I have a text file in this format
abc1=kdkaf
def2 = lakdfjads
aa3 =kdkakdfadsf
2323= kakdfadf
In Notepad++ I want to find and delete all the data after =
so that I can only have:
abc1
def2
aa3
2323
Search mode regular expression
[=].*
Replace with nothing.
It searches for "=" , followed by any character (.) any number of times (*), effectively "any sequence of characters", ending at the line break.
You can use a simple regex like this:
=.*
With a empty replacement string
Assuming you have alphabets only after the =， Try
\s*=\s*[a-zA-Z]*
Or this if you want to include CrLf too
\s*=\s*[a-zA-Z]*\r\n
After selecting 'Regular expression' in Search Mode
Related
I am trying to work on regular expressions. I have a mainframe file which has several fields. I have a flat file parser which distinguishes several types of records based on the first three letters of every line. How do I write a regular expression where the first three letters are 'CTR'.
Beginning of line or beginning of string?
Start and end of string
/^CTR.*$/
/ = delimiter
^ = start of string
CTR = literal CTR
$ = end of string
.* = zero or more of any character except newline
Start and end of line
/^CTR.*$/m
/ = delimiter
^ = start of line
CTR = literal CTR
$ = end of line
.* = zero or more of any character except newline
m = enables multi-line mode, this sets regex to treat every line as a string, so ^ and $ will match start and end of line
While in multi-line mode you can still match the start and end of the string with \A\Z permanent anchors
/\ACTR.*\Z/m
\A = means start of string
CTR = literal CTR
.* = zero or more of any character except newline
\Z = end of string
m = enables multi-line mode
As such, another way to match the start of the line would be like this:
/(\A|\r|\n|\r\n)CTR.*/
or
/(^|\r|\n|\r\n)CTR.*/
\r = carriage return / old Mac OS newline
\n = line-feed / Unix/Mac OS X newline
\r\n = windows newline
Note, if you are going to use the backslash \ in some program string that supports escaping, like the php double quotation marks "" then you need to escape them first
so to run \r\nCTR.* you would use it as "\\r\\nCTR.*"
^CTR
or
^CTR.*
edit:
To be more clear: ^CTR will match start of line and those chars. If all you want to do is match for a line itself (and already have the line to use), then that is all you really need. But if this is the case, you may be better off using a prefab substr() type function. I don't know, what language are you are using. But if you are trying to match and grab the line, you will need something like .* or .*$ or whatever, depending on what language/regex function you are using.
Regex symbol to match at beginning of a line:
^
Add the string you're searching for (CTR) to the regex like this:
^CTR
Example: regex
That should be enough!
However, if you need to get the text from the whole line in your language of choice, add a "match anything" pattern .*:
^CTR.*
Example: more regex
If you want to get crazy, use the end of line matcher
$
Add that to the growing regex pattern:
^CTR.*$
Example: lets get crazy
Note: Depending on how and where you're using regex, you might have to use a multi-line modifier to get it to match multiple lines. There could be a whole discussion on the best strategy for picking lines out of a file to process them, and some of the strategies would require this:
Multi-line flag m (this is specified in various ways in various languages/contexts)
/^CTR.*/gm
Example: we had to use m on regex101
Try ^CTR.\*, which literally means start of line, CTR, anything.
This will be case-sensitive, and setting non-case-sensitivity will depend on your programming language, or use ^[Cc][Tt][Rr].\* if cross-environment case-insensitivity matters.
^CTR.*$
matches a line starting with CTR.
Not sure how to apply that to your file on your server, but typically, the regex to match the beginning of a string would be :
^CTR
The ^ means beginning of string / line
There's are ambiguities in the question.
What is your input string? Is it the entire file? Or is it 1 line at a time? Some of the answers are assuming the latter. I want to answer the former.
What would you like to return from your regular expression? The fact that you want a true / false on whether a match was made? Or do you want to extract the entire line whose start begins with CTR? I'll answer you only want a true / false match.
To do this, we just need to determine if the CTR occurs at either the start of a file, or immediately following a new line.
/(?:^|\n)CTR/
(?i)^[ \r\n]*CTR
(?i) -- case insensitive -- Remove if case sensitive.
