I'm having some issues used sed to match a regex pattern. An example would be with the following lines:
spring-core-4.0.0 should be spring-core-4.1.0
spring-web-4.0.0 should be spring-web-4.1.0
I want the regex to match any characters in between spring and the version number. I'm not sure if I need to do something else for sed to remember what those characters are in each line.
sed -E 's/spring-([a-z]+)-4.0.0/spring-([a-z]+)-4.1.0/' file.txt
The current output is placing the regex pattern in the output, instead of matching it. Thanks for any tips.
$ sed 's/\(spring-[^-]*-\)4.0.0/\14.1.0/' file
spring-core-4.1.0
spring-web-4.1.0
Related
I have a file:
dynamicclaspath.cfg
VENDOR_JAR=/clear-as-1-d/apps/sterling/jar/struts/2_5_18/1_0_0/log4j-core-2.10.0.jar
VENDOR_JAR=/clear-as-1-d/apps/sterling/jar/log4j/2_17_1/log4j-core-2.10.0.jar
I want to replace any occurrence of log4j-core* with log4j-core-2.17.1.jar
I tried this but I know I'm missing a regex:
sed -i '/^log4j-core/ s/[-]* /log4j-core-2.17.1.jar/'
With your shown samples please try following sed program. Using -E option with sed to enable ERE(extended regular expressions) with it. In main program using substitute option to perform substitution. Using sed's capability to use regex and store matched values into temp buffer(capturing groups). Matching till last occurrence of / and then matching log4j-core till jar at last of value. While substituting it with 1st capturing group value(till last occurrence of /) followed by new value of log4j as per OP's requirement.
sed -E 's/(^.*\/)log4j-core-.*\.jar$/\1log4j-core-2.17.1.jar/' Input_file
Using sed
$ sed -E 's/(log4j-core-)[0-9.]+/\12.17.1./' input_file
VENDOR_JAR=/clear-as-1-d/apps/sterling/jar/struts/2_5_18/1_0_0/log4j-core-2.17.1.jar
VENDOR_JAR=/clear-as-1-d/apps/sterling/jar/log4j/2_17_1/log4j-core-2.17.1.jar
It depends on possible other contents in your input file how specific the search pattern must be.
sed 's/log4j-core-.*\.jar/log4j-core-2.17.1.jar/' inputfile
or
sed 's/log4j-core-[0-9.]*\.jar/log4j-core-2.17.1.jar/' inputfile
or (if log4j-core*.jar is always the last part of the line)
sed 's/log4j-core.*/log4j-core-2.17.1.jar/' inputfile
sed -i s'#2.10.0.jar$#2.17.1.jar#'g file
That seems to work.
I have a following line;
�5=?�#A00165:69:HKJ3YDMXX:1:1101:16812:7341 1:N:0:TCTTAAAG
and would like to remove characters, �5=?� in front of #. So the desired output looks as follows;
#A00165:69:HKJ3YDMXX:1:1101:16812:7341 1:N:0:TCTTAAAG
I used gnu sed (v4.8)with a following argument;
sed "s/.*#/#/"'
but this did not remove �5=?� thought it worked in the GNU sed live editor.
At this point, I really appreciate any help on this.
My system is 3.10.0-1160.71.1.el7.x86_64
Using sed, remove everything up to the first occurance of #
$ sed 's/^[^#]*//' input_file
#A00165:69:HKJ3YDMXX:1:1101:16812:7341 1:N:0:TCTTAAAG
This might work for you (GNU sed):
sed -E 's/(\o357\o277\o275)5=\?\1//g' file
This removes all occurrences of �5=?�.
N.B. To translate the octal strings use sed -n l file to display the file as is. The triplets \357\277\275 can be matched in the LHS of the substitute command by using \o357\o277\o275.
Original Text file:
.ABC0 (ABC0),
.EFG2 (EFG2),
.ZZZ3 (ZZZ3),
How to convert this part to
.ABC0 (abc0),
.EFG2 (efg2),
.ZZZ3 (zzz3),
with SED command easily?
There's issue to make it work.
echo ".ABC(ABC)," | sed -e 's/\(.*\.[A-Z]*\(\)\([A-Z]*\)\)/\1\L\2\E/ p'
You were almost there.
sed -e 's/\(\.[A-Z0-9]*\)\( ([A-Z0-9]*)\)/\1\L\2\E/g'
Remove the .* from the beginning, it matches as much as it can, i.e. it skips to the last occurrence of the pattern.
Include the digits in the character classes.
Use ( without a backslash to match a parenthesis literally.
This might work for you (GNU sed):
sed 's/([^)]*)/\L&/g' file
Replace the contents of parenthesis by its lowercase equivalent.
