I'm stumped with how to remove a portion of a string that has forward slashes and question marks in it.
Example: /diag/PeerManager/list?deviceid=RXMWANT8WFYJNF7K6DXXXJLJVN
and I need the output to be RXMWANT8WFYJNF7K6DXXXJLJVN
I've tried tr and sed but tr removes some of the characters I need in the output. sed is giving me trouble because of the forward slashes.
What's a quick method to remove the /diag/PeerManager/list?deviceid= portion of my string?
thanks!
echo "/diag/PeerManager/list?deviceid=RXMWANT8WFYJNF7K6DXXXJLJVN" | sed -n 's:/[a-zA-Z]/[a-zA-Z]/[a-zA-Z]?[a-zA-Z]=::p'
This should do the trick. I chose the colon as the delimiter as it will not cause any issues with the forward slash. This makes a lot of assumptions about the type of input it will be receiving, specifically that it will only contain three backslashes with lower and uppercase letters between them, a series of letters ending in a question mark, another series of letters ending in an equals sign. This then removes those items and prints the remaining characters (your device id).
This worked for me:
sed 's/.*deviceid=\([^&]*\).*/\1/'
Example:
$ echo '/diag/PeerManager/list?deviceid=RXMWANT8WFYJNF7K6DXXXJLJVN' | sed 's/.*deviceid=\([^&]*\).*/\1/'
RXMWANT8WFYJNF7K6DXXXJLJVN
This is not the most robust solution, but if you have a fixed set of input that will never change, it's probably good enough.
One way using awk, if there is only a single occurrence of an = on each line:
awk -F= '{ print $2 }' file.txt
Results:
RXMWANT8WFYJNF7K6DXXXJLJVN
Use Equals Sign as Field Delimiter
If you know that your GET query string will always have only one parameter (in this case, deviceid) then you can just use the equals sign as a field delimiter with the standard cut utility. For example:
$ echo '/diag/PeerManager/list?deviceid=RXMWANT8WFYJNF7K6DXXXJLJVN' |
cut -d= -f2-
RXMWANT8WFYJNF7K6DXXXJLJVN
How about:
$ echo /diag/PeerManager/list?deviceid=RXMWANT8WFYJNF7K6DXXXJLJVN | sed 's/^.*=//'
RXMWANT8WFYJNF7K6DXXXJLJVN
Related
I'm trying to write a script and one of the parts of the script requires me to concatenate some variables together to create a URL.
REPO_URL='https://github.com/Example/Repo.Game/'
FILENAME='Example.Game-linux.zip'
latest_version="$(curl -LIs "${REPO_URL}/releases/latest" | grep -i '^location:' | cut -d' ' -f2 | cut -d'/' -f8)"
echo "$latest_version"
echo "$FILENAME"
echo "$REPO_URL"
echo "${REPO_URL}releases/download/${latest_version}/${FILENAME}"
Output:
2.0.5164
Example.Game-linux.zip
https://github.com/Example/Repo.Game/
/Example.Game-linux.ziple/Repo.Game/releases/download/2.0.5164
My actual output:
2.0.5164
Oxide.Rust-linux.zip
https://github.com/OxideMod/Oxide.Rust/
/Oxide.Rust-linux.zipideMod/Oxide.Rust/releases/download/2.0.5164
It looks like some kind of overflow problem? I'm not exactly sure. I added abcabc to the filename and the output became
/Oxide.Rust-linux.zipabcabc/Oxide.Rust/releases/download/2.0.5164
Any help would be appreciated.
I resolved the problem by removing the carriage return value from the variable.
tr -d '\r' seems to have resolved it. I'm not sure where the variable came from and if anyone has advice on how to clean up this mess I would love some advice.
latest_version="$(curl -LIs "${REPO_URL}/releases/latest" | grep -i '^location:' | cut -d' ' -f2 | cut -d'/' -f8 | tr -d '\r')
You can use ANSI quoting, and variable substitution to remove control characters from variables without having to invoke sub-shells.
ANSI quoting uses the special format $'\*' to represent special characters. For example use $'\t' for tab, $'\n' for new-line and $'\r' for carriage-return.
Variable substitution uses extra characters at the end of the variable name to perform actions on the variable. For example
${variable//[pattern]/[substitution]} will replace all instances of [pattern] in ${variable} with [substitution].
${variable%[pattern]} will remove [pattern] from ${variable} if it is at the end.
By combining these two, you can remove carriage-return characters from the end of your variable like this:
echo ${variable%$'\r'}
Note: Variable substitution doesn't actually change the contents of the variable. To do that, you have to re-assign the result back to the variable:
variable="${variable%$'\r'}"
There is a cleaner way to get the version number, minus any trailing carriage-return, from github using sed.
latest_version =$(curl -LIs "${REPO_URL}/releases/latest" | sed -n 's/^Location:.*\/\([^\r]*\).*$/\1/p')
sed reads every line of input (STDIN by default) and performs operations on it defined by the action string parameter. The action string is a little tricky to explain in this case, but here goes:
The -n option suppresses the printing of each input line. Output will then only happen if it is explicitly stated in the action string.
