I have a list of lines:
<some_random_text="someval" my_val_="0.4" some_random_text_1="someval_">
<some_random_text="someval" my_val_="0.8" some_random_text_1="someval_">
<some_random_text="someval" my_val_="1.2" some_random_text_1="someval_">
and so on.
From each line, I want to return the numeric value given after my_val_. How can I do this in bash?
Within this very rigid structure, what you want to do is quite easy using sed:
sed 's/.*my_val_="\([0-9.]\{1,\}\)".*/\1/' file
or using extended regular expressions:
sed -r 's/.*my_val_="([0-9.]+)".*/\1/' file
This captures the part you're interested in (the digits and dots between the quotes) and uses them to replace the contents of the line.
As mentioned in the comments (thanks), the switch to enable extended regular expressions differs between versions of sed. Out of habit, I tend to use -r but some implementations (such as BSD sed on OSX) work with -E instead. Others work with either -r or -E but neither option is defined by the standard.
This could also be done in native bash (although I wouldn't recommend it...):
re='my_val_="([0-9.]+)"'
while read -r line; do
[[ $line =~ $re ]] && echo "${BASH_REMATCH[1]}"
done < file
=~ is the regex match operator. The captured digits and dots are stored in element 1 of the special array BASH_REMATCH.
The sed and bash approaches are subtly different, as the sed version will print all lines in the file, even if they don't match the pattern. If this is a problem, you can add the -n switch and a p at the end of the command to print matching lines:
sed -nr 's/.*my_val_="([0-9.]+)".*/\1/p' file
With grep:
grep -oP 'my_val_="\K[^"]*' filename
-o so that grep only prints only the match, -P so that Perl-compatible regexes are used.
The \K in the regex removes from the match everything that was matched by the part of the regex that came before it; this has the effect of a lookbehind: only non-quote characters that come directly after my_val_=" are matched.
Related
I have two regular expressions:
$ grep -E '\-\- .*$' *.sql
$ sed -E '\-\- .*$' *.sql
(I am trying to grep lines in sql files that have comments and remove lines in sql files that have comments)
The grep command works using this regex; however, the sed returns the following error:
sed: -e expression #1, char 7: unterminated address regex
What am I doing incorrectly with sed?
(The space after the two hyphens is required for sql comments if you are unfamiliar with MySql comments of this type)
You're trying to use:
sed -E '\-\- .*$' *.sql
Here sed command is not correct because you're not really telling sed to do something.
It should be:
sed -n '/-- /p' *.sql
and equivalent grep would be:
grep -- '-- ' *.sql
or even better with a fixed string search:
grep -F -- '-- ' *.sql
Using -- to separate pattern and arguments in grep command.
There is no need to escape - in a regex if it is outside bracket expression (or character class) i.e. [...].
Based on comments below it seems OP's intent is to remove commented section in all *.sql files that start with 2 hyphens.
You may use this sed for that:
sed -i 's/-- .*//g' *.sql
The problem here is not the regex, the problem is that sed requires a command. The equivalent of your grep would be:
sed -n '/\-\- .*$/p'
You suppress output for non-matching lines -n ... you search (wrap your regex in slashes) and you print p (after the last slash).
P.S.: As Anub pointed out, escaping the hyphens - inside the regex is unnecessary.
You are trying to use sed's \cregexpc syntax where with \-<...> you are telling sed the delimiter character you want use is a dash -, but you didn't terminate it where it should be: \-<...>- also add d command to delete those lines.
sed '\-\-\-.*$-d' infile
see man sed about that:
\cregexpc
Match lines matching the regular expression regexp. The c may be any character.
if default / was used this was not required so:
sed '/--.*$/d' infile
or simply:
sed '/^--/d' infile
and more accurately:
sed '/^[[:blank:]]*--/d' infile
I am trying to find a pattern of two consecutive lines, where the first line is a fixed string and the second has a part substring I like to replace.
This is to be done in sh or bash on macOS.
