I wanted to know exactly what is the difference between '>-' and '|-' especially in kubernetes yaml manifests
Newlines in folded block scalars (>) are subject to line folding, newlines in literal block scalars (|) are not.
Line folding replaces a single newline between non-empty lines with a space, and in the case of empty lines, reduces the number of newline characters between the surrounding non-empty lines by one:
a: > # folds into "one two\nthree four\n\nfive\n"
one
two
three
four
five
Line folding does not occur between lines when at least one line is more indented, i.e. contains whitespace at the beginning that is not part of the block's general indentation:
a: > # folds into "one\n two\nthree four\n\n five\n"
one
two
three
four
five
Adding - after either | or > will strip the newline character from the last line:
a: >- # folded into "one two"
one
two
b: >- # folded into "one\ntwo"
one
two
In contrast, | emits every newline character as-is, the sole exception being the last one if you use -.
Ok I got one main difference between > and | from here: https://docs.ansible.com/ansible/latest/reference_appendices/YAMLSyntax.html
Values can span multiple lines using | or >. Spanning multiple lines
using a “Literal Block Scalar” | will include the newlines and any
trailing spaces. Using a “Folded Block Scalar” > will fold newlines to
spaces; it’s used to make what would otherwise be a very long line
easier to read and edit. In either case the indentation will be
ignored.
Examples are:
include_newlines: |
exactly as you see
will appear these three
lines of poetry
fold_newlines: >
this is really a
single line of text
despite appearances
In fact the ">" is in my understanding, the equivalent of the escape characters '\' at the end of a bash script for example
If one can tell me what is the "-" used for in kubernetes yaml manifests it will complete my understanding :)
Related
I'm using the vscode vimplugin. I have a bunch of lines that look like:
Terry,169,80,,,47,,,22,,,6,,
I want to remove all the alphanumeric characters after the first comma so I get:
Terry,,,,,,,,,,,,,
In command mode I tried:
s/^.+\,[a-zA-Z0-9-]\+//g
But this does not appear to do anything. How can I get this working?
edit:
s/^[^,]\+,[a-zA-Z0-9-]\+//g
\+ is greedy; ^.\+, eats the entire line up to the last ,.
Instead of the dot (which means "any character") use [^,] which means "any but a comma". Then ^[^,]\+, means "any characters up to the first comma".
The problem with your requirement is that you want to anchor at the beginning using ^ so you cannot use flag g — with the anchor any substitution will be done once. The only way I can solve the puzzle is to use expressions: match and preserve the anchored text and then use function substitute() with flag g.
I managed with the following expression:
:s/\(^[^,]\+\)\(,\+\)\(.\+\)$/\=submatch(1) . submatch(2) . substitute(submatch(3), '[^,]', '', 'g')/
Let me split it in parts. Searching:
\(^[^,]\+\) — first, match any non-commas
\(,\+\) — any number of commas
\(.\+\)$ — all chars to the end of the string
Substituting:
\= — the substitution is an expression
See http://vimdoc.sourceforge.net/htmldoc/change.html#sub-replace-expression
submatch(1) — replace with the first match (non-commas anchored with ^)
submatch(2) — replace with the second match (commas)
substitute(submatch(3), '[^,]', '', 'g') — replace in the rest of the string
The last call to substitute() is simple, it replaces all non-commas with empty strings.
PS. Tested in real vim, not vscode.
I have a command like this, it is marking words to appear in an index in the document:
sed -i "s/\b$line\b/\\\keywordis\{$line\}\{$wordis\}\{$definitionis\}/g" file.txt
The problem is, it is finding matches within existing matches, which means its e.g. "hello" is replaced with \keywordis{hello}{a common greeting}, but then "greeting" might be searched too, and \keywordis{hello}{a common \keywordis{greeting}{a phrase used when meeting someone}}...
How can I tell sed to perform the replacement, but ignore text that is already inside curly brackets?
Curley brackets in this case will always appear on the same line.
How can I tell sed to perform the replacement, but ignore text that is already inside curly brackets?
First tokenize input. Place something unique, like | or byte \x01 between every \keywordis{hello}{a common greeting} and store that in hold space. Something along s/\\the regex to match{hello}{a common greeting}/\x01&\x01/g'.
Ten iterate over elements in hold space. Use \n to separate elements already parsed from not parsed - input from output. If the element matches the format \keywordis{hello}{a common greeting}, just move it to the front before the newline in hold space, if it does not, perform the replacement. Here's an example: Identify and replace selective space inside given text file , it uses double newline \n\n as input/output separator.
