I have a file that contains lines in a format similar to this...
/data/file.geojson?10,20,30,40
/data/file.geojson?bbox=-5.20751953125,49.05227025601607,3.0322265625,56.46249048388979
/data/file.geojson?bbox=-21.46728515625,45.99696161820381,19.2919921875,58.88194208135912
/data/file.geojson?bbox=-2.8482055664062496,54.38935426009769,-0.300750732421875,55.158473983815306
/data/file.geojson?bbox=-21.46728515625,45.99696161820381,19.2919921875,58.88194208135912
/data/file.geojson?bbox=-21.46728515625,45.99696161820381,19.2919921875,58.88194208135912
I've tried a combination of grep, sed, gawk, and |(pipes) to try and pattern match and then change the format to be more like this...
[10,40],[30,40],[30,20][10,20],
[-5.20751953125,56.46249048388979],[3.0322265625,56.46249048388979].....
Hopefully you get the idea from the first line so I don't have to type out all the examples manually!
I've got the hang of regex to match the co-ordinates. In fact the input file is the result of extracting from apache access logs. It might be easier to read/understand answers if they just match positive integer numbers, I will then be able to slot in a more complicated pattern to match the right range.
To be able to arrange the results like you which it is important to be able to access the last for values per line.
No pattern matching is required if you use awk. You can split the input strings by a set of delimiters and reassemble the resulting fields. 40 can be accessed as $(NF), 30 as $(NF-1) and so on.
awk -F'[?,=]' '
{printf "[%s,%s],[%s,%s],[%s,%s],[%s,%s]\n",
$(NF-3),$(NF),$(NF-1),$(NF),
$(NF-1),$(NF-2),$(NF-3),$(NF-2)
}' file
I'm using ?, , or = as the field delimiters. This makes it simple to access the columns of interest.
Output:
[10,40],[30,40],[30,20],[10,20]
[-5.20751953125,56.46249048388979],[3.0322265625,56.46249048388979],[3.0322265625,49.05227025601607],[-5.20751953125,49.05227025601607]
[-21.46728515625,58.88194208135912],[19.2919921875,58.88194208135912],[19.2919921875,45.99696161820381],[-21.46728515625,45.99696161820381]
[-2.8482055664062496,55.158473983815306],[-0.300750732421875,55.158473983815306],[-0.300750732421875,54.38935426009769],[-2.8482055664062496,54.38935426009769]
[-21.46728515625,58.88194208135912],[19.2919921875,58.88194208135912],[19.2919921875,45.99696161820381],[-21.46728515625,45.99696161820381]
[-21.46728515625,58.88194208135912],[19.2919921875,58.88194208135912],[19.2919921875,45.99696161820381],[-21.46728515625,45.99696161820381]
Btw, also sed can be used here:
sed -r 's/.*[?=]([^,]+),([^,]+),([^,]+),(.*)/[\1,\4],[\3,\4],[\3,\2],[\1,\2]/' file
The command is capturing the numbers at the end each in a separate capturing group and re-assembles them in the replacement part.