[ \r\n] -- ignore space and new lines
* -- 0 or more times the same
CTR - your starts with string.
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.
While porting many lines of code from one language to another I must replace all array access from the form of the function call x.get(y) to the square brackets notation x[y]. There are few text editors around that can do regular expression based replace.
What should be typed in the "text to find" field and what should be typed in the "replace with" field in this situation? Both x and y can vary, so the original code can have lines like:
... state.get(1);
... text.get(i);
... result.get(line);
after conversion:
... state[1];
... text[i];
... result[line];
You can search for \.get\((\w+)\) and replace with [$1].
The above pattern assumes only alphanumeric characters between the parentheses, but there are other alternatives:
.* (without checking ". matched newline") should match until the end of the line.
[^)]* should match characters that are not ). Would work for new lines.
In both cases, you may want to include the ; in your pattern.
Note that this is very fragile either way - you might encounter code like state.get(a.get(3 + sin(6))), and probably get incorrect results.
For Notepad++, I would write in Find what: ([0-9,a-z,A-Z,-,_]+).get\(([0-9,a-z,A-Z,-,_]+)\)
replace with \1[\2]
Input:
x.get(1);
text.get(i);
result.get(line);
Output:
x[1];
text[i];
result[line];
I'm very new to regular expression. I want to extract the following string
"109_Admin_RegistrationResponse_20130103.txt"
from this file content, the contents is selected per line:
01-10-13 10:44AM 47 107_Admin_RegistrationDetail_20130111.txt
01-10-13 10:40AM 11 107_Admin_RegistrationResponse_20130111.txt
The regular expression should not pick the second line, only the first line should return a true.
Your Regex has a lot of different mistakes...
Your line does not start with your required filename but you put an ^ there
missing + in your character group [a-zA-Z], hence only able to match a single character
does not include _ in your character group, hence it won't match Admin_RegistrationResponse
missing \ and d{2} would match dd only.
As per M42's answer (which I left out), you also need to escape your dot . too, or it would match 123_abc_12345678atxt too (notice the a before txt)
Your regex should be
\d+_[a-zA-Z_]+_\d{4}\d{2}\d{2}\.txt$
which can be simplified as
\d+_[a-zA-Z_]+_\d{8}\.txt$
as \d{2}\d{2} really look redundant -- unless you want to do with capturing groups, then you would do:
\d+_[a-zA-Z_]+_(\d{4})(\d{2})(\d{2})\.txt$
Remove the anchors and escape the dot:
\d+[a-zA-Z_]+\d{8}\.txt
I'm a newbie in php but i think you can use explode() function in php or any equivalent in your language.
$string = "01-09-13 10:17AM 11 109_Admin_RegistrationResponse_20130103.txt";
$pieces = explode("_", $string);
$stringout = "";
foreach($i = 0;$i<count($pieces);i++){
$stringout = $stringout.$pieces[$i];
}
I have a document that has a range of numbers like this:
0300010000000394001001,27
0300010000000394001002,0
0300010000000394002001,182
0300010000000394002002,51
0300010000000394003001,156
0300010000000394003002,40
I need to find the new line character and replace with a number of spaces depending on the string length.
If it has 24 characters like this - 0300010000000394001002,0 then I need to replace the new line character at the end with 5 blank spaces.
If it has 25 characters like this - 0300010000000394002002,51 then I need to replace the new line character at the end with 4 blank spaces and so on.
In my text editor I can use find and replace. I search for the line length by ^(.|\s){24}$ for 24 characters - but this will obviously replace the whole line and I only need to replace the new line character at the end.
I want to specify a new line character AFTER ^(.|\s){24}$. Is this possible?
It sounds like you need two things.
Multi-line Mode (See "Using ^ and $ as Start of Line and...")
Backreferencing
Most editors that support regex support these naturally, but you'll have to let us know what editor you're using for us to be specific. Without knowing what editor you're using, all I can say is that you want to do some combination of the following:
regex subst
----- -----
^(.{24})\n $1 <-- there are spaces here
^(.{24})^M \1 <-- there are spaces here
^(.{24})\s ^^^^^