I have multiple strings containing sql queries that outputs with \n and \t symbols in my logs.
Here is one for example
SELECT id, codcom, action, dc, tc FROM\n\t\t\t\t(SELECT\n\t\t\t\t\tid, codcom, action, dc, tc\n\t\t\t\tFROM\n\t\t\t\t\tattendance.inout\n\t\t\t\tWHERE\n\t\t\t\t\tdc between '2021-10-25' and '2021-10-31' and cod_ditta = 1 and codcom = 6293\n\t\t\t\tUNION\n\t\t\t\tSELECT\n\t\t\t\t\tid, codcom, action, dc, tc\n\t\t\t\tFROM\n\t\t\t\t\tattendance.inout_timb_ext\n\t\t\t\tWHERE\n\t\t\t\t\tdc between '2021-10-25' and '2021-10-31' and cod_ditta = 1 and codcom = 6293) timbs\n\t\t\t\tORDER BY dc , tc
is it possible to use a single regex to replace all the symbols that starts with the backslash symbol with a new line, tab, etcetera?
I know I could achieve the same result with multiple replaces. This question is more out of curiosity.
What I tried so far
So far I was able to select all the symbols that I want to replace with the regex (in the find section)
\\(\w)
But when it comes to replace I don't know how to replace it with the actual new line, tab, etc.
I tried this so far (in the replace section), without achieving the desired result:
\\$1
\\1
Not possible with \1 because the group will capture the \\t which is treated literally and is not \t (tab).
In sublime text, do it in two steps, use below expressions:
Step 1:
Find All (regex mode): (\\t)
Replace All: (space)
Step 2:
Find All (regex mode): (\\t)
Replace All: \n (newline)
Related
I have a textfile where I would like to replace all GUIDs with space.
I want:
92094, "970d6c9e-c199-40e3-80ea-14daf1141904"
91995, "970d6c9e-c199-40e3-80ea-14daf1141904"
87445, "f17e66ef-b1df-4270-8285-b3c15da366f7"
87298, "f17e66ef-b1df-4270-8285-b3c15da366f7"
96713, "3c28e493-015b-4b48-957f-fe3e7acc8412"
96759, "3c28e493-015b-4b48-957f-fe3e7acc8412"
94665, "87ac12a3-62ed-4e1d-a1a6-51ae05e01b1a"
94405, "87ac12a3-62ed-4e1d-a1a6-51ae05e01b1a"
To become:
92094,
91995,
87445,
87298,
96713,
96759,
94665,
94405,
How can i accomplish this in Sublime 3?
Ctrl+H
Find: "[\da-f-]{36}"
Replace: LEAVE EMPTY
Enable regex mode
Replace all
Explanation:
" : double quote
[ : start class character
\d : any digit
a-f : or letter from a to f
- : or a dash
]{36} : end class, 36 characters must be present
" : double quote
Result for given example:
92094,
91995,
87445,
87298,
96713,
96759,
94665,
94405,
Try doing a search for this pattern in regex search mode:
"[0-9a-z]{8}-[0-9a-z]{4}-[0-9a-z]{4}-[0-9a-z]{4}-[0-9a-z]{12}"
And then just replace with empty string. This should strip off the GUID, leaving you with the output you want.
Demo
Another regex solution involving a slightly different search-replace strategy where we don't care about the GUI format and simply get the first column:
Search for ([^,]*,).* (again don't forget to activate the regex mode .*).
Replace with $1.
Details about the regular expression
The idea here is to capture all first columns. A column here is defined by a sequence of
"some non-comma character": [^,]*
followed by a comma: [^,]*,
The first column can then be followed by anything .* (the GUI format doesn't matter): [^,]*,.*
Finally we need to capture the 1st column using group capturing: ([^,]*,).*
In the replace field we use a backreference $x which refers the the x-th capturing group.
Hope here is the right place to write ask this question.
I am preparing a script to import to a database using notepad++.
I have a huge file that has rows like that:
(10496, '69055-230', 'Rua', '5', 'Manaus', 'Parque 10 de Novembro',
'AM'),
INSERT INTO dne id, cep, tp_logradouro, logradouro, cidade,
bairro, uf VALUES
Is there a way using FIND/REPLACE to replace the ',' to ';' on every line before the INSERT statement?
I am not sure how to match the end of the line before a specific word.
The result would be
(10496, '69055-230', 'Rua', '5', 'Manaus', 'Parque 10 de Novembro',
'AM');
INSERT INTO dne id, cep, tp_logradouro, logradouro, cidade,
bairro, uf VALUES
Find what: ,(?=\s*INSERT)
Replace with: ;
Description
, matches a literal comma
(?=\s*INSERT) is a lookeahead that will assert for (but won't consume)
\s* any number of white spaces (including newlines)
INSERT as literal
If you also want to replace any commas before the end of the file, use
,(?=\h*\R\h*INSERT|\s*\z)
Note both expressions would fail if you have another instance of a comma followed by INSERT that shouldn't be replaced, but in that case you should specify it in the question.