Not all versions of sed support the + quantifier. The most compatible version would look like this :)
sed 's/.*[?=]\([^,]\{1,\}\),\([^,]\{1,\}+\),\([^,]\{1,\}\),\(.*\)/[\1,\4],[\3,\4],[\3,\2],[\1,\2]/' file
sed strips off items prior to numbers, then awk splits on comma and outputs in different order. Assuming data is in a file called "td.txt"
sed 's/^[^0-9-]*//' td.txt|awk -F, '{print "["$1","$4"],["$3","$4"],["$3","$2"],["$1","$2"],"}'
This might work for you (GNU sed):
sed -r 's/^.*\?[^-0-9]*([^,]*),([^,]*),([^,]*),([^,]*)/[\1,\4],[\3,\4],[\3,\2],[\1,\2]/' file
Or with more toothpicks:
sed 's/^.*\?[^-0-9]*\([^,]*\),\([^,]*\),\([^,]*\),\([^,]*\)/[\1,\4],[\3,\4],[\3,\2],[\1,\2]/' file
You can use the following to match:
(\/data\/file\.geojson\?(?:bbox=)?)([0-9.-]+),([0-9.-]+),([0-9.-]+),([0-9.-]+)
And replace with the following:
$1[$2,$3],[$4,$5]
See DEMO
Related
I have the following file I would like to clean up
cat file.txt
MNS:N+ GYPA*01 or GYPA*M
MNS:M+ GYPA*02 or GYPA*N
MNS:Mc GYPA*08 or GYP*Mc
MNS:Vw GYPA*09 or GYPA*Vw
MNS:Mg GYPA*11 or GYPA*Mg
MNS:Vr GYPA*12 or GYPA*Vr
My desired output is:
MNS:N+ GYPA*01 or GYPA*M
MNS:M+ GYPA*02 or GYPA*N
MNS:Mc GYPA*08 or GYP*Mc
MNS:Vw GYPA*09 or GYPA*Vw
MNS:Mg GYPA*11 or GYPA*Mg
MNS:Vr GYPA*12 or GYPA*Vr
I would like to remove everything between ":" and the first occurence of "or"
I tried sed 's/MNS:d*?or /MNS:/g' though it removes the second "or" as well.
I tried every option in https://www.geeksforgeeks.org/sed-command-in-linux-unix-with-examples/
to no avail. should I create alias sed='perl -pe'? It seems that sed does not properly support regex
perl should be more suitable here because we need Lazy match logic here.
perl -pe 's|(:.*?or +)(.*)|:\2|' Input_file
by using .*?or we are checking for the first nearest match for or string in the line.
This might work for you (GNU sed):
sed '/:.*\<or\>/{s/\<or\>/\n/;s/:.*\n//}' file
If a line contains : followed by the word or, then substitute the first occurrence of the word or with a unique delimiter (e.g.\n) and then remove everything between : and the unique delimiter.
Wrt I would like to remove everything between ":" and the first occurence of "or" - no you wouldn't. The first occurrence of or in the 2nd line of sample input is as the start of orweqqwe. That text immediately after : looks like it could be any set of characters so couldn't it contain a standalone or, e.g. MNS:2 or eqqwe or M+ GYPA*02 or GYPA*N
Given that and the fact it's apparently a fixed number of characters to be removed on every line, it seems like this is what you should really be using:
$ sed 's/:.\{14\}/:/' file
MNS:N+ GYPA*01 or GYPA*M
MNS:M+ GYPA*02 or GYPA*N
MNS:Mc GYPA*08 or GYP*Mc
MNS:Vw GYPA*09 or GYPA*Vw
MNS:Mg GYPA*11 or GYPA*Mg
MNS:Vr GYPA*12 or GYPA*Vr
If it is sure the or always occurs twice a line as provided example, please try:
sed 's/\(MNS:\).\+ or \(.\+ or .*\)/\1\2/' file.txt
Result:
MNS:N+ GYPA*01 or GYPA*M
MNS:M+ GYPA*02 or GYPA*N
MNS:Mc GYPA*08 or GYP*Mc
MNS:Vw GYPA*09 or GYPA*Vw
MNS:Mg GYPA*11 or GYPA*Mg
MNS:Vr GYPA*12 or GYPA*Vr
Otherwise using perl is a better solution which supports the shortest match as RavinderSingh13 answers.
ex supports lazy matching with \{-}:
ex -s '+%s/:\zs.\{-}or //g|wq' input_file
The pattern :\zs.\{-}or matches any character after the first : up to the first or.
I have lots of file containing following ipaddress, and i want to replace last digit of ip and look like i am having struggle to come up with correct regex
file1
IPADDR=10.30.2.26
NETMASK=255.255.0.0
GATEWAY=10.30.0.1
I want to replace 10.30.2.26 to 10.30.2.27 using sed but somehow i am missing something, i have tried following.
I have many file which i want to replace and last digit could be anything.
I have tried sed 's/[^IPADDR].$/7/g' file1
how do i match anything between ^IPADDR{anything}$ ?
In your regex, [^IPADDR] is a character class that search for any character except those listed between brackets. I'm not sure that's what you want.
