I'm trying to remove all instances of \n, from a string in javascript. Does anyone know the best way to do this?
I tried
mystring = mystring.replace(/\n/g,'');
Did you read MDN?
mystring = mystring.replace(/\n/g, '');
PS. You don't need to put i flag which stands for case insensitive, since there is just one line break sign :)
If you mean you need to get rid of \n\ literally then your regexp should be /\\n/g to remove the string \n.
You need to remove the " from your regex expression. You also do not need the "case insensitive" modifier, but just global (g). Use this instead.
mystring = mystring.replace(/\n/g, '')
And to be very verbose, you can replace all vertical whitespace characters with the following.
mystring = mystring.replace(/[\n\r\v]/g, '')
Related
For a test I need a string literal with trailing whitespace, like this (dots are spaces).
let expected = "
foo....a
.......b
.......
.......c
.......d
";
Unfortunately editors commonly strip trailing whitespace. Good in general, so I don't want to require people to disable it, but it breaks this code.
Is there a nice way to specify this string in such a way that it has the same value but doesn't have trailing whitespace in the code? By nice I mean it doesn't look too different from the existing code.
This is the best I can come up with:
let expected = "
foo....a
.......b
.......\
\n.......c
.......d
";
It's not great but I can't think of anything better. You could use something like format!(".....{s}", s="...") but that seems worse.
Thanks to Caesar this is a nicer solution!
let expected = "
foo....a
.......b
......\x20
.......c
.......d
";
You could do exactly what you used here on stackoverflow,
and just use dots:
let expected = "
foo....a
.......b
.......
.......c
.......d
".replace(".", " ");
If this weren't test code, and you really needed the string to be constant, you could use the lazy_static crate.
Does anyone know a method that allows you to replace all the characters in a word with a single character?
If not, can anyone suggest a way to basically print _ (underscore) the number of times which is the length of the string itself without using any loops or ifs in the code?
mystring = '_'*len(mystring)
Of course, I'm guessing at the name of your string variable and the character that you want to use.
Or, if you just want to print it out, you can:
print('_'*len(mystring))
import re
str = "abcdefghi"
print(re.sub('[a-z]','_',str))
I'm having issues when trying to remove the first space of a string if that string has 2 spaces in it. For example it should be turning "Fully Functional Method" into "FullyFunctional Method", but "Functional Method" should not be changed because it only has 1 space. I can't really think of a way to remove first space if the string contains 2 spaces.
I don't know exactly what you want to do, but you may search into RegExp and String.replace() to replace some stuff in a String.
Here is another link to understand the Characters, metacharacters, and metasequences.
var myPattern1:RegExp = / /g;
var str1:String = "This is a string that contains double spaces.";
trace(str1.replace(myPattern1, " "));
//this replaces all " " by " "...
//outputs : This is a string that contains double spaces.
Or in your case (I suppose) something like this
var myPattern2:RegExp = / /;
var str2:String = "Fully Functional Method";
trace(str2.replace(myPattern2, ""));
//If you omit the g, only the first space will be replaced by ""
//outputs : FullyFunctional Method
There is so much things you can do by using RegExp, that I will not explain this here...
Just check on the Adobe website...
This is a quick and efficient way to work on Strings.
I hope this will help.
Since you check at those links, you will understand that my example is pure rough and should be modified to have a FullyFunctional Method. :D
Do a linear scan through the string. Count the number of spaces and record the index of the first space, if any. If there are two spaces, return a string that is the concatenation of the characters up to but not including the first space, and the characters after the first space.
Keep it simple. It is possible to solve your problem with regex, but keep in mind that the worst case time complexity of finding a particular character in an unsorted set is always going to be O(N), so it won't be faster.
I am looking to remove special characters from a string using groovy, i'm nearly there but it is removing the white spaces that are already in place which I want to keep. I only want to remove the special characters (and not leave a whitespace). I am running the below on a PostCode L&65$$ OBH
def removespecialpostcodce = PostCode.replaceAll("[^a-zA-Z0-9]+","")
log.info removespecialpostcodce
Currently it returns L65OBH but I am looking for it to return L65 OBH
Can anyone help?
Use below code :
PostCode.replaceAll("[^a-zA-Z0-9 ]+","")
instead of
PostCode.replaceAll("[^a-zA-Z0-9]+","")
To remove all special characters in a String you can use the invert regex character:
String str = "..\\.-._./-^+* ".replaceAll("[^A-Za-z0-1]","");
System.out.println("str: <"+str+">");
output:
str: <>
to keep the spaces in the text add a space in the character set
String str = "..\\.-._./-^+* ".replaceAll("[^A-Za-z0-1 ]","");
System.out.println("str: <"+str+">");
output:
str: < >
So say I have a string with some underscores like hi_there.
Is there a way to auto-convert that string into "hi there"?
(the original string, by the way, is a variable name that I'm converting into a plot title).
Surprising that no-one has yet mentioned strrep:
>> strrep('string_with_underscores', '_', ' ')
ans =
string with underscores
which should be the official way to do a simple string replacements. For such a simple case, regexprep is overkill: yes, they are Swiss-knifes that can do everything possible, but they come with a long manual. String indexing shown by AndreasH only works for replacing single characters, it cannot do this:
>> s = 'string*-*with*-*funny*-*separators';
>> strrep(s, '*-*', ' ')
ans =
string with funny separators
>> s(s=='*-*') = ' '
Error using ==
Matrix dimensions must agree.
As a bonus, it also works for cell-arrays with strings:
>> strrep({'This_is_a','cell_array_with','strings_with','underscores'},'_',' ')
ans =
'This is a' 'cell array with' 'strings with' 'underscores'
Try this Matlab code for a string variable 's'
s(s=='_') = ' ';
If you ever have to do anything more complicated, say doing a replacement of multiple variable length strings,
s(s == '_') = ' ' will be a huge pain. If your replacement needs ever get more complicated consider using regexprep:
>> regexprep({'hi_there', 'hey_there'}, '_', ' ')
ans =
'hi there' 'hey there'
That being said, in your case #AndreasH.'s solution is the most appropriate and regexprep is overkill.
A more interesting question is why you are passing variables around as strings?
regexprep() may be what you're looking for and is a handy function in general.
regexprep('hi_there','_',' ')
Will take the first argument string, and replace instances of the second argument with the third. In this case it replaces all underscores with a space.
In Matlab strings are vectors, so performing simple string manipulations can be achieved using standard operators e.g. replacing _ with whitespace.
text = 'variable_name';
text(text=='_') = ' '; //replace all occurrences of underscore with whitespace
=> text = variable name
I know this was already answered, however, in my case I was looking for a way to correct plot titles so that I could include a filename (which could have underscores). So, I wanted to print them with the underscores NOT displaying with as subscripts. So, using this great info above, and rather than a space, I escaped the subscript in the substitution.
For example:
% Have the user select a file:
[infile inpath]=uigetfile('*.txt','Get some text file');
figure
% this is a problem for filenames with underscores
title(infile)
% this correctly displays filenames with underscores
title(strrep(infile,'_','\_'))