I'm trying to do something pretty simple but its appears more complicated than expected...
I've lines in a text file, separated by the comma and that I want to output to another file, without the first field.
Input:
echo file1,item, 12345678 | awk -F',' '{OFS = ";";$1=""; print $0}'
Output:
;item; 12345678
As you can see the spaces before 12345678 are kind of merged into one space only.
I also tried with the cut command:
echo file1,item, 12345678 | cut -d, -f2-
and I ended up with the same result.
Is there any workaround to handle this?
Actually my entire script is as follows:
cat myfile | while read l_line
do
l_line="'$l_line'"
v_OutputFile=$(echo $l_line | awk -F',' '{print $1}')
echo $(echo $l_line | cut -d, -f2-) >> ${v_OutputFile}
done
But stills in l_line all spaces but one are removed. I also created the quotes inside the file but same result.
it has nothing to do with awk. quote the string in your echo:
#with quotes
kent$ echo 'a,b, c'|awk -F, -v OFS=";" '{$1="";print $0}'
;b; c
#without quotes
kent$ echo a,b, c|awk -F, -v OFS=";" '{$1="";print $0}'
;b; c
The problem is with your invocation of the echo command you're using to feed awk the test data above. The shell is looking at this command:
echo file1,item, 12345678
and treating file1,item, and 12345678 as two separate parameters to echo. echo just prints all its parameters, separated by one space.
If you were to quote the whitespace, as follows:
echo 'file1,item, 12345678'
the shell would interpret this as a single parameter to feed to echo, so you'd get the expected result.
Update after edit to OP - having seen your full script, you could do this entirely in awk:
awk -F, '{ OFS = "," ; f = $1 ; sub("^[^,]*,","") ; print $0 >> f }' myfile
Related
I have a text file with the following structure:
bla1
bla2
bla3
bla4
bla5
So you can see that some lines of text are preceeded by an empty line.
I understand that sed has the concept of two buffers, a pattern space buffer and a hold space buffer, so I'm guessing these need to come in to play here, but I'm unclear how to specify them to accomplish what I need.
In my contrived example above, I'd expect to see the following lines outputted:
bla3
bla5
sed is for doing s/old/new on individual lines, that is all. Any time you start talking about buffers or doing anything related to multi-lines comparisons you're using the wrong tool.
You could do this with awk:
$ awk -v RS= -F'\n' 'NR>1{print $1}' file
bla3
bla5
but it would fail to print the first non-empty line if the first line(s) in the file were empty so this may be what you want if you want lines of all space chars considered to be empty lines:
$ awk 'NF && !p{print} {p=NF}' file
bla3
bla5
and this otherwise:
$ awk '($0!="") && (p==""){print} {p=$0}' file
bla3
bla5
All of the above will work even if there are multiple empty lines preceding any given non-empty line.
To see the difference between the 3 approaches (which you won't see given the sample input in the question):
PS1> printf '\nfoo\n \nbar\n\netc\n' | cat -E
$
foo$
$
bar$
$
etc$
PS1> printf '\nfoo\n \nbar\n\netc\n' | awk -v RS= -F'\n' 'NR>1{print $1}'
etc
PS1> printf '\nfoo\n \nbar\n\netc\n' | awk 'NF && !p{print} {p=NF}'
foo
bar
etc
PS1> printf '\nfoo\n \nbar\n\netc\n' | awk '($0!="") && (p==""){print} {p=$0}'
foo
etc
You can use the hold buffer easily to print the line before the blank like this:
sed -n -e '/^$/{x; p;}' -e h input
But I don't see an easy way to use it for your use case. For your case, instead of using the hold buffer, you could do:
sed -n -e '/^$/ba' -e d -e :a -e n -e p input
But I would do this with awk.
awk 'NR!=1{print $1}' RS= FS=\\n input-file
awk 'p;{p=/^$/}' file
above command does these for each line:
if p is 1, print line;
if line is empty, set p to 1.
if lines consisting of one or more spaces are also considered empty:
awk 'p;{p=!NF}' file
to print non-empty lines each coming right after an empty line, you can use this:
awk 'p*!(p=/^$/)' file
if p is 1 and this line is not empty (1*!(0) = 1*1 = 1), print this line;
otherwise (1*!(1) = 1*0 = 0, 0*anything = 0), don't print anything.
note that this one may not work with all awks, a portable version of this would look like:
awk 'p*(/./);{p=/^$/}' file
if lines consisting of one or more spaces are also considered empty:
awk 'p*NF;{p=!NF}' file
see them online here, and here.