You can use an address instead to find lines starting with IPADDR(/^IPADDR/) and apply the substitution command on it:
sed '/^IPADDR/s/[0-9]$/7/' file
You may use the following command:
sed -r 's/(^IPADDR=[0-9.]+)([0-9]$)/\17/g' file
Prints:
IPADDR=10.30.2.27
NETMASK=255.255.0.0
GATEWAY=10.30.0.1
Let say I have millions of string in a text file in this format:
st=expand&c=22&t=button&k=fun HTTP
This is a string we can look at as a hash with keys st, c, t and k. Some of the strings in the text file might not have a given &KEY=VALUE present and might thus look like this:
st=expand&k=fun HTTP
How would one use sed to change the string to following
expand,,,fun
that is, even thought the key=value isn't present, we still add a comma. We can assume that we have a fixed key set [st,c,t,k].
What I've tried is something like (just an idea!!)
sed 's/\(st=\|c=\|t=\|k=\)\([\(^\&\|HTTP\)])\(\&\|HTTP\)/\3,/g' big_file
but obviously, if c isn't there, it isn't adding a comma since it doesn't find any. Any ideas how to approach this? Using awk might also be acceptable (or any other fast text-processing utility)
Thanks!
Input data example
st=expand&c=22&t=button&k=fun HTTP
c=22&t=button&k=fun HTTP
st=expand&c=22&t=party&k=fun HTTP
st=expand&c=22&k=fun HTTP
st=expand HTTP
HTTP
Output data
expand,22,button,fun
,22,button,fun
expand,22,party,fun
expand,22,,fun
expand,,,
,,,
You can use this sed:
sed -E 's/(st=([^& ]*)|)(.*c=([^& ]*)|)(.*t=([^& ]*)|)(.*k=([^& ]*)|) HTTP/\2,\4,\6,\8/' file
expand,22,button,fun
,22,button,fun
expand,22,party,fun
expand,22,,fun
expand,,,
,,,
Sed Demo
RegEx Demo
Whenever you have name=value pairs in your input data, it's simplest and clearest and usually most efficient to create a name->value array and then print the values by name in whatever order you want, e.g.:
$ cat tst.awk
BEGIN { FS="[&= ]"; OFS="," }
{
delete n
for (i=1;i<NF;i+=2) {
n[$i] = $(i+1)
}
print n["st"], n["c"], n["t"], n["k"]
}
$ awk -f tst.awk file
expand,22,button,fun
,22,button,fun
expand,22,party,fun
expand,22,,fun
expand,,,
,,,
Another pattern for sed to try:
sed -r "s/(st=(\w+))?(&?c=(\w+))?(&t=(\w+))?(&k=(\w+))?( HTTP)/\2,\4,\6,\8/g" big_file
expand,22,button,fun
,22,button,fun
expand,22,party,fun
expand,22,,fun
expand,,,
REGEX 101 DEMO
How about something like this? It's not perfectly strict, but as long as your data follows the format you described on every line, it will work.
Regex:
^(?:st=([^&\n]*))?&?(?:c=([^&\n]*))?&?(?:t=([^&\n]*))?&?(?:k=([^&\n]*))? HTTP$ (must be run once per line or with multi-line and global options enabled)
Substitution:
\1,\2,\3,\4
Try it here: https://regex101.com/r/nE1oP7/2
EDIT: If you are using sed, you will need to change the non-capturing groups to regular ones ((?:) to ()) and update the backreferences accordingly (\2,\4,\6,\8). Demo: http://ideone.com/GNRNGp
I am attempting to parse (with sed) just First Last from the following DN(s) returned by the DSCL command in OSX terminal bash environment...
CN=First Last,OU=PCS,OU=guests,DC=domain,DC=edu
I have tried multiple regexs from this site and others with questions very close to what I wanted... mainly this question... I have tried following the advice to the best of my ability (I don't necessarily consider myself a newbie...but definitely a newbie to regex..)
DSCL returns a list of DNs, and I would like to only have First Last printed to a text file. I have attempted using sed, but I can't seem to get the correct function. I am open to other commands to parse the output. Every line begins with CN= and then there is a comma between Last and OU=.
Thank you very much for your help!