Because, as you noted, replacements can have overlapping words with the patterns you are searching for, I believe the simplest would be after each replacement shuffling the pattern space like for ready output and starting the process all over for the current line.
Then on the end, shuffle the hold space to remove \x01 and newline and any leftovers and output.
Overall, it's Latex. I believe it would be simpler to do it manually.
By "eating" the string from the back and placing it in front of input/output separator inside pattern space, I simplified the process. The following program:
sed '
# add our input/output separator - just a newline
s/^/\n/
: loop
# l1000
# Ignore any "\keywords" and "{stuff}"
/^\([^\n]*\)\n\(.*\)\(\\[^{}]*\|{[^{}]*}\)$/{
s//\3\1\n\2/
b loop
}
# Replace hello followed by anthing not {}
# We match till the end because regex is greedy
# so that .* will eat everything.
/^\([^\n]*\)\n\(.*\)hello\([{}]*\)$/{
s//\\keywordis{hello}{a common greeting}\3\1\n\2/
b loop
}
# Hello was not matched - ignore anything irrelevant
# note - it has to match at least one character after newline
/^\([^\n]*\)\n\(.*\)\([^{}]\+\)$/{
s//\3\1\n\2/
b loop
}
s/\n//
' <<<'
\keywordis{hello}{hello} hello {some other hello} another hello yet
'
outputs:
\keywordis{hello}{hello} \keywordis{hello}{a common greeting} {some other hello} another \keywordis{hello}{a common greeting} yet
I've got an application that has no useful api implemented, and the only way to get certain information is to parse string output. This is proving to be very painful...
I'm trying to achieve this in bash on SLES12.
Given I have the following strings:
QMNAME(QMTKGW01) STATUS(Running)
QMNAME(QMTKGW01) STATUS(Ended normally)
I want to extract the STATUS value, ie "Ended normally" or "Running".
Note that the line structure can move around, so I can't count on the "STATUS" being the second field.
The closest I have managed to get so far is to extract a single word from inside STATUS like so
echo "QMNAME(QMTKGW01) STATUS(Running)" | sed "s/^.*STATUS(\(\S*\)).*/\1/"
This works for "Running" but not for "Ended normally"
I've tried switching the \S* for [\S\s]* in both "grep -o" and "sed" but it seems to corrupt the entire regex.
This is purely a regex issue, by doing \S you requested to match non-white space characters within (..) but the failing case has a space between which does not comply with the grammar defined. Make it simple by explicitly calling out the characters to match inside (..) as [a-zA-Z ]* i.e. zero or more upper & lower case characters and spaces.
sed 's/^.*STATUS(\([a-zA-Z ]*\)).*/\1/'
Or use character classes [:alnum:] if you want numbers too
sed 's/^.*STATUS(\([[:alnum:] ]*\)).*/\1/'
sed 's/.*STATUS(\([^)]*\)).*/\1/' file
Output:
Running
Ended normally
Extracting a substring matching a given pattern is a job for grep, not sed. We should use sed when we must edit the input string. (A lot of people use sed and even awk just to extract substrings, but that's wasteful in my opinion.)
So, here is a grep solution. We need to make some assumptions (in any solution) about your input - some are easy to relax, others are not. In your example the word STATUS is always capitalized, and it is immediately followed by the opening parenthesis (no space, no colon etc.). These assumptions can be relaxed easily. More importantly, and not easy to work around: there are no nested parentheses. You will want the longest substring of non-closing-parenthesis characters following the opening parenthesis, no mater what they are.
With these assumptions:
$ grep -oP '\bSTATUS\(\K[^)]*(?=\))' << EOF
> QMNAME(QMTKGW01) STATUS(Running)
> QMNAME(QMTKGW01) STATUS(Ended normally)
> EOF
Running
Ended normally
Explanation:
Command options: o to return only the matched substring; P to use Perl extensions (the \K marker and the lookahead). The regexp: we look for a word boundary (\b) - so the word STATUS is a complete word, not part of a longer word like SUBSTATUS; then the word STATUS and opening parenthesis. This is required for a match, but \K instructs that this part of the matched string will not be returned in the output. Then we seek zero or more non-closing-parenthesis characters ([^)]*) and we require that this be followed by a closing parenthesis - but the closing parenthesis is also not included in the returned string. That's a "lookahead" (the (?= ... ) construct).