I'm trying to replace some regex line in a apache file.
i define:
OLD1="[0-9]*.[0-9]+"
NEW1="[a-z]*.[0-9]"
when i'm executing:
sed -i 's/$OLD1/$NEW1/g' demo.conf
there's no change.
This is what i tried to do
sed -i "s/${OLD1}/${NEW1}/g" 001-kms.conf
sed -i "s/"$OLD1"/"$NEW1"/g" 001-kms.conf
sed -i "s~${OLD1}~${NEW1}~g" 001-kms.conf
i'm expecting that the new file will replace $OLD1 with $NEW1
OLD1="[0-9]*.[0-9]+"
Because the [ * . are all characters with special meaning in sed, we need to escape them. For such simple case something like this could work:
OLD2=$(<<<"$OLD1" sed 's/[][\*\.]/\\&/g')
It will set OLD2 to \[0-9\]\*\.\[0-9\]+. Note that it doesn't handle all the possible cases, like OLD1='\.\[' will convert to OLD2='\\.\\[ which means something different. Implementing a proper regex to properly escape, well, other regex I leave as an exercise to others.
Now you can:
sed "s/$OLD2/$NEW1/g"
Tested with:
OLD1="[0-9]*.[0-9]+"
NEW1="[a-z]*.[0-9]"
sed "s/$(sed 's/[][\*\.]/\\&/g' <<<"$OLD1")/$NEW1/g" <<<'XYZ="[0-9]*.[0-9]+"'
will output:
XYZ="[a-z]*.[0-9]"
you need matching on exact string
You would need something that can match on exact string [0-9]*.[0-9]+ which sed does not support well.
Therefore instead I am using this pipeline replacing one character at a time (it also is easier to read I think):echo "[0-9]*.[0-9]+" | sed 's/0/a/' | sed 's/9/z/' | sed 's/+//'
You would have to cat your files or use find with execute to then apply this pipe.
I had tried following (from other SO answers):
- sed 's/\<exact_string/>/replacement/'doesn't work as \< and \> are left and right word boundaries respectively.
- sed 's/(CB)?exact_string/replacement/'found in one answer but nowhere in documentation
I used Win 10 bash, git bash, and online Linux tools with the same results.
when I thought matching was on the pattern rather than exact string
Replacement cannot be a regex - at most it can reference parts of the regex expression which matched. From man sed:
s/regexp/replacement/
Attempt to match regexp against the pattern space. If successful, replace that portion matched with replacement. The replacement may contain the special character & to refer to that portion of the pattern space which matched, and the special escapes \1 through \9 to refer to the corresponding matching sub-expressions in the regexp.
Additionally you have to escape some characters in your regex (specifically . and +) unless you add option -E for extended regex as per comment under your question. (N.B. only if you want to match on the full-stop . rather than it meaning any character)
$ echo "01.234--ZZZ" | sed 's/[0-9]*\.[0-9]\+/REPLACEMENT/g'
REPLACEMENT--ZZZ
I have an input stream of many lines which look like this:
path/to/file: example: 'extract_me.proto'
path/to/other-file: example: 'me_too.proto'
path/to/something/else: example: 'and_me_2.proto'
...
I'd like to just extract the *.proto filenames from these lines, and I have tried:
[INPUT] | sed 's/^.*\([a-zA-Z0-9_]+\.proto\).*$/\1/'
I know that part of my problem is that .* is greedy and I'm going to get things like e.proto and o.proto and 2.proto, but I can't even get that far... it just outputs with the same lines as the input. Any help would be greatly appreciated.
I find it helpful to use extended regex for this purpose (-r) in which case you need not escape your brackets.
sed -r 's/^.*[^a-zA-Z0-9_]([a-zA-Z0-9_]+\.proto).*$/\1/'
The addition of [^a-zA-Z0-9_] forces the .* to not be greedy.
Since you tag your command with linux, I'll assume you have GNU grep. Pick one of
grep -oP '\w+\.proto' file
grep -o "[^']+\\.proto" file
one way to do it:
sed 's/^.*[^a-zA-Z0-9_]\([a-zA-Z0-9_]\+\.proto\).*$/\1/'
escaped the + char
put a negation before the alphanum+underscore to delimit the leading chars
another way: use single quote delimitation, after all it's here for that:
sed "s/^.*'\([a-zA-Z0-9_]\+\.proto\)'.*\$/\1/"
Use this sed:
sed "s/^.*'\([a-zA-Z0-9_]\+\.proto\).*$/\1/"
+ - Extended-RegEx. So, you need to escape to get special meaning. The preceding item will be matched one or more times.
Another way:
sed "s/^.*'\([^']\+\.proto\)'.*$/\1/"
With GNU sed:
sed -E "s/.*'([^']+)'$/\1/"