If I had a regex tool at hand that would operate on the entire text, this would be easy for me. However, all I find is bash's simple text replacement - which doesn't work with regex, and sed, which is line oriented.
I suspect that I can use sed in a way where it first finds a matching first line, and only then looks to replace the following line if its pattern also matches, but I cannot figure this out.
Or are there other tools present on macOS that would let me do a regex-based search-and-replace over an entire file or a string? Maybe with Python (v2.7 and v3 is installed)?
Here's a sample text and how I like it modified:
keyA
value:474
keyB
value:474 <-- only this shall be replaced (follows "keyB")
keyC
value:474
keyB
value:474
Now, I want to find all occurances where the first line is "keyB" and the following one is "value:474", and then replace that second line with another value, e.g. "value:888".
As a regex that ignores line separators, I'd write this:
Search: (\bkeyB\n\s*value):474
Replace: $1:888
So, basically, I find the pattern before the 474, and then replace it with the same pattern plus the new number 888, thereby preserving the original indentation (which is variable).
You can use
sed -e '/keyB$/{n' -e 's/\(.*\):[0-9]*/\1:888/' -e '}' file
# Or, to replace the contents of the file inline in FreeBSD sed:
sed -i '' -e '/keyB$/{n' -e 's/\(.*\):[0-9]*/\1:888/' -e '}' file
Details:
/keyB$/ - finds all lines that end with keyB
n - empties the current pattern space and reads the next line into it
s/\(.*\):[0-9]*/\1:888/ - find any text up to the last : + zero or more digits capturing that text into Group 1, and replaces with the contents of the group and :888.
The {...} create a block that is executed only once the /keyB$/ condition is met.
See an online sed demo.
Use a perl one-liner with -0777 to scan over multiple lines:
$ # inline edit:
$ perl -0777 -i -pe 's/\bkeyB\s*value):\d*/$1:888/' file.txt
$ # to stdout:
$ cat file.txt | perl -0777 -pe 's/\bkeyB\s*value):\d*/$1:888/'
In plain bash:
#!/bin/bash
keypattern='^[[:blank:]]*keyB$'
valpattern='(.*):'
replacement=888
while read -r; do
printf '%s\n' "$REPLY"
if [[ $REPLY =~ $keypattern ]]; then
read -r
if [[ $REPLY =~ $valpattern ]]; then
printf '%s%s\n' "${BASH_REMATCH[0]}" "$replacement"
else
printf '%s\n' "$REPLY"
fi
fi
done < file
I'm trying to write a bash script to enclose words contained in a file with single quotes.
Word - Hello089
Result - 'Hello089',
I tried the following regex but it doesn't work. This works in Notepad++ with find and replace. I'm not sure how to tweak it to make it work in bash scripting.
sed "s/(.+)/'$1',/g" file.txt > result.txt
Replacement backreferences (also called placeholders) are defined with \n syntax, not $n (this is perl-like backreference syntax).
Note you do not need groups here, though, since you want to wrap the whole lines with single quotation marks. This is also why you do not need the g flags, they are only necessary when you plan to find multiple matches on the same line, input string.
You can use the following POSIX BRE and ERE (the one with -E) solutions:
sed "s/..*/'&',/" file.txt > result.txt
sed -E "s/.+/'&',/" file.txt > result.txt
In the POSIX BRE (first) solution, ..* matches any char and then any 0 or more chars (thus emulating .+ common PCRE pattern). The POSIX ERE (second) solution uses .+ pattern to do the same. The & in the right-hand side is used to insert the whole match (aka \0). Certainly, you may enclose the whole match with capturing parentheses and then use \1, but that is redundant:
sed "s/\(..*\)/'\1',/" file.txt > result.txt
sed -E "s/(.+)/'\1',/" file.txt > result.txt
See the escaping, capturing parentheses in POSIX BRE must be escaped.
