检查行是否以数字开头，然后使用Bash将这些行放在单独的文件中

Question

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 1

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

Answer 2

Without using regex you can use glob as this:

[[ $line = [0-9]* ]] && [[ $line = *".tests."* ]]

• [0-9]* matches a string start start with digits
• *".tests."* matches a string that contains .tests.

Answer 3

you can use

if [[ ${line} =~ ^[0-9] && ${line} =~ ".tests." ]] ; then

or

if [[ ${line} == [0-9]* && ${line} == *".tests."* ]] ; then