If sed/awk is not mandatory, you can do it with grep:
grep -A 1 '^$' input.txt | grep -v -E '^$|--'
You can use sed to match a range of lines and do sub-matches inside the matches, like so:
# - use the "-n" option to omit printing of lines
# - match lines between a blank line (/^$/) and a non-blank one (/^./),
# then print only the line that contains at least a character,
# i.e, the non-blank line.
sed -ne '
/^$/,/^./ {
/^./{ p; }
}' input.txt
tested by gnu sed, your data in 'a':
$ sed -nE '/^$/{N;s/\n(.+)/\1/p}' a
bla3
bla5
add -i option precedes -n to real editing
I've got a file with following records:
depots/import/HDN1YYAA_15102018.txt;1;CAB001
depots/import/HDN1YYAA_20102018.txt;2;CLI001
depots/import/HDN1YYAA_20102018.txt;32;CLI001
depots/import/HDN1YYAA_25102018.txt;1;CAB001
depots/import/HDN1YYAA_50102018.txt;1;CAB001
depots/import/HDN1YYAA_65102018.txt;1;CAB001
depots/import/HDN1YYAA_80102018.txt;2;CLI001
depots/import/HDN1YYAA_93102018.txt;2;CLI001
When I execute following oneliner awk:
cat lignes_en_erreur.txt | awk 'FS=";"{ if(NR==1){print $1}}END {}'
the output is not the expected:
depots/import/HDN1YYAA_15102018.txt;1;CAB001
While I am suppose get only the frist column:
If I run it through all the records:
cat lignes_en_erreur.txt | awk 'FS=";"{ if(NR>0){print $1}}END {}'
then it will start filtering only after the second line and I get the following output:
depots/import/HDN1YYAA_15102018.txt;1;CAB001
depots/import/HDN1YYAA_20102018.txt
depots/import/HDN1YYAA_20102018.txt
depots/import/HDN1YYAA_25102018.txt
depots/import/HDN1YYAA_50102018.txt
depots/import/HDN1YYAA_65102018.txt
depots/import/HDN1YYAA_80102018.txt
depots/import/HDN1YYAA_93102018.txt
Does anybody knows why awk is skiping the first line only.
I tried deleting first record but the behaviour is the same, it will skip the first line.
First, it should be
awk 'BEGIN{FS=";"}{ if(NR==1){print $1}}END {}' filename
You can omit the END block if it is empty:
awk 'BEGIN{FS=";"}{ if(NR==1){print $1}}' filename
You can use the -F command line argument to set the field delimiter:
awk -F';' '{if(NR==1){print $1}}' filename
Furthermore, awk programs consist of a sequence of CONDITION [{ACTIONS}] elements, you can omit the if:
awk -F';' 'NR==1 {print $1}' filename
You need to specify delimiter in either BEGIN block or as a command-line option:
awk 'BEGIN{FS=";"}{ if(NR==1){print $1}}'
awk -F ';' '{ if(NR==1){print $1}}'
cut might be better suited here, for all lines
$ cut -d';' -f1 file
to skip the first line
$ sed 1d file | cut -d';' -f1
to get the first line only
$ sed 1q file | cut -d';' -f1
however at this point it's better to switch to awk
if you have a large file and only interested in the first line, it's better to exit early
$ awk -F';' '{print $1; exit}' file
I have following lines in a file. Please note I have intentionally kept the extra hash between 2 and 0 in the 2nd line.
File name : test.txt
Name#|#Age#|#Dept
AC#|#2#0#|#Science
BC#|#22#|#Commerce
I am using awk to get the data in Dept column
awk -F "#|#" -v c="Dept" 'NR==1{for (i=1; i<=NF; i++) if ($i==c){p=i; break}; next} {print $p}' "test.txt" >> result.txt
The result.txt shows me the following
|
Commerce
The first line is coming as pipe because if the extra # in the first line.
Can anyone help on this
Currently the meaning of the delimiter set is: match # or #. The pipe | character in this case acts as an OR statement; instead try using:
awk -F '#[|]#' ...
Putting | into a character class [ ... ] awk will match it literally.
If you desire to extract Dept in your content, here's a good choice you can choose from,
awk -F'#' 'NR>1{print $NF}' test.txt
output:
Science
Commerce
I have lines like this:
Volume.Free_IBM_LUN59_28D: 2072083693568
I would like to get only IBM_LUN59_28D from this line using awk.
