awk command to accept two variables as parameters and return a value
awk, unix
Solution
You have to use `awk's` variable passing before the awk script begins to avoid hairy quoting, plus fix other problems:
#!/user/bin/sh
READ_FILE=/export/home/user/command_file.txt
VA1=A
VA2=B
GET_VALUE=$(awk -v var="$VA1" -v var1="$VA2" '$1 ~ var && $2 ~ var1 {print $3}' $READ_FILE)
echo $GET_VALUE
Problem
I have a file which has 50 rows. Each row is made up of three columns. The first two columns are the variables and this will be passed as parameters to return the 3rd column's value. for ex.. command_file.txt is the file and it contains ``` A B 10 C D 20 E F 30 G H 50 I J 70 ... ``` I have a script with the following command. ``` #!/user/bin/sh READ_FILE=/export/home/user/command_file.txt VA1=A VA2=B GET_VALUE=`awk '/ -v var="$VA1" '$1 ~ var' -v var1="$VA2" '$1 ~ var1''/ $READ_FILE l awk '{print $3}'` echo $GET_VALUE ``` When I call this script passing A and B as parameters I expect a value of 10 to be returned.But it returned errors. But if I hard code the value on the below command it will work. ``` GET_VALUE=`awk '/A B'/ $READ_FILE lawk '{print $3}'` ``` Any suggestions? Thanks.