I am trying to search for a pattern and from the results i am extracting just the second column. The command works well in command line but not inside a bash script.
#!/bin/bash
set a = grep 'NM_033356' test.txt | awk '{ print $2 }'
echo $a
It doesnt print any output at all.
Input
NM_033356 2
NM_033356 5
NM_033356 7
Your code:
#!/bin/bash
set a = grep 'NM_033356' test.txt | awk '{ print $2 }'
echo $a
Change it to:
#!/bin/bash
a="$(awk '$1=="NM_033356"{ print $2 }' test.txt)"
echo "$a"
Code changes are based on your sample input.
.......
a="$(awk '/NM_033356/ { print $2 }' test.txt)"
Try this:
a=`grep 'NM_033356' test.txt | awk '{ print $2 }'`
Related
I am facing some issue when I am reading the 3rd word(a hex string) of each line in a text file and compare it with a hex number. Can some one please help me on it.
#!/bin/bash
A=$1
cat $A | while read a; do
a1=$(echo \""$a"\" | awk '{ print $3 }')
#echo $a > cut -d " " -f 3
echo $a1
(("$a1" == 0x10F7))
echo $?
done
But when I use below, the comparison happens correctly,
a1= 0xADCAFE
(( "$a1" == 0x10F7 ))
echo $?
Then why it is showing issue when I read like below,
a1=$(echo \""$a"\" | awk '{ print $3 }')
or> a1=$(echo $a | awk '{ print $3 }')
echo $a prints intended hex value, but comparison does not happen.
Regards,
Running Awk inside a while read loop is an antipattern. Just do the loop in Awk; it's good at that.
awk '$3 == 4343' "$1"
If you want to compare against a string whose value is "0x10F7" then it's
awk '$3 == "0x10F7"' "$1"
If you want to match either, case insensitively etc, a regex is a good way to do that.
awk '$3 ~ /^(0x10[Ff]7|4343)$/' "$1"
Notice how the $1 in double quotes is handled by the shell, and gets replaced by a (properly quoted!) copy of the script's first command-line argument before Awk runs, while the Awk script in single quotes has its own namespace, so $3 is an Awk variable which refers to the third field in the current input line.
Either way, avoid the useless use of cat and always always always quote variables which contain file names with double quotes.
That's literal double quotes. You seem to have tried both a dangerous bare $a and a doubly double-quoted "\"$a\"" where the simple "$a" would be what you actually want.
Thank you all for your responses, Now my script is working fine. I was trying to match two files, below script does the purpose
#!/bin/bash
A=$1
B=$2
dos2unix -f "$A"
dos2unix -f "$B"
rm search_match.txt search_data_match.txt search_nomatch.txt search_data_nomatch.txt
while read line;do
search_word=$(echo $line | awk '{ print $1 }')
grep "$search_word" $B >> temp_file.txt
while read var;do
file1_hex=$(echo $line | awk '{ print $2 }')
file2_hex=$(echo $var | awk '{ print $3 }')
(("$file1_hex" == "$file2_hex"))
zero=$(echo $?)
if [ "$zero" -eq 0 ] ; then
echo $line >> search_match.txt
echo $var >> search_data_match.txt
else
echo $line >> search_nomatch.txt
echo $var >> search_data_nomatch.txt
fi
done < "temp_file.txt"
rm temp_file.txt
done < "$A"
I am making a bash script. I have to get 3 variables
VAR1=$(cat /path to my file/ | grep "string1" | awk '{ print $2 }'
VAR2=$(cat /path to my file/ | grep "string2" | awk '{ print $2 }'
VAR3=$(cat /path to my file/ | grep "string3" | awk '{ print $4 }'
My problem is that if I write
echo $VAR1
echo $VAR2
echo $VAR3
I can see values correctly
But when I try to write them in one line like this
echo "VAR1: $VAR1 VAR2: $VAR2 VAR3: $VAR3"
Value from $VAR3 is written at the beginning of output overwritting values of $VAR1 and $VAR2
I expect my explanation had been clear. Any doubt please let me know
Thanks and regards.
