Check if line starts with digit then put those lines in a separate file using Bash

I would like to check in bash if line starts with digit then put those lines in a separate file. I tried ^[[0-9]] but it doesn’t work.

Here is the code which I tried:

myFailedFile=/tmp/1.txt

myTestsFile:
177711 TESTING ...yahoo.tests.calendar.resources.scheduler.1
854756 TESTING ...yahoo.tests.calendar.resources.scheduler.2
* 2102637 DONE ...yahoo.tests.mail.contacts.add.3

while read line
do
    if ( [[ ${line} = "^[[0-9]]"* ]] && [[ ${line} = *".tests."* ]] ); then
        echo -e "${line}r" >> ${myFailedFile}
    fi
done <"${myTestsFile}"

Expected output of myFailedFile:
177711 TESTING ...yahoo.tests.calendar.resources.scheduler.1
854756 TESTING ...yahoo.tests.calendar.resources.scheduler.2

Answer

The correct operator to use regex in Bash script is =~. Moreoever you don’t need to double [ in range of characters. Try this:

while read line
do
    if ( [[ ${line} =~ ^[0-9] ]] && [[ ${line} = *".tests."* ]] ); then
        echo -e "${line}r" >> ${myFailedFile}
    fi
done <"${myTestsFile}"

Edit:

But you don’t need a Bash loop for that job. You can do it with a sed one-liner:

sed '/^[0-9].*.tests./!d' "${myTestsFile}" > myFailedFile

Explanations(from right to left):

• !d: do not delete
• /^[0-9].*.tests./: all lines that start with one or more digits and that contain .tests. string