I have 100K rows file and every row contain date 12/13/2019. I want to replace that date with 12/20/2019. But when I am entering the command like :%s/12/13/2019/12/20/2019/g. It gives an error that couldn't find pattern.
Format for date is (MM/DD/YYYY)
A substitution is made of several parts:
:<range>s/<search>/<replace>/<flags>
Between those parts, you have / as default separator. Since / separates the <search> part from the <replace> part, any / in your search pattern is going to be interpreted as a separator, leading to undesirable results.
One solution is to escape your slashes with an anti-slash:
:%s/12\/13\/2019/12\/20\/2019/g
Another one (my favourite) is to use an alternative separator:
:%s#12/13/2019#12/20/2019#g
Reference:
:help :s
:help pattern-delimiter
Related
I would like to mark in Notepad++ the sql scripts in a text log. The sql files have this format in the text:
AAAAAAAA.BBBBBBBBBBB.sql
So what I execute is this sentence in search menu:
\w*.sql
As I should get BBBBBBBBBBB.sql. The point is that in some script names there are dashes (-), and when that happens I dont get the whole name, but just the end after the last dash.
For example, in:
AAAAAAAA.BBBBB-CCCCCCC.sql
I would like to get BBBBB-CCCCCCC.sql, but I just get CCCCCCC.sql
Is there any possible formula to get them?
If the match can not start and end with a hyphen:
\w+(?:-\w+)*\.sql
\w+ Match 1+ word characters
(?:-\w+)* Optionally match - and 1+ word characters
\.sql Match .sql
See a regex demo.
Note that in your pattern the \w* can also match 0 occurrences and that the . can match any character if it is not escaped.
Another option could be using a character class to match either - or a word character, but this would also allow to mix and match like --a--.sql
[\w-]+\.sql
See another regex demo.
I'm fairly new to Vim and I haven't been able to find on this site how to search and replace with a varying part of a string. I need to apply a global edit to all times "SetTag("...")" appears with ... being any word. My edit is to add one more word after the second quotation mark. example: SetTag("err" + __LINE__ with the bolded part being what I need to add. Can anyone let me know how this is possible with a vim search command? Thanks!
nb: I assume "word" is any sequence of characters other than a doublequote character. Modify as needed.
:%s/SetTag("\([^"]*\)")/SetTag("\1" + __LINE__)/
the escaped parentheses grab the sub-match; the \1 in the replacement string is replaced by that sub-match.
I am translating a game, and the game's text box only supports 50 characters max per line. Is there a way to use a formula to split the entire sentence every 50 characters or whole word (49, 48, 47, etc)?
I am currently working with this formula.
=JOIN(CHAR(10),SPLIT(REGEXREPLACE(A1, "(.{50})", "/$1"),"/"))
The problem with this code, is that it splits at exactly 50 characters (one time), and will split in the middle of the word.
So again, my goal is to have it not split on the 50th character IF the 50th character is in the middle of the word, and for the rule to apply for the rest of the lines too because it only applies on the first line.
Please take a look at this test google sheet to get an example of what I am talking about.
If it's impossible to do it on Google Sheets, I don't mind moving to Excel provided I get a functioning code.
For the record, I did ask in Google's product forums 2 days ago, and still haven't received an answer.
=REGEXREPLACE(A1, "(.{1,50})\b", "$1" & CHAR(10))
{50} matches exactly 50 times, but what you need is 50 or less.
\b is word boundary that matches between alphanumeric and non-alphanumeric character.
= REGEXEXTRACT(A1,"(?ism)^"&REPT("([\w\d'\(\),. ]{0,49}\s)", ROUNDUP(LEN(A1)/50,0))&"([\w\d'\(\),. ]{0,49})$")
Tested with various expressions and works as intended. Note that only these characters [a-zA-Z0-9_'(),.] are allowed, Which means - and other characters not mentioned will not work. If you need them, add them inside the REPT expression and finishing regexp formula. Otherwise, This will work perfectly.
You are pretty close. I'm not an expert in Sheets, so not sure if this is the best way, but your Regex is wrong for what you want.
Also, you need to be certain that you don't use a split character that might appear in the phrase itself. However, using CHAR(10) for the replace character allows you to insert LF without going through the JOIN SPLIT sequence.
replace any line feeds, carriage returns and spaces with a single space
Match strings that start with a non-Space character followed by up to 49 more characters which are followed by a space or the end of the string.
replace the capture group with the capturing group followed by the CHAR(10) (and delete the space following).
There will be extra CHAR(10) at the end which you can strip off.
EDIT Regex changed slightly due to a difference in behavior between Google's RE and what I am used to (probably has to do with how a non-backtracking regex works). The problem showed up on your example:
=regexreplace(REGEXREPLACE(REGEXREPLACE(A1 & " ","[\r\n\s]+"," "),"(\S.{0,49})\s","$1" & char(10)),"\n+\z","")
I have below string in csv files
Part Number WP1166496 (AP6005317) replaces 1166496, 1156976.
Expected Output -
Part Number WP1166496 replaces 1166496, 1156976.
I want to replace (AP6005317) this with blanks.
As there are many rows with different values.
So how can I replace this string with brackets to blanks value.
I don't know how to achieve this exactly in Microsoft Excel.
If you look for find and replace feature, most probably you can see option to replace with regular expressions.
Use regular expression option and replace \(.*\) with (simple space). This will solve your problem.
Note : This is tested and verified in LibreOffice Calc.
I have a very large number (a couple hundred digits long), and I'd like to use vim to add commas to the number in the appropriate manner, i.e. after each group of three digits, moving from right to left. How can I do this efficiently?
Taken from here
Substitue command that adds commas in the right spot.
:%s/\(\d\)\(\(\d\d\d\)\+\d\#!\)\#=/\1,/g
This uses a zero width lookahead to match any number that isn't followed by groups of three numbers followed by one number. (or 3n+1 numbers)
So the numbers that match in are marked with ^. These are then replaced with a comma after it the match.
31415926
^ ^
Which replaces to
31,415,926
A friend of mine suggests using the printf program: ciw<C-r>=system("printf \"%'d\" ".shellescape(#"))<CR>.
This is one way of doing it:
s/\d\{-1,}\ze\(\d\{3}\)\+\s/&,/g
Notes:
\{-1,} is saying match at least 1 but in a non-greedy way (Vim doesn't seem to support the usual \+\? syntax; also, for quantifiers, you just need to escape the opening curly brace)
\ze is saying match the pattern behind this but don't store the match in & (equivalent to positive look-ahead)
\(\d\{3}\)\+\> matches groups of 3 digits that ends with word-nonword boundary (word in this sense means alphanumerical + underscore).
Alternatively, you can use \s for space/tab, or \D for non-digit instead of \>, whichever fits your needs better
The way that I used is to create a macro that adds one single comma, and then invoke the macro a whole bunch of times, like qahhi,<ESC>hq#a#a#a#a…