I have the following file called testfile, with the following contents:
{"items":[{"ogit\/_created on":1413388512511,"\/environmentType":"PROD","\/soxRelevant":"true","\/environmentName":"dbVertical"}]}
I used the following awk command to get the values for soxRelevant, environment type and environment name.
cat test file | tr -d '"' | awk 'BEGIN {RS=","; FS=":"; ORS=",";} /soxRelevant/ {print $2}; /environmentType/ {print $2}; /environmentName/ {print $2};'
The output was as follows
PROD,true,dbVertical,
However I want the soxRelevant output first followed by environment type then environment name, as specified in the awk command:
I want the output to be:
true, PROD, dbVertical
How do I do this?
You could push the elements to an array then print them in the END block:
awk 'BEGIN {RS=ORS=","; FS=":"}
/soxRelevant/ {a[1]=$2}
/environmentType/ {a[2]=$2}
/environmentName/ {a[3]=$2}
END{for(n=1;n<=3;++n)print a[n]}'
Related
I am new to writing shell scripts
I am trying to write an AWK command which does exactly the below
cut --complement -c $IGNORE_RANGE file.txt > tmp
$IGNORE_RANGE can be of any value say, 1-5 or 5-10 etc
i cannot use cut since i am in AIX and AIX does not support --complement, is there any way to achieve this using AWK command
Example:
file.txt
abcdef
123456
Output
cut --complement -c 1-2 file.txt > tmp
cdef
3456
cut --complement -c 4-5 file.txt > tmp
abcf
1236
cut --complement -c 1-5 file.txt > tmp
f
6
Could you please try following, written and tested with shown samples. We have range variable of awk which should be in start_of_position-end_of_position and we could pass it as per need.
awk -v range="4-5" '
BEGIN{
split(range,array,"-")
}
{
print substr($0,1,array[1]-1) substr($0,array[2]+1)
}
' Input_file
OR to make it more clear in understanding wise try following:
awk -v range="4-5" '
BEGIN{
split(range,array,"-")
start=array[1]
end=array[2]
}
{
print substr($0,1,start-1) substr($0,end+1)
}
' Input_file
Explanation: Adding detailed explanation for above.
awk -v range="4-5" ' ##Starting awk program from here creating range variable which has range value of positions which we do not want to print in lines.
BEGIN{ ##Starting BEGIN section of this program from here.
split(range,array,"-") ##Splitting range variable into array with delimiter of - here.
start=array[1] ##Assigning 1st element of array to start variable here.
end=array[2] ##Assigning 2nd element of array to end variable here.
}
{
print substr($0,1,start-1) substr($0,end+1) ##Printing sub-string of current line from 1 to till value of start-1 and then printing from end+1 which basically means will skip that range of characters which OP does not want to print.
}
' Input_file ##Mentioning Input_file name here.
You can do this in awk:
awk -v st=1 -v en=2 '{print substr($0, 1, st-1) substr($0, en+1)}' file
cdef
3456
Or:
awk -v st=4 -v en=5 '{print substr($0, 1, st-1) substr($0, en+1)}' file
abcf
1236
I would like to modify several shell variables within awk:
echo "$LINE_IN" | awk '/pattern1/ {print $0; WRITTEN=1; REC=$REC+1}' >> $FILE1
I tried to put eval, but still does not work:
eval $( echo "$LINE_IN" | awk '/pattern1/ {print $0; WRITTEN=1; REC=$REC+1}' >> $FILE1 )
Any suggestion?
I would like to use k-shell script, thanks!
Count the hits when you are finished:
echo "${LINE_IN}" | grep -E 'pattern1' > "${FILE1}"
REC=$(wc -l < "${FILE1}")
if (( REC > 0 )); then
WRITTEN=1
fi
When you really want to use awk, you must let awk write the results to stdout and parse stdout:
echo "${LINE_IN}" | awk '/echo/ {print $0 > "x3"; WRITTEN=1; REC++}
END { print "WRITTEN=" WRITTEN; print "REC=" REC}'
WRITTEN=1
REC=6
And when you want the variables really set, wrap it:
source (echo "${LINE_IN}" | awk '/echo/ {print $0 > "x3"; WRITTEN=1; REC++}
END { print "WRITTEN=" WRITTEN; print "REC=" REC}')
Note: Get used to using lowercase variable names like written, file and rec.
