
string removestring("AAA");
You could simply run something like this multiple times on your string:
yourstring.erase(yourstring.find(removestring), removestring.length());
Of course you will have to check that string::find actually finds an occurence before using string::erase.
Here is the code. It's not very much efficient, but works well, and is a tiny code.
string h = "AAAbbbAAAB";
int pos = h.find("AAA");
while(pos!=-1)
{
h.replace(pos, 3, "");
pos = h.find("AAA");
}
cout << h << endl;
It only works if you know the pattern. If it doesn't what you want, maybe you're looking for a pattern matching algorithm, like KMP.
This question already has answers here:
Get values between curly braces c#
(3 answers)
Closed 7 years ago.
how to get the value between first { and last } from a string which have multiple {}.
eg string: ".....[object:{ ..{...{..}...}..}]"
My approach using C#:
line="abcd..efg..[object:{ ab{..c{d.}.e.}f....g}]"
string p = ".*\\[Object:{([A-Za-z{}]*)}\\]";
Regex r = new Regex(p);
Match m=r.match(line);
string value=m.Groups[1].Value.ToString();
Result should be:
value= ab{..c{d.}.e.}f....g
{.*}
or
(?<={).*(?=})
This should do the trick for you.See demo
string strRegex = #"{.*}";
Regex myRegex = new Regex(strRegex, RegexOptions.Multiline);
string strTargetString = #".....[object:{ ..{...{..}...}..}]";
foreach (Match myMatch in myRegex.Matches(strTargetString))
{
if (myMatch.Success)
{
// Add your code here
}
}