You don't even need a regular expression for that.
Select Extended in Search Mode
Replace ,\nINSERT INTO with ;\nINSERT INTO
This matches , at the end of a line just before INSERT INTO at the beginning of the next line. Keep in mind that \n will match only in a Linux/Unix/Mac OS X file. For Windows use \r\n, for Mac OS Classic \r (reference).
Using sublim text or notepad++, click CTRL+h and replace all ")INSERT," by ");INSERT"
I expect that the INSERT statements will all have the form:
INSERT INTO table col1, col2, col3, ...
VALUES (val1, val2, val3, ...),
^^ what you want to replace
Assuming that the only place that ), will be observed is the end of the VALUES line, then you can just can just do the following replacement:
Find: ),$
Replace: );$
You can do this replacement with the regex option enabled.
I have a text file
#sp_id int,
#sp_name varchar(120),
#sp_gender varchar(10),
#sp_date_of_birth varchar(10),
#sp_address varchar(120),
#sp_is_active int,
#sp_role int
Here, I want to get only the first word from each line. How can I do this? The spaces between the words may be space or tab etc.
Here is what I suggest:
Find what: ^([^ \t]+).*
Replace with: $1
Explanation: ^ matches the start of line, ([^ \t]+) matches 1 or more (due to +) characters other than space and tab (due to [^ \t]), and then any number of characters up to the end of the line with .*.
See settings:
In case you might have leading whitespace, you might want to use
^\s*([^ \t]+).*
I did something similar with this:
with open('handles.txt', 'r') as handles:
handlelist = [line.rstrip('\n') for line in handles]
newlist = [str(re.findall("\w+", line)[0]) for line in handlelist]
This gets a list containing all the lines in the document,
then it changes each line to a string and uses regex to extract the first word (ignoring white spaces)
My file (handles.txt) contained info like this:
JoIyke - personal twitter link;
newMan - another twitter handle;
yourlink - yet another one.
The code will return this list:
[JoIyke, newMan, yourlink]
Find What: ^(\S+).*$
Replace by : \1
You can simply use this to get the first word.Here we are capturing the first word in a group and replace the while line by the captured group.
Find the first word of each line with /^\w+/gm.
I've got a CSV file with some 600 records where I need to replace some [CRLF] with a [space] but only when the [CRLF] is positioned between two ["] (quotation marks). When the second ["] is encountered then it should skip the rest of the line and go to the next line in the text.
I don't really have a starting point. Hope someone comes up with a suggestion.
Example:
John und Carol,,Smith,,,J.S.,,,,,,,,,,,,,+11 22 333 4444,,,,,"streetx 21[CRLF]
New York City[CRLF]
USA",streetx 21,,,,New York City,,,USA,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,Normal,,My Contacts,[CRLF]
In this case the two [CRLF] after the first ["] need to be replaced with a space [ ]. When the second ["] is encountered, skip the end of the line and go to next line.
Then again, now on the next line, after the first ["] is encountered replace all [CRLF] until the second ["] is encountered. The [CRLF]s vary in numbers.
In the CSV-file the amount of commas [,] before (23) and after (65) the 2 quotation marks ["] is constant.
So maybe a comma counter could be used. I don't know.
Thanks for feedback.
This will work using one regex only (tested in Notepad++):
Enter this regex in the Find what field:
((?:^|\r\n)[^"]*+"[^\r\n"]*+)\r\n([^"]*+")
Enter this string in the Replace with field:
$1 $2
Make sure the Wrap around check box (and Regular expression radio button) are selected.
Do a Replace All as many times as required (until the "0 occurrences were replaced" dialog pops up).
Explanation:
(
(?:^|\r\n) Begin at start of file or before the CRLF before the start of a record
[^"]*+ Consume all chars up to the opening "
" Consume the opening "
[^\r\n"]*+ Consume all chars up to either the first CRLF or the closing "
) Save as capturing group 1 (= everything in record before the target CRLF)
\r\n Consume the target CRLF without capturing it
(
[^"]*+ Consume all chars up to the closing "
" Consume the closing "
) Save as capturing group 2 (= the rest of the string after the target CRLF)
Note: The *+ is a possessive quantifier. Use them appropriately to speed up execution.