Thanks
You can use sub to do substitutions on each input line, as per the following transcript:
pax> echo 'Volume.Free_IBM_LUN59_28D: 2072083693568' | awk '
...> {
...> sub (".*Free_", "");
...> sub (":.*", "");
...> print
...> }'
IBM_LUN59_28D
That command crosses multiple lines for readability but, if you're operating on a file and not too concerned about readability, you can just use the compressed version:
awk '{sub(".*Free_","");sub(":.*","");print}' inputFile
If you're amenable to non-awk solutions, you could also use sed:
sed -e 's/.*Free_//' -e 's/:.*//' inputFile
Note that both those solutions rely on your (somewhat sparse) test data. If your definition of "like" includes preceding textual segments other than Free_ or subsequent characters other than :, some more work may be needed.
For example, if you wanted the string between the first _ and the first :, you could use:
awk '{sub("[^_]*_","");sub(":.*","");print}'
With sed:
sed 's/[^_]*_\(.*\):.*/\1/'
Search for sequence of non _ characters followed by _ (this will match Volume.Free_), then another sequence of characters (this will match IBM_LUN59_28D, we group this for future use), followed by : and any char sequence. Substitute with the saved pattern (\1). That's it.
Sample:
$ echo "Volume.Free_IBM_LUN59_28D: 2072083693568" | sed 's/[^_]*_\(.*\):.*/\1/'
IBM_LUN59_28D
Here is one awk
awk -F"Free_" 'NF>1{split($2,a,":");print a[1]}'
Eks:
echo "Volume.Free_IBM_LUN59_28D: 2072083693568" | awk -F"Free_" 'NF>1{split($2,a,":");print a[1]}'
IBM_LUN59_28D
It divides the line by Free_.
If line then have more than one field NF>1 then:
Split second field bye : and print first part a[1]
With awk:
echo "$val" | awk -F: '{print $1}' | awk -F. '{print $2}' | awk '{print substr($0,6)}'
where the given string is in $val.
I'm trying different commands to process csv file where the separator is the pipe | character.
While those commands do work when the comma is a separator, it throws an error when I replace it with the pipe:
awk -F[|] "NR==FNR{a[$2]=$0;next}$2 in a{ print a[$2] [|] $4 [|] $5 }" OFS=[|] file1.csv file2.csv
awk "{print NR "|" $0}" file1.csv
I tried, "|", [|], /| to no avail.
I'm using Gawk on windows. What I'm I missing?
You tried "|", [|] and /|. /| does not work because the escape character is \, whereas [] is used to define a range of fields, for example [,-] if you want FS to be either , or -.
To make it work "|" is fine, are you sure you used it this way? Alternativelly, escape it --> \|:
$ echo "he|llo|how are|you" | awk -F"|" '{print $1}'
he
$ echo "he|llo|how are|you" | awk -F\| '{print $1}'
he
$ echo "he|llo|how are|you" | awk 'BEGIN{FS="|"} {print $1}'
he
But then note that when you say:
print a[$2] [|] $4 [|] $5
so you are not using any delimiter at all. As you already defined OFS, do:
print a[$2], $4, $5
Example:
$ cat a
he|llo|how are|you
$ awk 'BEGIN {FS=OFS="|"} {print $1, $3}' a
he|how are
For anyone finding this years later: ALWAYS QUOTE SHELL METACHARACTERS!
I think gawk (GNU awk) treats | specially, so it should be quoted (for awk). OP had this right with [|]. However [|] is also a shell pattern. Which in bash at least, will only expand if it matches a file in the current working directory:
$ cd /tmp
$ echo -F[|] # Same command
-F[|]
$ touch -- '-F|'
$ echo -F[|] # Different output
-F|
$ echo '-F[|]' # Good quoting
-F[|] # Consistent output
So it should be:
awk '-F[|]'
# or
awk -F '[|]'
awk -F "[|]" would also work, but IMO, only use soft quotes (") when you have something to actually expand (or the string itself contains hard quotes ('), which can't be nested in any way).
Note that the same thing happens if these characters are inside unquoted variables.
If text or a variable contains, or may contain: []?*, quote it, or set -f to turn off pathname expansion (a single, unmatched square bracket is technically OK, I think).
If a variable contains, or may contain an IFS character (space, tab, new line, by default), quote it (unless you want it to be split). Or export IFS= first (bearing the consequences), if quoting is impossible (eg. a crazy eval).
Note: raw text is always split by white space, regardless of IFS.
Try to escape the |
echo "more|data" | awk -F\| '{print $1}'
more
You can escape the | as \|
$ cat test
hello|world
$ awk -F\| '{print $1, $2}' test
hello world