Using SED to find a word between 2 back slashes - sed

It is maybe a noob question but i cannot find an answer.
How can i use SED to find Santa83_4. The name Santa83_4 is an example and can be anything with any charcater.
....\randomtext\randomtext\name\Santa83_4\rate\randomtext\randomtext...
Thanks is advance


grep approach:
s='...\randomtext\randomtext\name\Santa83_4\rate\randomtext\randomtext..'
echo $s | grep -Po '\\name\\\K[^\\]+'
sed approach:
echo $s | sed -n 's/.*\\name\\\([^\\]*\).*/\1/p'
The output(for both approaches):
Santa83_4

Related

Using sed to eliminate a specific string

I appreciate your help with this problem. I like to eliminate everything that is not a specific pattern from a string.
For example, below I like to eliminate everything that is not "5TTGTC".
But as seen here ^5TTGTC is not right. I used different combinations of ^(), ^{}, ^[], but none gave me what I am looking for. Appreciate your feedback!
echo ".,..,...+5TTGTC...+5TTGCC.+5TTGTC,,.,.,,.,+5ttgtc,.,,.,.+5TTGTC.+5TTGTC,..+5TTGTC" | sed 's/^5TTGTC//g'
Thanks in advance

You may use the following command if you want case sensitivity:
echo ".,..,...+5TTGTC...+5TTGCC.+5TTGTC,,.,.,,.,+5ttgtc,.,,.,.+5TTGTC.+5TTGTC,..+5TTGTC" | sed -r 's/(5TTGTC)|[,.A-Za-z+0-9]/\1/g'
The code above prints:
5TTGTC5TTGTC5TTGTC5TTGTC5TTGTC
The regular expression used above uses alternation to capture what you are interested in.
We match and capture what we are interested in (5TTGCC) and we match everything that is not the substring, in this case characters ,.A-Za-z+0-9.
You can check the behaviour of the regex here.
As pointed out by #EdMorton, the command can be simplified to:
echo ".,..,...+5TTGTC...+5TTGCC.+5TTGTC,,.,.,,.,+5ttgtc,.,,.,.+5TTGTC.+5TTGTC,..+5TTGTC" | sed -r 's/(5TTGTC)|./\1/g'
You can try this here.
For compatibility across sed versions the -r flag can be replaced by the -E flag.

You don't make it very clear what you are trying to achieve.
One way to get where you are trying to go could be the -o option in grep.
echo ".,..,...+5TTGTC...+5TTGCC.+5TTGTC,,.,.,,.,+5ttgtc,.,,.,.+5TTGTC.+5TTGTC,..+5TTGTC" | grep -o '5TTGTC'
Output:
5TTGTC
5TTGTC
5TTGTC
5TTGTC
5TTGTC
You can then change 5TTGTC into a pattern, e.g. grep -o '[0-9]TT[AG]GTC'

With any sed:
$ echo ".,..,...+5TTGTC...+5TTGCC.+5TTGTC,,.,.,,.,+5ttgtc,.,,.,.+5TTGTC.+5TTGTC,..+5TTGTC" |
sed 's/#//g; s/5TTGTC/#/g; s/[^#]//g; s/#/5TTGTC/g'
5TTGTC5TTGTC5TTGTC5TTGTC5TTGTC
With any awk:
$ echo ".,..,...+5TTGTC...+5TTGCC.+5TTGTC,,.,.,,.,+5ttgtc,.,,.,.+5TTGTC.+5TTGTC,..+5TTGTC" |
awk -v str='5TTGTC' '{gsub(str,"\n"); gsub(/[^\n]/,""); gsub(/\n/,str)}1'
5TTGTC5TTGTC5TTGTC5TTGTC5TTGTC

Get a column using sed and modify it

I need to modify the 5 to 9 column directly in each line from a file.
Currently i'm doing this in a while loop, getting each column by line.
For example a line looks like:
echo "m.mustermann#muster.com;surnanme;givenname;displayname;1111;2222;3333;44(#44;(5555"
line_9=$(echo $line | awk -F "[;]" '{print $9}' | sed 's/[^0-9+*,]*//g')
Is there a possibility to do that with "sed -i" instead of awk
Thanks for any help

I'm not sure it can be done generally in sed, but you could definitely do it in awk:
… | awk -F";" '{ gsub("[^0-9]*","",$9); print $9 }'
If you really want to do it with sed, the expression will look something like:
… | sed -e 's,\(^[^;]*;[^;]*;[^;]*;[^;]*;[^;]*;[^;]*;[^;]*;[^;]*;[0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)\(.*\),\1\2\3\4\5\6\7\8\9,'