The s/[pattern]/[substitution]/p construct says whenever you find [pattern], replace it with [substitution] and print it. Our [pattern] is ^Location:.*\/\(.*\)$, and our [substitution] is \1.
The expression ^ matches the beginning of the line.
The expression . means any single character, and the expression .* means any number of characters (including zero). This will match the largest possible string, so, for example .*/ will match abc/def/ in the string abc/def/ghi.
The expression \/ just escapes the forward slash (because we are using backslash as a delimiter, we have to escape it).
The expression \([pattern]\) says any time you find [pattern], remember it. in our case, it will remember whatever matches [^\r].
The expression [{chars}] matches any one of the characters in {chars}. [^{chars}] matches any character that is not in {chars}. so [^\r]* matches any number of characters that is not a carriage return.
The expression $ matches the end of a line.
The expression \1 is replaced by the first remembered pattern.
So altogether, our action string says:
If you find a line that starts with Location:, followed by any number of characters, followed by a /, followed by any number of characters that are not a carriage return (which will be remembered), followed by any number of characters, followed by an end of line, then print the remembered characters.
I have data in variable like json -
v={k1:v1,k2:v2,k3:v3,k4:v4,k5:v5,k6:v6,k7:v7,k8:v8};
where key and value could be any values.
I need to split this into multiple lines after every 10 character..which i did by
echo "${v}" | sed -r 's/.{10}/&\n/g'
This does the split as per sed . But now i need to make sure split should happen only after comma character found after every 10 characters...so that out put should have meaningful lines ..
output should be ..
k1:v1,k2:v2,
.....
Whole idea is not break lines in between
Thanks
You may use
sed -r 's/.{10}[^,]*,/&\n/g'
See the sed demo online.
The .{10}[^,]*, pattern matches
.{10} - any 10 chars
[^,]* - 0 or more chars other than ,
, - a comma.
The &\n replacement pattern replaces with the whole match (&) and appends a newline to it.
If actually you just want to add a newline after every 2nd comma then that's this in GNU sed and some other seds:
$ echo "$v" | sed 's/,[^,]*,/&\n/g'
k1:v1,k2:v2,
k3:v3,k4:v4,
k5:v5,k6:v6,
k7:v7,k8:v8
or this for portability across all seds in all shells:
sed 's/,[^,]*,/&\
/g'
or this using any awk:
awk '{gsub(/,[^,]*,/,"&\n")}1'
Sorry and thanks for feed back here is the expected output-
{k1:v1,k2:v2,
k3:v3,k4:v4,
k5:v5,k6:v6,
k7:v7,k8:v8}
Rule should be look for first 10 characters then first comma when found add a new line after that pattern.
Wiktor code snippt does this actually. However , i am happy to see to achieve this with other ways. Thanks all.
I trying to find and replace items using bash. I was able to use sed to grab out some of the characters, but I think I might be using it in the wrong matter.
I am basically trying to remove the characters after ";" and before "," including removing ","
sed -e 's/\(;\).*\(,\)/\1\2/'
That is what I used to replace it with nothing. However, it ends up replacing everything in the middle so my output came out like this:
cmd2="BMC,./socflash_x64 if=B600G3_BMC_V0207.ima;,reboot -f"
This is the original text of what I need to replace
cmd2="BMC,./socflash_x64 if=B600G3_BMC_V0207.ima;X,sleep 120;after_BMC,./run-after-bmc-update.sh;hba_fw,./hba_fw.sh;X,sleep 5;DB,2;X,reboot -f"
Is there any way to make it look like this output?
./socflash_x64 if=B600G3_BMC_V0207.ima;sleep 120;./run-after-bmc-update.sh;./hba_fw.sh;sleep 5;reboot -f
Ff there is any way to make this happen other than bash I am fine with any type of language.
Non-greedy search can (mostly) be simulated in programs that don't support it by replacing match-any (dot .) with a negated character class.
Your original command is
sed -e 's/\(;\).*\(,\)/\1\2/'
You want to match everything in between the semi-colon and the comma, but not another comma (non-greedy). Replace .* with [^,]*
sed -e 's/\(;\)[^,]*\(,\)/\1\2/'
You may also want to exclude semi-colons themselves, making the expression
sed -e 's/\(;\)[^,;]*\(,\)/\1\2/'
Note this would treat a string like "asdf;zxcv;1234,qwer" differently, since one would match ;zxcv;1234, and the other would match only ;1234,
In perl:
perl -pe 's/;.*?,/;/g;' -pe 's/^[^,]*,//' foo.txt
will output:
./socflash_x64 if=B600G3_BMC_V0207.ima;sleep 120;./run-after-bmc-update.sh;./hba_fw.sh;sleep 5;2;reboot -f
The .*? is non greedy matching before the comma. The second command is to remove from the beginning to the comma.
