Replace regex \\+n from string in swift - swift

i want to remove every string starts with one or more \ and follow with n.
For example:
input: {\n\n abc \\nb\\\ncc}
expect output: { abc bcc}
In javascript it works with regex /\\+n/g
But it doesn't work in swift:
str.replacingOccurrences(of: "\\+n", with: "", options: .regularExpression)

To remove all ns that contain one or more backslashes in front, and all newlines with any amount of backslashes in front, you can use
#"\\+n|\\*\n"#
Note the # before and after the double quotes mean that the literal is a raw string literal where backslashes are treated as literal backslashes and do not form string escape sequences, like \n, \t, \r, etc.
Here, the pattern means
\\+n - one or more backslashes followed with n
| - or
\\*\n - zero or more backslashes followed with a newline.
Note that here, \n is a regex escape matching a newline, it is not an LF, \x0A character.

Related

Sed replace strings starts with special characters

I'm trying to replace strings with sed in ; php_value[date.timezone] = Europe/Riga
i tried something like this:
sed -i 's/; php_value[date.timezone] = Europe/\Riga/; php_value[date.timezone] = America/\Sao_Paulo/g' file
Output:
sed: -e expression #1, char 47: extra characters after command
You can use
sed -i 's/; php_value\[date\.timezone] = Europe\/Riga/; php_value[date.timezone] = America\/Sao_Paulo/g' file
See the online demo.
NOTE:
[ and . are special regex metacharacters and need to be escaped to match literal [ and ., hence, \[ and \. in the regex part
/ is a regex delimiter char here, and should also be escaped. To escape /, use \/. Well, if you use another regex delimiter char, you will have no need escaping /, e.g.
sed -i 's,; php_value\[date\.timezone] = Europe/Riga,; php_value[date.timezone] = America/Sao_Paulo,g' file
See the commas as regex delimiters here.

replace regex value in powershell

From the below line:
[1;32mget[0m /dir/dir/dir {} [oiid:197,uiid:7522] [32m200[0m (676ms)
convert it to:
get /dir/dir/dir {} [oiid:197,uiid:7522] 200 (676ms)
any suggestions on regex please the value 200 can be any number and the value get can be any http method.
I have tried
$_.replace("[[\d+m]","").replace('[[1;\d+m]',"").replace('[[\d+]m]',"")
} | Set-Content $newfilepath```
You can do the following using -replace operator and a regex:
$string = '[1;32mget[0m /dir/dir/dir {} [oiid:197,uiid:7522] [32m200[0m (676ms)'
$string -replace '\[\d.*?m'
\[ matches [ and note that it needs to be escaped for literal match because [ is special to regex. \d is a digit. .*? matches as few characters as possible until m is matched.
The String class Replace() method does not support regex. So you cannot use regex expressions like \d inside.

Replace every " within string

I have lines in a text file which looks like this example:
"2009217",2015,3,"N","N","2","UPPER DARBY FIREFIGHTERS "PAC"","","","","7235 WEST CHESTER PIKE","","UPPER DARBY","PA","19082","","6106220269",4245.0100,650.0000,.0000
I want to replace every double quote in multiple partial strings similar to this "UPPER DARBY FIREFIGHTERS "PAC""across the whole file.
So the result should be as below for each instance of the recurring double quotes:
"2009217",2015,3,"N","N","2","UPPER DARBY FIREFIGHTERS PAC","","","","7235 WEST CHESTER PIKE","","UPPER DARBY","PA","19082","","6106220269",4245.0100,650.0000,.0000
I came to this sed line:
cat file.txt | sed "s/\([^,]*,[^,]*,[^,]*,[^,]*,[^,]*,[^,]*,\)\([^,]*\),\(.*\)/\1\2\3/"
But now I don't know how to replace the double quote within \2.
Is that possible with sed?
I would personally use awk for that because it is more readable:
#!/usr/bin/env awk
BEGIN {
# Use ',' as the input and output field delimiter
FS=OFS=","
}
{
# Iterate through all fields. (NF is the number of fields.)
for(i=1;i<=NF;i++) {
# If the field starts and ends with a '"'
if($i ~ /^".*"$/) {
# Replace all '""
gsub(/"/,"",$i)
# Wrap in '"' again
$i = "\"" $i "\""
}
}
}
print
This might work for you (GNU sed):
sed -r ':a;s/^((([^",]*,)*("[^",]*",([^",]*,)*)*)"[^",]*)"([^,])/\1\6/;ta' file
This removes extra double quotes from strings surrounded by double quotes and delimited by ,'s.
It does this by eliminating properly constructed double quotes strings and non-quoted strings (in this example numbers) and then removes double quotes that are not followed by ,
[^",]*, # non double quoted strings
"[^",]*", # properly quoted strings
(([^",]*,)*("[^",]*",([^",]*,)*)*) # eliminate all properly constructed strings
"[^",]*"([^,]) # improper double quotes
^
|

Perl, Split string by specific pattern

I found how to split a string by whitespaces, but that only takes into an account a single character. In my case, I have comments pasted into a file that includes newlines and whitespaces. I have them separated by this string: [|]
So I need to split my $string into an array for example, where $string =
This is a comment.
This is a newline.
This is the end[|]This is second comment.
This is second newline.
[|]Last comment
Gets split into $array[0], $array[1], and $array[2] which include the newlines and whitespaces. Separated by [|]
Every example I find on the web uses a single character, such as space or newline, to split strings. In my case I have to use a more specific identifier, which is why I selected [|] but having troubles splitting it by this.
I have tried to limit it to parse by a single '|' character with this code:
my #words = split /|/, $string;
foreach my $thisline (#words) {
print "This line = '" . $thisline . "'\n";
But this seems to split the entire string, character-by-character into #words.
[, |, and ] are all special characters in regular expressions -- | is used to separate options, and […] are used to specify character sets. Using an unquoted | makes the expression match the empty string (more specifically: the empty string or the empty string), causing it to match and split on every character boundary. These characters must be escaped to use them literally in an expression:
my #words = split /\[\|\]/, $string;
Since all the lines makes this visually confusing, you should probably use m{} quotes instead of //, and \Q…\E to quote a range of characters instead of a separate backslash for each one. (This is functionally identical, it's just a little easier to read.)
my #words = split m{\Q[|]\E}, $string;

Ignoring escape characters in perl

my %result = "\\path\tfolder\file.txt";
How can I ignore the \t escape sequence without prepending a '\'. Is there something like:
my %result = r"\\path\tfolder\file.txt";
The above doesn't work.
Single quotes process two escape sequences: \\ and \', so you would have to double the leading double-backslash but not the others:
my $result = '\\\\server\toppath\files';
To get what you want, you could use a here-document at the cost of some syntactic bulk.
chomp(my $result = <<'EOPath');
\\server\toppath\files
EOPath
Note the change of sigil from % to $ because a string is a scalar, and hashes are for associations.