Linux bash script: how to search by column but return full string?
I have a tab delimited file with data like this:
col1 col2 col3
I wrote a bash script that allows you to search for a file using this code:
echo -en "Search term: "
read search
data=`cat data.data | egrep -i "$search"`
This works great for searching the entire file, but now I only want to search on a specific column (which the user can select).
I know the command cut
and can search on a column using this:
cat data.data | cut -f$col | egrep -i "$search"
But then only that column is displayed, so if I use this method, I need to somehow return the rest of the row.
How can I search by column in a file, but return full rows for the results?
awk is easier for this:
data=$(awk -v col=$col -v term="$term" 'toupper($col)==toupper(term)' file)
col - column number
term - search query
You can also pass in a field separator -F
if needed.
You can pass two variables to awk: the column number and the search term.
awk -vcol="$col" -vsearch="$search" '$col ~ search' data.data
If the value $col
is 2, then $2
awk will match the second column. The operator is ~
used to match a regular expression pattern. The string will be printed if the column matches the regular expression.
Testing:
$ cat data.data
col1 col2 col3
$ col=2
$ search=l2
$ awk -vcol="$col" -vsearch="$search" '$col ~ search' data.data
col1 col2 col3
$ search=l3
$ awk -vcol="$col" -vsearch="$search" '$col ~ search' data.data
# no output
If you want to make the pattern match case sensitive, you have two options: convert everything to upper or lower case ( tolower($col) ~ tolower(search)
), or if you're using GNU awk, set the IGNORECASE variable:
$ search=L2
$ awk -vIGNORECASE=1 -vcol="$col" -vsearch="$search" '$col ~ search' data.data
col1 col2 col3