i need to use variable in instead of direct date.
cat file | awk -F, '{ if ($1>"2012-08-20 11:30" && $1<"2012-08-22 16:00") print }'
thanks in advance
Based on your shown code, could you please try following and let me know if this helps you.(In lack of samples I haven't tested it)
awk -v date1="2012-08-20 11:30" -v date2="2012-08-22 16:00" -F, '($1>date1 && $1<date2)' Input_file
In case your variables are coming from shell to awk then following could help you on same, you could change date subtraction order as per your need too:
date1="2012-08-20 11:30"
date2="2012-08-22 16:00"
awk -v date_1="$date1" -v date_2="$date2" -F, '($1>date_1 && $1<date_2)' Input_file
Related
I want to simply use the two for loop variables in my awk code but I can't. Please help or guide me in the right direction.
for i in {30,60,100};
do
for j in {7,8};
do
awk -v x=$i -v y=$j '{if ($NF <=x) print $0}' S_$i.txt > S_$i_$j.txt;
done;
done
This was the error I received.
awk: fatal: cannot open file S_.txt for reading (No such file or directory). I saw this error.
S_$i_$j.txt is trying to access a variable named $i_. Use S_${i}_${j}.txt instead but also always quote your shell variables so it should really be:
awk -v x="$i" -v y="$j" '{if ($NF <= x) print $0}' "S_${i}.txt" > "S_${i}_${j}.txt"
or more awkishly:
awk -v x="$i" -v y="$j" '$NF <= x' "S_${i}.txt" > "S_${i}_${j}.txt"
and note that you never use y inside your awk script so it could just be:
awk -v x="$i" '$NF <= x' "S_${i}.txt" > "S_${i}_${j}.txt"
but then it's not clear why you'd want to create 2 copies of your output with each inner loop.
Whatever you're doing, though, could almost certainly be done much faster with a single call to awk than calling it multiple times within shell loops!
The problem you asked about has absolutely nothing to do with for loop variables in my awk code btw, it's all shell fundamentals.
Thanks for your quick response.
However, I tried the following and it worked:
for i in {30,60,100};
do
for j in {7,8};
do
awk -v x=$i -v y=$j '{if ($NF <=x) print $0}' "S_"$j".txt" > "S_"$j"_"$i".txt";
done;
done;
Additionally, I realized that S_30.txt didn't exist. So when I changed it to "S_"$j".txt" it worked fine. My bad on that one.
Hoping someone can help me make my awk commands more efficient please!
Let's say my text file has around 30 lines of this type of thing:
ENTIRE:11.3.28.4.0
OSVER:Solaris11
VARFREE:3G
I'm assigning these to variables in a bash script like this:
ENTIRE=$(awk -F\: '$1 ~ /ENTIRE/ {print $2}' $HOSTFILE)
RELEASE=$(awk -F\: '$1 ~ /RELEASE/ {print $2}' $HOSTFILE)
OSVER=$(awk -F\: '$1 ~ /OSVER/ {print $2}' $HOSTFILE)
Because I have around 30 of these, it means awk is run 30 times, which is slow, and clearly not the best way.
Can anyone suggest how I can build these into one awk command please?
Thanks in advance!
You don't need awk at all. If modifying the original file isn't an option, use a while loop and the declare command to define each variable.
while IFS=: read name value; do
declare "$name=$value"
done < "$HOSTFILE"
An example:
$ IFS=: read name value <<< "foo:bar"
$ declare "$name=$value"
$ echo "$foo"
bar
I have a .dat file with | separator and I want to change the value of the column which is defined by a number passed as argument and stored in a var. My code is
awk -v var="$value" -F'|' '{ FS = OFS = "|" } $1=="$id" {$"\{$var}"=8}1'
myfile.dat > tmp && mv tmp myfiletemp.dat
This changes the whole line to 8, obviously doesn't work. I was wondering what is the right way to write this part
{$"\{$var}"=8}1
For example, if I want to change the fourth column to 8 and I have value=4, how do I get {$4=8}?
The other answer is mostly correct, but just wanted to add a couple of notes, in case it wasn't totally clear.
Referring to a variable with a $ in front of it turns it in to a reference to the column. So i=3; print $i; print i will print the third column and then the number 3.
Putting all your variables in the command line will avoid any problems with trying to include bash variables inside your single-quoted awk code, which won't work.
You can let awk do the output to the specific file instead of relying on bash to redirect output and move files.
The -F option on the command line specifies FS for you, so no need to redeclare it in your code.
Here's how I would do this:
#!/bin/bash
column=4
value=8
id=1
awk -v col="$column" -v val="$value" -v id="$id" -F"|" '
BEGIN {OFS="|"}
{$1==id && $col=val; print > "myfiletemp.dat"}
' myfile.dat
you can refer to the awk variable directly by it's name, slight rewrite of your script with correct reference to column number var...
awk -F'|' -v var="$value" 'BEGIN{OFS=FS} $1=="$id"{$var=8}1'
should work as long as $value is a number. If id is another bash variable, pass it the same way as an awk variable
awk -F'|' -v var="$value" -v id="$id" 'BEGIN{OFS=FS} $1==id{$var=8}1'
Not only can you use a number in a variable by putting a $ in front of it, you can also use put a $ in front of an expression!
$ date | tee /dev/stderr | awk '{print $(2+2)}'
Mon Aug 3 12:47:39 CDT 2020
12:47:39
I have a file which contains data like below.
appid=TestApp
version=1.0.1
We want to parse the file and capture the value assigned to appid field.
I have tried with awk command as below
awk '/appid=/{print $1}' filename.txt
However it outputs the whole line
appid=TestApp
but we required only
TestApp
Please let me know how I can achieve this using awk/grep/sed shell commands.
You need to change the field separator:
awk -F'=' '$1 ~ /appid/ {print $2}' filename.txt
or with an exact match
awk -F'=' '$1 == "appid" {print $2}' filename.txt
outputs
TestApp
There's about 20 different ways to do this but it's usually a good idea when you have name = value statements in a file to simply build an array of those assignments and then just print whatever you care about using it's name, e.g.:
$ cat file
appid=TestApp
version=1.0.1
$
$ awk -F= '{a[$1]=$2} END{print a["appid"]}' file
TestApp
$ awk -F= '{a[$1]=$2} END{print a["version"]}' file
1.0.1
$ awk -F= '{a[$1]=$2} END{for (i in a) print i,"=",a[i]}' file
appid = TestApp
version = 1.0.1
If you are in the shell already then simply sourcing the file will let you get what you want.
. filename.txt
echo $appid
I know I can redirect awk's print output to another file from within a script, like this:
awk '{print $0 >> "anotherfile" }' 2procfile
(I know that's dummy example, but it's just an example...)
But what I need is to redirect output to another file, which has a dynamic name like this
awk -v MYVAR"somedinamicdata" '{print $0 >> "MYWAR-SomeStaticText" }' 2procfile
And the outpus should be redirected to somedinamicdata-SomeStaticText.
I know I can do it via:
awk '{print $0 }' 2procfile >> "$MYVAR-somedinamicdata"
But the problem is that it's a bigger awk script, and I have to output to several files depending on certain conditions (and this awk script is called from another bash, and it passes some dynamic variable via the -v switch... and son on.
Is it possible anyhow?
Thanks in advance.
i think
awk -v MYVAR="somedinamicdata" '{print $0 >> (MYVAR "-SomeStaticText") }' 2procfile
should do it. String concatenation in awk is just put one after another.