Update:
This more general version of the regex will work with any line break sequence (\r\n, \r or \n):
((?:^|[\r\n]+)[^"]*+"[^\r\n"]*+)[\r\n]+([^"]*+")
Maybe do it in three steps (assuming you have 88 fields in the CSV, because you said there are 23 commas before, and 65 after each second ")
Step 1: replace all CR/LF with some character not anywhere in the file, like ~
Search: \r\n Replace: ~
Step 2: replace all ~ after every 88th 'comma group' (or however many fields in CSV) with \r\n -- to reinsert the required CSV linebreaks:
Search: ((?:[^,]*?,){88})~ Replace: $1\r\n
Step 3: replace all remaining ~ with space
Search ~ Replace: <space>
In this case the source data is generated by the export function in GMail for your contacts.
After the modification outlined below (without RegEx) the result can be used to tidy up your contacts database and re-import it to GMail or to MS Outlook.
Yes, I am standing on the shoulders of #alan and #robinCTS. Thank you both.
Instructions in 5 steps:
use Notepad++ / find replace / extended search mode / wrap around = on
-1- replace all [CRLF] with a unique set characters or a string (I used [~~])
find: \r\n and replace with: ~~
The file contents are now on one line only.
-2- Now we need to separate the header line. For this move to where the first record starts exactly before the 88th. comma (including the word after the 87th. comma [,]) and enter the [CRLF] manually by hitting the return key. There are two lines now: header and records.
-3- now find all [,~~] and replace with [,\r\n] The result is one record per line.
-4- remove the remaining [~~] find: ~~ and replace with: [ ] a space.
The file is now clean of unwanted [CRLF]s.
-5- Save the file and use it as intended.
I have a list (in a .txt file) which I'd like to quickly convert to JavaScript Syntax, so I want to take the following:
AliceBlue
AntiqueWhite
Aqua
Aquamarine
Azure
Beige
Bisque
Black
BlanchedAlmond
and convert it to an array literal...
var myArray = ["AliceBlue", "AntiqueWhite", ... ]
I have the list in notepad++ and I need a reg expression to add the " at the start of the line and ", at the end and remove the line break... does anyone have a quick fix to do this? I'm terrible with RegEx.
I often have to perform such tasks so to know how to do this would be a great benefit to me. Many thanks
You won't be able to do it in a single replacement; you'll have to perform a few steps. Here's how I'd do it:
Find (in regular expression mode):
(.+)
Replace with:
"\1"
This adds the quotes:
"AliceBlue"
"AntiqueWhite"
"Aqua"
"Aquamarine"
"Azure"
"Beige"
"Bisque"
"Black"
"BlanchedAlmond"
Find (in extended mode):
\r\n
Replace with (with a space after the comma, not shown):
,
This converts the lines into a comma-separated list:
"AliceBlue", "AntiqueWhite", "Aqua", "Aquamarine", "Azure", "Beige", "Bisque", "Black", "BlanchedAlmond"
Add the var myArray = assignment and braces manually:
var myArray = ["AliceBlue", "AntiqueWhite", "Aqua", "Aquamarine", "Azure", "Beige", "Bisque", "Black", "BlanchedAlmond"];
One simple way is replace \n(newline) with ","(double-quote comma double-quote) after this append double-quote in the start and end of file.
example:
AliceBlue
AntiqueWhite
Aqua
Aquamarine
Beige
Replcae \n with ","
AliceBlue","AntiqueWhite","Aqua","Aquamarine","Beige
Now append "(double-quote) at the start and end
"AliceBlue","AntiqueWhite","Aqua","Aquamarine","Beige"
If your text contains blank lines in between you can use regular expression \n+ instead of \n
example:
AliceBlue
AntiqueWhite
Aqua
Aquamarine
Beige
Replcae \n+ with "," (in regex mode)
AliceBlue","AntiqueWhite","Aqua","Aquamarine","Beige
Now append "(double-quote) at the start and end
"AliceBlue","AntiqueWhite","Aqua","Aquamarine","Beige"
Put your cursor at the begining of line 1.
click Edit>ColumnEditor. Put " in the text and hit enter.
Repeat 2 but put the cursor at the end of line1 and put ", and hit enter.
In notepad++, for placing any thing before value
Press CTRL+H
Replace ^ with ' (sign you want append at the start)
Select search mode as Regular Expression
Click Replace All
In notepad++, for placing any thing After value
Press CTRL+H
Replace $ with ' (sign you want to append at the end)
Select search mode as Regular Expression
Click Replace All
Ex: After performing above steps
AHV01 replaced with 'AHV01'
Happy Learning!!
Thanks.
Place your cursor at the end of the text.
Press SHIFT and ->. The cursor will move to the next line.
Press CTRL-F and type , in "Replace with:" and press ENTER.
You will need to put a quote at the beginning of your first text and the end of your last.
I am using Notepad 8.1.9.2 64bit on Windows10, the replacement procedures can be finished in one step, try this:
Find what: (.+)\r\n
Replace with: "\1",
Note: Wrap around and regular express option is selected.
And then you still need to add bracket manually in your code
Thanks!