See the online sed demo.
s="Hello089";
sed "s/..*/'&',/" <<< "$s"
# => 'Hello089',
sed -E "s/.+/'&',/" <<< "$s"
# => 'Hello089',
$1 is expanded by the shell before sed sees it, but it's the wrong back reference anyway. You need \1. You also need to escape the parentheses that define the capture group. Because the sed script is enclosed in double quotes, you'll need to escape all the backslashes.
$ echo "Hello089" | sed "s/\\(.*\\)/'\1',/g"
'Hello089',
(I don't recall if there is a way to specify single quotes using an ASCII code instead of a literal ', which would allow you to use single quotes around the script.)
I have the following string /path/to/my-jar-1.0.jar for which I am trying to write a bash regex to pull out my-jar.
Now I believe the following regex would work: ([^\/]*?)-\d but I don't know how to get bash to run it.
The following: echo '/path/to/my-jar-1.0.jar' | grep -Po '([^\/]*?)-\d' captures my-jar-1
In BASH you can do:
s='/path/to/my-jar-1.0.jar'
[[ $s =~ .*/([^/[:digit:]]+)-[[:digit:]] ]] && echo "${BASH_REMATCH[1]}"
my-jar
Here "${BASH_REMATCH[1]}" will print captured group #1 which is expression inside first (...).
You can do this as well with shell prefix and suffix removal:
$ path=/path/to/my-jar-1.0.jar
# Remove the longest prefix ending with a slash
$ base="${path##*/}"
# Remove the longest suffix starting with a dash followed by a digit
$ base="${base%%-[0-9]*}"
$ echo "$base"
my-jar
Although it's a little annoying to have to do the transform in two steps, it has the advantage of only using Posix features so it will work with any compliant shell.
Note: The order is important, because the basename cannot contain a slash, but a path component could contain a dash. So you need to remove the path components first.
grep -o doesn't recognize "capture groups" I think, just the entire match. That said, with Perl regexps (-P) you have the "lookahead" option to exclude the -\d from the match:
echo '/path/to/my-jar-1.0.jar' | grep -Po '[^/]*(?=-\d)'
Some reference material on lookahead/lookbehind:
http://www.perlmonks.org/?node_id=518444
output=`grep -R -l "${images}" *`
new_output=`regex "slide[0-9]" $output`
Basically $output is a string like this:
slides/_rels/slide9.xml.rels
The number in $output will change. I want to grab "slide9" and put that in a variable. I was hoping new_output would do that but I get a command not found for using regex. Any other options? I'm using a bash shell script.
Well, regex is not a program like grep. ;)
But you can use
grep -Eo "(slide[0-9]+)"
as a simple approach. -o means: show only the matching part, -E means: extended regex (allows more sophisticated patterns).
Reading I want to grab "slide9" and put that in a variable. I assume you want what matches your regexp to be the only thing put in $new_output? If so, then you can change that to:
new_output=`egrep -R -l "${images}" * | sed 's/.*\(slide[0-9]+\).*/\1/'`
Note no setting of output= is required (unless you use that for something else)
If you need $output to use elsewhere then instead use:
output=`grep -R -l "${images}" *`
new_output=`echo ${ouput} | sed 's/.*\(slide[0-9]+\).*/\1/'`
sed's s/// command is similar to perls s// command and has an equivalent in most languages.
Here I'm matching zero or more characters .* before and after your slide[0-9]+ and then remembering (backrefrencing) the result \( ... \) in sed (the brackets may or may not need to be escaped depending on the version of sed). We then replace that whole match (i.e the whole line) with \1 which expands to the first captured result in this case your slide[0-9]+ match.
In these situations using awk is better :
output="`grep -R -l "main" codes`"
echo $output
tout=`echo $output | awk -F. '{for(i=1;i<=NF;i++){if(index($i,"/")>0){n=split($i,ar,"/");print ar[n];}}}'`
echo $tout
This prints the filename without the extension. If you want to grab only slide9 than use the solutions provided by others.
Sample output :
A#A-laptop ~ $ bash try.sh
codes/quicksort_iterative.cpp codes/graham_scan.cpp codes/a.out
quicksort_iterative graham_scan a