How to print files names from file where awk is selecting values?
I have a .txt file having files names as
z1.cap
z2.cap
z3.cap
z4.cap
Sample data present in these files are like shown below,
OTR 25896 PAT210 $TREMD DEST
OFR 21475 NAT102 #TREMD DEST
then I'm using below code to print desired values from files.
while read file_name
do
echo "progressing with file :${file_name}"
cat ${file_name} | grep "PAT210" | awk -F' ' '$5 == "(DEST" { print $file_name, $1}' | uniq >> OUTPUT_FILE
Now I want output which consists of 2 fields like,
z1.cap OTR
z2.cap OFR
and so on...
But i'm getting ouputs like,
- OTR
- OFR
Any help is aprreciated, Thanks.
linux shell unix
asked Nov 21 '18 at 11:35
karthikkarthik
64
add a comment |
I have a .txt file having files names as
z1.cap
z2.cap
z3.cap
z4.cap
Sample data present in these files are like shown below,
OTR 25896 PAT210 $TREMD DEST
OFR 21475 NAT102 #TREMD DEST
then I'm using below code to print desired values from files.
while read file_name
do
echo "progressing with file :${file_name}"
cat ${file_name} | grep "PAT210" | awk -F' ' '$5 == "(DEST" { print $file_name, $1}' | uniq >> OUTPUT_FILE
Now I want output which consists of 2 fields like,
z1.cap OTR
z2.cap OFR
and so on...
But i'm getting ouputs like,
- OTR
- OFR
Any help is aprreciated, Thanks.
linux shell unix
asked Nov 21 '18 at 11:35
karthikkarthik
64
I am trying to understand what you want by reading the code that has problems. Can you explain why you findz2.cap OFRwhen the code looks forPAT210?
– Walter A
Nov 21 '18 at 21:58
Inside theawkcode, shell variables are not replaced. For a filename input you can use FILENAME, in other cases useawk -v awkvariable=$shellvariable '...awkcode...'.
– Walter A
Nov 21 '18 at 22:01
@WalterA I am also at a loss as to what the OP is trying to achieve. It took some back and forth to get the question into its current state. Decided to leave well-enough alone.
– Niall Cosgrove
Nov 22 '18 at 20:30
add a comment |
I have a .txt file having files names as
z1.cap
z2.cap
z3.cap
z4.cap
Sample data present in these files are like shown below,
OTR 25896 PAT210 $TREMD DEST
OFR 21475 NAT102 #TREMD DEST
then I'm using below code to print desired values from files.
while read file_name
do
echo "progressing with file :${file_name}"
cat ${file_name} | grep "PAT210" | awk -F' ' '$5 == "(DEST" { print $file_name, $1}' | uniq >> OUTPUT_FILE
Now I want output which consists of 2 fields like,
z1.cap OTR
z2.cap OFR
and so on...
But i'm getting ouputs like,
- OTR
- OFR
Any help is aprreciated, Thanks.
linux shell unix
asked Nov 21 '18 at 11:35
karthikkarthik
64
I have a .txt file having files names as
z1.cap
z2.cap
z3.cap
z4.cap
Sample data present in these files are like shown below,
OTR 25896 PAT210 $TREMD DEST
OFR 21475 NAT102 #TREMD DEST
then I'm using below code to print desired values from files.
while read file_name
do
echo "progressing with file :${file_name}"
cat ${file_name} | grep "PAT210" | awk -F' ' '$5 == "(DEST" { print $file_name, $1}' | uniq >> OUTPUT_FILE
Now I want output which consists of 2 fields like,
z1.cap OTR
z2.cap OFR
and so on...
But i'm getting ouputs like,
- OTR
- OFR
Any help is aprreciated, Thanks.
linux shell unix
linux shell unix
asked Nov 21 '18 at 11:35
karthikkarthik
64
asked Nov 21 '18 at 11:35
karthikkarthik
64
edited Nov 21 '18 at 12:43
karthik
asked Nov 21 '18 at 11:35
karthikkarthik
64
asked Nov 21 '18 at 11:35
karthikkarthik
64
asked Nov 21 '18 at 11:35
karthikkarthik
64
64
I am trying to understand what you want by reading the code that has problems. Can you explain why you findz2.cap OFRwhen the code looks forPAT210?
– Walter A
Nov 21 '18 at 21:58
Inside theawkcode, shell variables are not replaced. For a filename input you can use FILENAME, in other cases useawk -v awkvariable=$shellvariable '...awkcode...'.
– Walter A
Nov 21 '18 at 22:01
@WalterA I am also at a loss as to what the OP is trying to achieve. It took some back and forth to get the question into its current state. Decided to leave well-enough alone.