Something like:
echo $cmd2 | tr ';' '\n' | cut -d',' -f2- | tr '\n' ';' ; echo
result is:
./socflash_x64 if=B600G3_BMC_V0207.ima;sleep 120;./run-after-bmc-update.sh;./hba_fw.sh;sleep 5;2;reboot -f;
however, I thing your requirements are a few more complex, because 'DB,2' seems a particular case. After "tr" command, insert a "grep" or "grep -v" to include/exclude these cases.
I would like to display the last word in these lines I tried to look for example the word value but no answer, so I thought to look for the words between quotes but my file contains other words between quotes that I have I need not actually want to display the values of the select tag knowing that my html file is.
grep '*' hosts.html | awk '{print $NF}'
For example:
value='www.visit-tunisia.com'>www.visit-tunisia.com
value='www.watania1.tn'>www.watania1.tn
value='www.watania2.tn'>www.watania2.tn
I would have
www.visit-tunisia.com
www.watania1.tn
www.watania2.tn
You need to set the field separator to > you do this with the -F option:
$ awk -F'>' '{print $NF}' hosts.html
www.visit-tunisia.com
www.watania1.tn
www.watania2.tn
Note: I'm not sure what you are trying to achieve by grep '*' hosts.html?
Interpreting the comment liberally, you have input lines which might contain:
value='www.visit-tunisia.com'>www.visit-tunisia.com
value='www.watania1.tn'>www.watania1.tn
value='www.watania2.tn'>www.watania2.tn
and you would like the names which are repeated on a line as the output:
www.visit-tunisia.com
www.watania1.tn
www.watania2.tn
This can be done using sed and capturing parentheses.
sed -n -e "s/.*'\([^']*\)'.*\1.*/\1/p"
The -n says "don't print unless I say to do so". The s///p command prints if the substitute works. The pattern looks for a stream of 'anything' (.*), a single quote, captures what's inside up to the next single quote ('\([^']*\)') followed by any text, the captured text (the first \1), and anything. The replacement text is what was captured (the second \1).
Example:
$ cat data
www and wotnot
value='www.visit-tunisia.com'>www.visit-tunisia.com
blah
value='www.watania1.tn'>www.watania1.tn
hooplah
value='www.watania2.tn'>www.watania2.tn
if 'nothing' is required, nothing will be done.
$ sed -n -e "s/.*'\([^']*\)'.*\1.*/\1/p" data
www.visit-tunisia.com
www.watania1.tn
www.watania2.tn
nothing
$
Clearly, you can refine the [^']* part of the match if you want to. I used double quotes around the expression since the pattern matches on single quotes. Life is trickier if you need to allow both single and double quotes; at that point, I'd put the script into a file and run sed -f script data to make life easier.
sed 's/.*>\(.*\)/\1/g' your_file
I have lines of data that looks like this:
sp_A0A342_ATPB_COFAR_6_+_contigs_full.fasta
sp_A0A342_ATPB_COFAR_9_-_contigs_full.fasta
sp_A0A373_RK16_COFAR_10_-_contigs_full.fasta
sp_A0A373_RK16_COFAR_8_+_contigs_full.fasta
sp_A0A4W3_SPEA_GEOSL_15_-_contigs_full.fasta
How can I use sed to delete parts of string after 4th column (_ separated) for each line.
Finally yielding:
sp_A0A342_ATPB_COFAR
sp_A0A342_ATPB_COFAR
sp_A0A373_RK16_COFAR
sp_A0A373_RK16_COFAR
sp_A0A4W3_SPEA_GEOSL
cut is a better fit.
cut -d_ -f 1-4 old_file
This simply means use _ as delimiter, and keep fields 1-4.
If you insist on sed:
sed 's/\(_[^_]*\)\{4\}$//'
This left hand side matches exactly four repetitions of a group, consisting of an underscore followed by 0 or more non-underscores. After that, we must be at the end of the line. This is all replaced by nothing.
sed -e 's/\([^_]*\)_\([^_]*\)_\([^_]*\)_\([^_]*\)_.*/\1_\2_\3_\4' infile > outfile
Match "any number of not '_'", saving what was matched between \( and \), followed by '_'. Do this 4 times, then match anything for the rest of the line (to be ignored). Substitute with each of the matches separated by '_'.
Here's another possibility:
sed -E -e 's|^([^_]+(_[^_]+){3}).*$|\1|'
where -E, like -r in GNU sed, turns on extended regular expressions for readability.
Just because you can do it in sed, though, doesn't mean you should. I like cut much much better for this.
AWK likes to play in the fields:
awk 'BEGIN{FS=OFS="_"}{print $1,$2,$3,$4}' inputfile
or, more generally:
awk -v count=4 'BEGIN{FS="_"}{for(i=1;i<=count;i++){printf "%s%s",sep,$i;sep=FS};printf "\n"}'
sed -e 's/_[0-9][0-9]*_[+-]_contigs_full.fasta$//g'
Still the cut answer is probably faster and just generally better.
Yes, cut is way better, and yes matching the back of each is easier.
I finally got a match using the beginning of each line:
sed -r 's/(([^_]*_){3}([^_]*)).*/\1/' oldFile > newFile