I am trying to add the filename to the end of each line as a new field. It works except instead of getting the filename I get -.
Base file:
070323111|Hudson
What I want:
070323111|Hudson|20150106.csv
What I get:
070323111|Hudson|-
This is my code:
mv $1 $1.bak
cat $1.bak | awk '{print $0 "|" FILENAME}' > $1
- is the way to present the filename when there is not such info. Since your are doing cat $1.bak | awk ..., awk is not reading from a file but from stdin.
Instead, just do:
awk '...' file
in your case:
awk '{print $0 "|" FILENAME}' $1.bak > $1
From man awk:
FILENAME
The name of the current input file. If no files are specified on the
command line, the value of FILENAME is “-”. However, FILENAME is
undefined inside the BEGIN rule (unless set by getline).
I have a function
xyz()
{
x=$1*2
echo x
}
then I want to use it to replace a particular column in a csv file by awk.
File input.csv:
abc,2,something
def,3,something1
I want output like:
abc,4,somthing
def,6,something1
Command used:
cat input.csv|awk -F, -v v="'"`xyz "$2""'" 'BEGIN {FS=","; OFS=","} {$2=v1; print $0}'
Open file input.csv, calling function xyz by passing file 2nd filed as argument and result is stored back to position 2 of file, but is not working!
If I put constant in place of $2 while calling function it works:
Please help me to do this.
cat input.csv|awk -F, -v v="'"`xyz "14""'" 'BEGIN {FS=","; OFS=","} {$2=v1; print $0}'
This above line of code is working properly by calling the xyz function and putting the result back to 2nd column of file input.csv, but with only 14*2, as 14 is taken as constant.
There's a back-quote missing from your command line, and a UUOC (Useless Use of Cat), and a mismatch between variable v on the command line and v1 in the awk program:
cat input.csv|awk -F, -v v="'"`xyz "$2""'" 'BEGIN {FS=","; OFS=","} {$2=v1; print $0}'
^ Here ^ Here ^ Here
That should be written using $(…) instead:
awk -F, -v v="'$(xyz "$2")'" 'BEGIN {FS=","; OFS=","} {$2=v; print $0}' input.csv
This leaves you with a problem, though; the function xyz is invoked once by the shell before you start your awk script running, and is never invoked by awk. You simply can't do it that way. However, you can define your function in awk (and on the fly):
awk -F, 'BEGIN { FS = ","; OFS = "," }
function xyz(a) { return a * 2 }
{ $2 = xyz($2); print $0 }' \
input.csv
For your two-line input file, it produces your desired output.
I am struggling with an awk problem in my bash shell script. In the below snippet of code i am passing a variable var_awk for regular expression in awk. The idea is to get lines above a regular expression but the below echo is not displaying any data
echo `ls -ltr $date*$f* | /usr/xpg4/bin/awk -v reg=$var_awk '/reg/ {print $0}'`
I am unable to reg for regex though when i do print reg it is printing but when not doing regex as expected.
if [ $GE == "HBCA" ] || [ $GE == "HBUS" ] || [ $GE == "HBEU" ]; then
for f in `ls -ltr $date*GEN*REVAL*log|grep -v LPD | awk '{split($9,a,"_")}{print a[3]}'`; do
echo $f
var_awk="$date"_RESET_CALC_"$f"
echo $var_awk
echo `ls -ltr $date*$f* | /usr/xpg4/bin/awk -v reg=$var_awk '/reg/ {print $0}'`
You cannot use variable in regex that way. You need to do:
/usr/xpg4/bin/awk -v reg="$var_awk" '$0~reg{ print $0 }'
or simply
/usr/xpg4/bin/awk -v reg="$var_awk" '$0~reg'
Inside / / your variable reg will be used as a literal word.
Quote your shell variables.
try this:
...whatever you had already..|awk -v reg="$var_awk" '$0~reg'
it is better to wrap shell variable with quotes, e.g. if your var has spaces.
/pattern/ in awk is called regex constant. It cannot be used with variable, that's why it is called constant. We need to use dynamic regex here in this example.