I think all of the regular expression answers provided so far are buggy, insofar as they do not properly handle quoted ',' characters in the common name. For example, consider a distinguishedName like:
CN=Doe\, John,CN=Users,DC=example,DC=local
Better to use a real library able to parse the components of a distinguishedName. If you're looking for something quick on the command line, try piping your DN to a command like this:
echo "CN=Doe\, John,CN=Users,DC=activedir,DC=local" | python -c 'import ldap; import sys; print ldap.dn.explode_dn(sys.stdin.read().strip(), notypes=1)[0]'
(depends on having the python-ldap library installed). You could cook up something similar with PHP's built-in ldap_explode_dn() function.
Two cut commands is probably the simplest (although not necessarily the best):
DSCL | cut -d, -f1 | cut -d= -f2
First, split the output from DSCL on commas and print the first field ("CN=First Last"); then split that on equal signs and print the second field.
Using sed:
sed 's/^CN=\([^,]*\).*/\1/' input_file
^ matches start of line
CN= literal string match
\([^,]*\) everything until a comma
.* rest
http://www.gnu.org/software/gawk/manual/gawk.html#Field-Separators
awk -v RS=',' -v FS='=' '$1=="CN"{print $2}' foo.txt
I like awk too, so I print the substring from the fourth char:
DSCL | awk '{FS=","}; {print substr($1,4)}' > filterednames.txt
This regex will parse a distinguished name, giving name and val a capture groups for each match.
When DN strings contain commas, they are meant to be quoted - this regex correctly handles both quoted and unquotes strings, and also handles escaped quotes in quoted strings:
(?:^|,\s?)(?:(?<name>[A-Z]+)=(?<val>"(?:[^"]|"")+"|[^,]+))+
Here is is nicely formatted:
(?:^|,\s?)
(?:
(?<name>[A-Z]+)=
(?<val>"(?:[^"]|"")+"|[^,]+)
)+
Here's a link so you can see it in action:
https://regex101.com/r/zfZX3f/2
If you want a regex to get only the CN, then this adapted version will do it:
(?:^|,\s?)(?:CN=(?<val>"(?:[^"]|"")+"|[^,]+))
I have a file that looks like this:
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
I want it to look like this
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
I thought I could use sed to do this but I can't figure out how to store something in a buffer and then modify it.
Am I even using the right tool?
Thanks
You don't have to get tricky with regular expressions and replacement strings: use sed's p command to print the line intact, then modify the line and let it print implicitly
sed 'p; s/\.png//'
Glenn jackman's response is OK, but it also doubles the rows which do not match the expression.
This one, instead, doubles only the rows which matched the expression:
sed -n 'p; s/\.png//p'
Here, -n stands for "print nothing unless explicitely printed", and the p in s/\.png//p forces the print if substitution was done, but does not force it otherwise
That is pretty easy to do with sed and you not even need to use the hold space (the sed auxiliary buffer). Given the input file below:
$ cat input
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
you should use this command:
sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
The result:
$ sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
This commands is just a replacement command (s///). It matches anything starting with #" followed by non-period chars ([^.]*) and then by .png",. Also, it matches all non-period chars before .png", using the group brackets \( and \), so we can get what was matched by this group. So, this is the to-be-replaced regular expression:
#"\([^.]*\)\.png",
So follows the replacement part of the command. The & command just inserts everything that was matched by #"\([^.]*\)\.png", in the changed content. If it was the only element of the replacement part, nothing would be changed in the output. However, following the & there is a newline character - represented by the backslash \ followed by an actual newline - and in the new line we add the #" string followed by the content of the first group (\1) and then the string ",.
This is just a brief explanation of the command. Hope this helps. Also, note that you can use the \n string to represent newlines in some versions of sed (such as GNU sed). It would render a more concise and readable command:
sed 's/#"\([^.]*\)\.png",/&\n#"\1",/' input
I prefer this over Carles Sala and Glenn Jackman's:
sed '/.png/p;s/.png//'
Could just say it's personal preference.
or one can combine both versions and apply the duplication only on lines matching the required pattern
sed -e '/^#".*\.png",/{p;s/\.png//;}' input