Remove line from file in bash script using sed command

To remove a line containing $record from the file:

grep -v "$record" student_records.txt >delete.me && mv -f delete.me student_records.txt

In the above, $record is treated as a regular expression. This means, for example, that a period is a wildcard. If that is unwanted, add the -F option to grep to specify that $record is to be treated as a fixed string.

Comments

Consider these two line:

r=$(grep -h "$record"  student_records.txt|cut -d"," -f1)  #find the record that needs to be deleted
echo $line -->> This proves that previous command works

The first line defines a shell variable r. The second line prints the shell variable line, a variable which was unaffected by the previous command. Consequently, the second line is not a successful test on the first.

sed -i '/^$r/d' student_records.txt  -->> this does not work

Obseve that the expression $r appears inside single-quotes. The shell does not alter any inside single quotes. Consequently, $r will remain a dollar sign followed by an r. Since a dollar sign matches the end of a line, this expression will match nothing. The following would work better:

sed -i "/^$r/d" student_records.txt

Unlike the grep command, however, the above sed command is potentially dangerous. It would be easy to construct a value of r that would cause sed to do surprising things. So, don't use this approach unless you trust the process by which you obtained r.

What if more than one line matches record?

If there is more than one line that matches record, then the following would generate an unterminated address regex error from sed:

r=$(grep -h "$record"  student_records.txt|cut -d"," -f1)
sed -i "/^$r/d" student_records.txt

This error is an example of the surprising results that can happen when a shell variable is expanded into a sed command.

By contrast, this approach would remove all matching lines:

grep -v "$record" student_records.txt >delete.me && mv -f delete.me student_records.txt