Rambert
It seems to me that $VAR3 contains \r which in some shells will move the cursor to the beginning of the line. Use printf instead:
printf "VAR1: %s VAR2: %s VAR3: %s\n" "$VAR1" "$VAR2" "$VAR3"
Also note that the way you extract the values is highly inefficient and can be reduced to one call to awk:
read -r var1 var2 var3 _ < <(awk '/string1/ { a=$2 }
/string2/ { b=$2 }
/string3/ { c=$4 }
END { print(a, b, c) }' /path/to/file)
printf "VAR1: %s VAR2: %s VAR3: %s\n" "$var1" "$var2" "$var3"
A nitpick is that uppercase variable names are reserved for environment variables, so I changed all to lowercase.
<(...) is a process substitution and will make ... write to a "file" and return the file name:
$ echo <(ls)
/dev/fd/63
And command < file is a redirection changing standard input of command to be comming from the file file.
You could write :
cat /path to my file/ | grep "string1" | awk '{ print $2 }'
as
awk '/string1/{print $2}' /path/to/file
In other words you could do with awk alone what you intended to do with cat, grep & awk
So finally get :
VAR1=$(awk '/string1/{print $2}' /path/to/file) #mind the closing ')'
Regarding the issue you face, it looks like you have carriage returns or \r in your variables. In bash echo will not interpret escape sequences without the -e option, but the printf option which [ #andlrc ] pointed out is a good try though as he mentioned in his [ answer ]
which in some shells will move the cursor to the beginning
Notes :
Another subtle point to keep in mind is to avoid using upper case variable names like VAR1 for user scripts. So replace it with var1 or so
When assigning values to variable spaces are not allowed around =, so
VAR1="Note there are no spaces around = sign"
is the right usage
I am trying to get the md5 of every individual line item and dump each md5 into a line on the next file (.md5). The below script echos everything to the screen. How do I redirect the echo output to the .md5 file.
more email/test |
while
read line;
md5=`md5sum $line | awk '{ print $1 }'`
do echo -n $md5;
done < .md5
Using bash
Try:
while IFS= read -r line;
do
md5sum $line | awk '{ print $1 }'
done <email/test >.md5
Using awk
The bash loop is unnecessary:
awk '{ "md5sum " $0 | getline; print $1}' email/test >.md5
I have the following stored in a file named tmp.txt
user/config/jars/content-config-factory-3.2.0.0.jar
I need to store this word to a variable -
$variable=content-config-factory
I have written the following
while read line
do
var=$(echo $line | awk 'BEGIN{FS="\/"; OFS=" "} {print $NF}' )
var=$(echo $var | awk 'BEGIN{FS="-"; OFS=" "} {print $(1)}' )
echo $var
done < tmp.txt
This returns the result "content" instead of "content-config-factory".
Can anyone please tell me how to extract a word between two characters from a string efficiently.
An awk solution would be like
awk -F/ '{sub("-[^-]+$", "", $NF); print $NF}
Test
$ echo "user/config/jars/content-config-factory-3.2.0.0.jar" | awk -F/ '{sub("-[^-]+$", "", $NF); print $NF}'
content-config-factory
You can try this way also and get your expected result
variable=$(sed 's:.*/\(.*\)-.*:\1:' FileName)
echo $variable
OutPut :
content-config-factory
You could use grep,
grep -oP '(?<=/)[^/]*(?=-\d+\.)' file
Example:
$ var=$(echo 'user/config/jars/content-config-factory-3.2.0.0.jar' | grep -oP '(?<=/)[^/]*(?=-\d+\.)')
$ echo "$var"
content-config-factory
Here is a simplified version of my problem.
if (echo "AA BB CC" | awk '{ print $1 $2 }' | grep -q "B"); then
echo $2
fi
I would like to make $2 available in bash, so I can use it elsewhere in the script.
Can that be done?
Update
I realized that I had simplified the problem too much. The awk expression should have been awk '{ print $1 $2 }' instead of just awk '{ print $2 }' which I originally posted.
You can use set:
set -- `echo "AA BB CC" | awk '{print $2}'`
case $1 in *B*) echo $1;; esac
... or if you used the awk just to split the output, let set do that part as well:
set -- `echo "AA BB CC"`
case $2 in *B*) echo $2;; esac
Remember the output of awk, test it for the regular expression and print it:
output=$( echo "AA BB CC" | awk '{ print $2 }' )
if grep -q B <<< "$output" ; then echo "$output" ; fi
You can capture stdout into a variable by using the backtick operator, e.g.
a=`echo foo`
echo $a
For your example, it would be something like:
a=`echo "AA BB CC" | awk '{ print $2 }' | grep -q "B"`
echo $a