– Niall Cosgrove
Nov 22 '18 at 20:30
add a comment |
I am trying to understand what you want by reading the code that has problems. Can you explain why you findz2.cap OFRwhen the code looks forPAT210?
– Walter A
Nov 21 '18 at 21:58
Inside theawkcode, shell variables are not replaced. For a filename input you can use FILENAME, in other cases useawk -v awkvariable=$shellvariable '...awkcode...'.
– Walter A
Nov 21 '18 at 22:01
@WalterA I am also at a loss as to what the OP is trying to achieve. It took some back and forth to get the question into its current state. Decided to leave well-enough alone.
– Niall Cosgrove
Nov 22 '18 at 20:30
I am trying to understand what you want by reading the code that has problems. Can you explain why you find
z2.cap OFR when the code looks for PAT210?– Walter A
Nov 21 '18 at 21:58
I am trying to understand what you want by reading the code that has problems. Can you explain why you find
z2.cap OFR when the code looks for PAT210?– Walter A
Nov 21 '18 at 21:58
Inside the
awk code, shell variables are not replaced. For a filename input you can use FILENAME, in other cases use awk -v awkvariable=$shellvariable '...awkcode...'.– Walter A
Nov 21 '18 at 22:01
Inside the
awk code, shell variables are not replaced. For a filename input you can use FILENAME, in other cases use awk -v awkvariable=$shellvariable '...awkcode...'.– Walter A
Nov 21 '18 at 22:01
@WalterA I am also at a loss as to what the OP is trying to achieve. It took some back and forth to get the question into its current state. Decided to leave well-enough alone.
– Niall Cosgrove
Nov 22 '18 at 20:30
@WalterA I am also at a loss as to what the OP is trying to achieve. It took some back and forth to get the question into its current state. Decided to leave well-enough alone.
– Niall Cosgrove
Nov 22 '18 at 20:30
add a comment |
1 Answer
1
active
oldest
votes
To access the filename that awk is currently processing use the builtin variable FILENAME
To bind other shell variables from your shell to variables in awk use:
awk -v var1=$shvar1 -v var2=$shvar2 'your awk code using var1 and var2'
Assuming files.txt contains your list of files and with zero understanding of what exactly you are trying to achieve:
for file_name in $(cat files.txt)
do
echo "progressing with file :${file_name}"
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $file_name | uniq >> OUTPUT_FILE
done
I removed the cat and incorporated the grep into your awk. The cat was unnecessary since awk can read the file itself.
You can remove the for loop entirely by saying
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $(<files.txt) | uniq >> OUTPUT_FILE
The $(<files.txt) will send each filename to awk.
answered Nov 21 '18 at 13:01
Niall CosgroveNiall Cosgrove
1,1681923
Here, we need to traverse all the 4 files. i.e, z5.cap, z6.cap and so on. if you remove cat how we are traversing?
– karthik
Nov 21 '18 at 13:09
The way you have written it,read file_namegets the names from user input. I can modify it to read the names from the .txt file if you like.
– Niall Cosgrove
Nov 21 '18 at 13:12
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
To access the filename that awk is currently processing use the builtin variable FILENAME
To bind other shell variables from your shell to variables in awk use:
awk -v var1=$shvar1 -v var2=$shvar2 'your awk code using var1 and var2'
Assuming files.txt contains your list of files and with zero understanding of what exactly you are trying to achieve:
for file_name in $(cat files.txt)
do
echo "progressing with file :${file_name}"
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $file_name | uniq >> OUTPUT_FILE
done
I removed the cat and incorporated the grep into your awk. The cat was unnecessary since awk can read the file itself.
You can remove the for loop entirely by saying
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $(<files.txt) | uniq >> OUTPUT_FILE
The $(<files.txt) will send each filename to awk.
answered Nov 21 '18 at 13:01
Niall CosgroveNiall Cosgrove
1,1681923
Here, we need to traverse all the 4 files. i.e, z5.cap, z6.cap and so on. if you remove cat how we are traversing?
– karthik
Nov 21 '18 at 13:09
The way you have written it,read file_namegets the names from user input. I can modify it to read the names from the .txt file if you like.
– Niall Cosgrove
Nov 21 '18 at 13:12
add a comment |
To access the filename that awk is currently processing use the builtin variable FILENAME
To bind other shell variables from your shell to variables in awk use:
awk -v var1=$shvar1 -v var2=$shvar2 'your awk code using var1 and var2'
Assuming files.txt contains your list of files and with zero understanding of what exactly you are trying to achieve:
for file_name in $(cat files.txt)
do
echo "progressing with file :${file_name}"
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $file_name | uniq >> OUTPUT_FILE
done
I removed the cat and incorporated the grep into your awk. The cat was unnecessary since awk can read the file itself.