I have tried to create regex for the below:
STRING sou_u02_mlpv0747_CCF_ASB001_LU_FW_ALERT|/opt/app/medvhs/mvs/applications/cm_vm5/fwhome/UnifiedLogging|UL_\d{8}_CCF_ASB001_LU_sou_u02_mlpv0747_Primary.log.csv|FATAL|red|1h||fw_alert
REGEX----> /^[^#]\w+\|[^\|]+\|\w+\|\w+\|\w*\|\w*\|([^\|]+|)\|\w*$/
I am unable to figure out the mistake here.
I created the above by referring another regex which working fine and given below
/^[^#]\w+\|[^\|]+\|([^\|]+|)\|[rm]\|(in|out|old|new|arch|missing)\|\w+\|([0-9-,]+|)\|\w*\|\w*$/
sou_u02_mlpv0747_CCF_ASB001_LU_ODR|/opt/app/medvhs/mvs/applications/cm_vm5/components/CCF_ASB001_LU/SPOOL/ODR||r|out|30m|0400-1959|30m|gprs_in_stag
can some one please help me. Any leads would be highly apprciated.
Let's start from a brief look at your source text (the first that you included).
It is composed of "sections" separated with | char.
This char (|) must be matched by \|. Remember about the preceding
backslash, otherwise, a "bare" | would mean the alternative separator
(you used it in one place).
And now take a look at each section (between |):
Some of them contain only a sequence of word chars (and can be matched
by \w+).
Other sections, however, contain also other chars, e.g. slashes,
backslash, braces and dots, so each such section is actually a sequence
of chars other than "|" and must be matched by [^|]+ (here,
between [ and ], the vertical bar may be unescaped).
Now let's write each section and its "type":
sou_u02_..._FW_ALERT - word chars.
/opt/app/.../UnifiedLogging - other chars (because of slashes).
UL_\d{8}_..._Primary.log.csv - other chars (because of \d{8}
and dots).
FATAL|red|1h - 3 sections composed of word chars.
An empty section, between 2 consecutive | chars.
fw_alert - word chars.
And now, how to match these groups, and the separating |:
Point 1: \w+\| - word chars and (escaped) vertical bar.
Point 2 and 3 (together): (?:[^|]+\|){2} - a non-capturing
group - (?:...), containing a sequence of "other" chars - [^|]+
and a vertical bar - \|, occurring 2 times {2}.
Point 4 (three "word char" groups): (?:\w+\|){3} - similiar to
the previous point.
Point 5: Just as in your solution - ([^|]+|)\|, a capturing group -
(...), with 2 alternatives ...|.... The first alternative is
[^|]+ (a sequence of "other" chars), and the second alternative
is empty. After the capturing group there is \| to match the vertical
bar.
Point 6: \w+ - word chars. This time no \|, as this is the last
section.
The regex assembled so far must be:
prepended with a ^ (start of string) and
appended with a $ (end of string).
So the whole regex, matching your source text can be:
^\w+\|(?:[^|]+\|){2}(?:\w+\|){3}([^|]+|)\|\w+$
Actually, the only capturing group can be written another way,
as ([^|]*) - without alternatives, but with * as the
repetition count, allowing also empty content.
Your choice, which variant to apply.
The third field
UL_\d{8}_CCF_ASB001_LU_sou_u02_mlpv0747_Primary.log.csv
Contains a backslash, \, braces { } and dots .. None of these can be matched by \w
Note also that there is no need to escape a pipe | inside a characters class: [^|]+ is fine
I know there is a similar question in SO How can I replace mutliple empty lines with a single empty line in bash?. But my question is can this be implemented by just using the sed command?
Thanks
Give this a try:
sed '/^$/N;/^\n$/D' inputfile
Explanation:
/^$/N - match an empty line and append it to pattern space.
; - command delimiter, allows multiple commands on one line, can be used instead of separating commands into multiple -e clauses for versions of sed that support it.
/^\n$/D - if the pattern space contains only a newline in addition to the one at the end of the pattern space, in other words a sequence of more than one newline, then delete the first newline (more generally, the beginning of pattern space up to and including the first included newline)
You can do this by removing empty lines first and appending line space with G command:
sed '/^$/d;G' text.txt
Edit2: the above command will add empty lines between each paragraph, if this is not desired, you could do:
sed -n '1{/^$/p};{/./,/^$/p}'
Or, if you don't mind that all leading empty lines will be stripped, it may be written as:
sed -n '/./,/^$/p'
since the first expression just evaluates the first line, and prints it if it is blank.
Here: -n option suppresses pattern space auto-printing, /./,/^$/ defines the range between at least one character and none character (i.e. empty space between newlines) and p tells to print this range.