For a version with sed (posix) only
line_9="$(echo $line | sed 'H;x;s/^\(.\)\(\([^;]*;\)\{8\}\)\([^;]*\)/\2\1\4\1/;h;s/\(\n\).*\1/\1/;x;s/.*\(\n\)\(.*\)\1.*/\2/;s/[^0-9+*,]*//g;G;s/\(.*\)\(\n\)\(.*\)\2/\3\1/;h;s/.*//;x' )"

sed/awk : match a pattern and return everything between the end of the pattern and a semicolon

I have a line:
<random junk>TYPE=snp;<more random junk>
and I need to return everything between the end of TYPE= and the ; (in this case snp but it could be any of a number of text strings.
I tried various sed / awk solutions but I can't seem to get it working. I have the feeling this is a simple problem so, sorry about that.

This seems to work:
sed 's/.*TYPE=\(.*\);.*/\1/'
EDIT:
Ah, so there can be semicolons in the random junk. Try this:
sed 's/.*TYPE=\([^;]*\);.*/\1/'

requires GNU grep:
grep -Po '(?<=TYPE=)[^;]+'
meaning: preceded by "TYPE=", find some non-semicolon characters

One way using GNU sed:
sed -r 's/.*TYPE=([^;]+).*/\1/' file.txt

Since you also tagged this awk:
$ text='<random junk>TYPE=snp;<more random junk>'
$ echo "$text" | awk -FTYPE= '{sub(/;.*/,"",$2); print $2}'
snp
$ text='foo=bar;baz=fnu;TYPE=snp;XAI=0;XAM=0'
$ echo "$text" | awk -FTYPE= '{sub(/;.*/,"",$2); print $2}'
snp
(Only using the variable to keep the lines from wrapping.)
Or, to parse this as set of variable=value pairs rather than just a string of text:
$ echo "$text" | awk -vRS=";" -F= '$1=="TYPE" {print $2}'
snp

You can also do this in pure bash, if you want:
$ t="red=blue;TYPE=snp;XAI=0.0037843;XAM=0.0170293;XAS=0.013245;XRI=0;XRM=0"
$ t=${t#*TYPE=}
$ t=${t%%;*}
$ echo $t
snp

Replace token in text by the filename

I have a bunch of textfiles that contain the token -filename-. I need to replace it by the real path and filename of the file.
Thats what I have so far:
grep -lr -e '-filename-' *.txt | xargs sed -i
's/-filename-/therealname/g'
Is there a way to replace therealname with the name of the file?

Just do a bit more bash-fu
for x in *.txt; do
sed -i "s/-filename-/$x/g" $x;
done
Of course, the newlines are just for clarity. Feel free to cram that into one line.

like
for f in $(grep...) ; do sed -i "s,-filename-,$f,g" $f ; done
you mean?
With xargs it will be something like this.
grep ... | xargs -I% -n 1 sed -i "s,-filename-,%,g" %

for f in YOUR_FILE_LIST_MASK ; do
sed -i "s:-filename-:${f}" ${f}
done
can do it.

Trim text using sed

How do I remove the first and the last quotes?
echo "\"test\"" | sed 's/"//' | sed 's/"$//'
The above is working as expected, But I guess there must be a better way.

You can combine the sed calls into one:
echo "\"test\"" | sed 's/"//;s/"$//'
The command you posted will remove the first quote even if it's not at the beginning of the line. If you want to make sure that it's only done if it is at the beginning, then you can anchor it like this:
echo "\"test\"" | sed 's/^"//;s/"$//'
Some versions of sed don't like multiple commands separated by semicolons. For them you can do this (it also works in the ones that accept semicolons):
echo "\"test\"" | sed -e 's/^"//' -e 's/"$//'

Maybe you prefer something like this:
echo '"test"' | sed 's/^"\(.*\)"$/\1/'

if you are sure there are no other quotes besides the first and last, just use /g modifier
$ echo "\"test\"" | sed 's/"//g'
test
If you have Ruby(1.9+)
$ echo $s
blah"te"st"test
$ echo $s | ruby -e 's=gets.split("\"");print "#{s[0]}#{s[1..-2].join("\"")+s[-1]}"'
blahte"sttest
Note the 2nd example the first and last quotes which may not be exactly at the first and last positions.
example with more quotes
$ s='bl"ah"te"st"tes"t'
$ echo $s | ruby -e 's=gets.split("\"");print "#{s[0]}#{s[1..-2].join("\"")+s[-1]}"'
blah"te"st"test