You can remove the for loop entirely by saying
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $(<files.txt) | uniq >> OUTPUT_FILE
The $(<files.txt) will send each filename to awk.
answered Nov 21 '18 at 13:01
Niall CosgroveNiall Cosgrove
1,1681923
Here, we need to traverse all the 4 files. i.e, z5.cap, z6.cap and so on. if you remove cat how we are traversing?
– karthik
Nov 21 '18 at 13:09
The way you have written it,read file_namegets the names from user input. I can modify it to read the names from the .txt file if you like.
– Niall Cosgrove
Nov 21 '18 at 13:12
add a comment |
To access the filename that awk is currently processing use the builtin variable FILENAME
To bind other shell variables from your shell to variables in awk use:
awk -v var1=$shvar1 -v var2=$shvar2 'your awk code using var1 and var2'
Assuming files.txt contains your list of files and with zero understanding of what exactly you are trying to achieve:
for file_name in $(cat files.txt)
do
echo "progressing with file :${file_name}"
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $file_name | uniq >> OUTPUT_FILE
done
I removed the cat and incorporated the grep into your awk. The cat was unnecessary since awk can read the file itself.
You can remove the for loop entirely by saying
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $(<files.txt) | uniq >> OUTPUT_FILE
The $(<files.txt) will send each filename to awk.
answered Nov 21 '18 at 13:01
Niall CosgroveNiall Cosgrove
1,1681923
To access the filename that awk is currently processing use the builtin variable FILENAME
To bind other shell variables from your shell to variables in awk use:
awk -v var1=$shvar1 -v var2=$shvar2 'your awk code using var1 and var2'
Assuming files.txt contains your list of files and with zero understanding of what exactly you are trying to achieve:
for file_name in $(cat files.txt)
do
echo "progressing with file :${file_name}"
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $file_name | uniq >> OUTPUT_FILE
done
I removed the cat and incorporated the grep into your awk. The cat was unnecessary since awk can read the file itself.
You can remove the for loop entirely by saying
awk -F' ' '($5 == "DEST") && ($3=="PAT210") { print FILENAME, $1}' $(<files.txt) | uniq >> OUTPUT_FILE
The $(<files.txt) will send each filename to awk.
answered Nov 21 '18 at 13:01
Niall CosgroveNiall Cosgrove
1,1681923
edited Nov 22 '18 at 20:24
answered Nov 21 '18 at 13:01
Niall CosgroveNiall Cosgrove
1,1681923
answered Nov 21 '18 at 13:01
Niall CosgroveNiall Cosgrove
1,1681923
answered Nov 21 '18 at 13:01
Niall CosgroveNiall Cosgrove
1,1681923
1,1681923
Here, we need to traverse all the 4 files. i.e, z5.cap, z6.cap and so on. if you remove cat how we are traversing?
– karthik
Nov 21 '18 at 13:09
The way you have written it,read file_namegets the names from user input. I can modify it to read the names from the .txt file if you like.
– Niall Cosgrove
Nov 21 '18 at 13:12
add a comment |
Here, we need to traverse all the 4 files. i.e, z5.cap, z6.cap and so on. if you remove cat how we are traversing?
– karthik
Nov 21 '18 at 13:09
The way you have written it,read file_namegets the names from user input. I can modify it to read the names from the .txt file if you like.
– Niall Cosgrove
Nov 21 '18 at 13:12
Here, we need to traverse all the 4 files. i.e, z5.cap, z6.cap and so on. if you remove cat how we are traversing?
– karthik
Nov 21 '18 at 13:09
Here, we need to traverse all the 4 files. i.e, z5.cap, z6.cap and so on. if you remove cat how we are traversing?
– karthik
Nov 21 '18 at 13:09
The way you have written it,
read file_name gets the names from user input. I can modify it to read the names from the .txt file if you like.– Niall Cosgrove
Nov 21 '18 at 13:12
The way you have written it,
read file_name gets the names from user input. I can modify it to read the names from the .txt file if you like.– Niall Cosgrove
Nov 21 '18 at 13:12
add a comment |
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I am trying to understand what you want by reading the code that has problems. Can you explain why you find
z2.cap OFRwhen the code looks forPAT210?– Walter A
Nov 21 '18 at 21:58
Inside the
awkcode, shell variables are not replaced. For a filename input you can use FILENAME, in other cases useawk -v awkvariable=$shellvariable '...awkcode...'.– Walter A
Nov 21 '18 at 22:01
@WalterA I am also at a loss as to what the OP is trying to achieve. It took some back and forth to get the question into its current state. Decided to leave well-enough alone.
– Niall Cosgrove
Nov 22 '18 at 20:30