I have been researching this for quite some time but cannot seem to find an answer. Perhaps someone here can help.
I am trying to use sed to replace words in yml / yaml files. Since some of the words are included in the names I want to only replace words that appear after the colon (':').
For example. If the .yml file includes:
en:
label_some_tracker: A tracker
label_all_tracker: All trackers
label_attachment_type_trackers: Select trackers.
tracker_plural: trackers
and I want to replace all occurrences of tracker with issue in all values. The pattern:
s/tracker/issue/
also changes the names of the fields, which breaks my code.
I can reduce the size of the problem somewhat by including terms for all possible variants of a word. For example:
s/trackers/issues/
s/tracker/issue/
but that doesn't deal with all situations.
I have tried inserting a space before the search term:
s/ tracker/ issue/
but that matches names where the search term is at the beginning of the line.
If I search for whole words then it still seems to pick up the names because ':' and '_' are 'non word' characters.
If I try to put spaces at the beginning and end of the search term but then it misses words that are at the end of a line or words patterns with punctuation marks before the training space.
The only sure way seems to be to only replace words after a colon (':') but I cannot seem to figure out how to do that with sed.
Does anyone here know how?
With GNU sed:
sed -E 's/(:.*)tracker/\1issue/g' file
Output:
en:
label_some_tracker: A issue
label_all_tracker: All issues
label_attachment_type_trackers: Select issues.
tracker_plural: issues
Replace second occurance:
sed 's/tracker/issue/2' file
Related
How do I prevent newlines in the readme.md files (GitHub)?
We can always write the whole thing in one line to prevent it. But is there an exclusive tag/option to prevent the same, especially for tags that create newlines (headings) like span in html?
Doesn't a space followed by a backslash do the concatenation you want? It does for me. That way I can break a paragraph into one sentence per line.
I have a very long comment I want to add to a Postgres table.
Since I do not want a very long single line as a comment I want to split it into several lines.
Is this possible? \n does not work since Postgres does not use the backslash as an escape character.
Just write a multi-line string:
COMMENT ON TABLE foo IS 'This
comment
is stored
in multiple lines';
You can also embed \n escape sequences in “extended” string constants that start with E:
COMMENT ON TABLE foo IS E'A comment\nwith three\nlines.';
You can use automatic concatenation of adjacent string literals together with E'\n' escape sequences for linebreaks:
COMMENT ON TABLE foo IS E''
'This comment is stored in multiple lines. But only some'
'end with linebreaks like this one.\n'
'You can even create empty lines to simulate paragraphs:'
'\n\n'
'This would be the second paragraph, then.';
Details:
Note the initial E'' at the end of the first line. This is essential to make all the adjacent string literals that follow it use the extended string literal syntax, providing us with the option to write \n for a linebreak. Of course, that E could also be placed into the second line instead, at the start of the real string: E'This comment …'. Me putting it into the first line is just source code aesthetics … character alignment and stuff.
I consider this solution slightly better than multi-line strings (proposed in another answer here) because it allows to fit the comment into the typical line width limit and the indentation requirements of source files. Useful when you keep your SQL in well-formatted files under version control, that is, treating it just as any other source code. When including indentation into multi-line strings, on the other hand, this results in lots of additional whitespace in the live table comment.
Note for OP: When you say "I do not want a very long single line as a comment", it is not clear if you don't want that long line in your .sql source code file, or if you don't want it in the table comment of the live table, such as when seen in a database admin tool. It does not really matter, as this solution gives you tools for both purposes: use adjacent string literals to fit your line into the source code file, without affecting line breaks in the live table comment; and use \n to create line breaks and empty lines in the live table comment.
I want to search a file for a specific string and then place a comment at the beginning of that string. But I need an answer that avoids regex, global changes, and all the other fancy stuff.
I wrote this line:
sed -i.bak '/PermitRootLogin no/# PermitRootLogin no/' ./sshd_config
but I get an error:
sed: -e expression #1, char 21: comments don't accept any addresses
I assume the issue is that I need to escape the # character, but I'm not finding any resources on how to do that, or even mentioning it. I've tried various combinations of putting ^ or \ or \^ in front of the # but I'm jut not getting it right.
Please note I am intentionally repeating the text to be replaced. I would like the most simple possible solution to this question: how to replace "XYX" with "# XYZ" in the most obvious possible way.
As indicated in the comments by #mlt , you could try adding an s at the beginning your sed command. Straight from his comment:
s/PermitRootLogin....
I see that you said you're intentionally repeating the test to be replaced. If by that you mean, you want it to be the same, maybe consider grouping your matched text. I understand you may have meant that you just want it hand typed. Anyway, here is how to match the grouped text and add the comment character:
s/(PermitRootLogin)/# \1/
The parens indicated that the matched text should be consider a group, the \1 indicates that you want to put that matched group there.
I hope this was helpful. Happy coding! Leave a comment if you have any questions.
Imagine I have a file that has the following type of line:
FIXED_DATA1 VARIABLE_DATA FIXED_DATA2
I want to change the fixed data and leave the variable data as is. For various reasons, using two sed operations to replace the fixed data will not work. For instance, the fixed fields might be double-quotes, and the line has other areas containing them, thus really the regex is written to match a pattern in the variable data and the fixed data.
If I'm bent on using sed, is there a way to change both fixed data fields at once while leaving the variable field unchanged?
Thanks.
You need to partition the line into the three pieces, replace the outer two and leave the middle alone:
sed 's/^FIX1 \(.*\) FIX2$/New \1 End/'
You can make the beginning and end matches more complex as needed.
I'm using regular expression lib icucore via RegKit on the iPhone to
replace a pattern in a large string.
The Pattern i'm looking for looks some thing like this
| hello world (P1)|
I'm matching this pattern with the following regular expression
\|((\w*|.| )+)\((\w\d+)\)\|
This transforms the input string into 3 groups when a match is found, of which group 1(string) and group 3(string in parentheses) are of interest to me.
I'm converting these formated strings into html links so the above would be transformed into
Hello world
My problem is the trailing space in the third group. Which when the link is highlighted and underlined, results with the line extending beyond the printed characters.
While i know i could extract all the matches and process them manually, using the search and replace feature of the icu lib is a much cleaner solution, and i would rather not do that as a result.
Many thanks as always
Would the following work as an alternate regular expression?
\|((\w*|.| )+)\s+\((\w\d+)\)\| Where inserting the extra \s+ pulls the space outside the 1st grouping.
Though, given your example & regex, I'm not sure why you don't just do:
\|(.+)\s+\((\w\d+)\)\|
Which will have the same effect. However, both your original regex and my simpler one would both fail, however on:
| hello world (P1)| and on the same line | howdy world (P1)|
where it would roll it up into 1 match.
\|\s*([\w ,.-]+)\s+\((\w\d+)\)\|
will put the trailing space(s) outside the capturing group. This will of course only work if there always is a space. Can you guarantee that?
If not, use
\|\s*([\w ,.-]+(?<!\s))\s*\((\w\d+)\)\|
This uses a lookbehind assertion to make sure the capturing group ends in a non-space character.