Shell Programming and Scripting

Hi,

Can anyone help me in understanding how the below code works?

Code:

echo "texxt" | sed 's/[^x]//'

gives output exxt, ideally it should give xxt. as this should remove the chars which is not x.

Code:

echo 'x_a_b_a_c_a_d' | sed 's/.*\(a\)/\1/'

gives output as a_d, which should be 'a' as it's the only character in the section

Code:

echo 'foo-bar-1.4.5-19.spkg' | sed 's/-[^-]*$//'

and how the above code works?

try

Code:

echo "texxt" | sed 's/[^x]//g'

the output is normal. You ask to replace all characters up to *the last* a in string. Remember the sed regex are "greedy". They take all they can.

Same as above. That pattern will take all non - characters following a - up to the end of the string. The only part of the string that match that pattern is from the last - up to the end.

Code:

s/[^x]//g

This means "replace the first character which isn't a 'x' with nothing". Since that's done as soon as the first 't' is replaced, sed is finished. If you want to remove all characters leading up to the first 'x', use

Code:

s/[^x]*//g

In the second snipped:

Code:

echo 'x_a_b_a_c_a_d' | sed 's/.*\(a\)/\1/'

you're replacing as many characters as possible from the start until the last possible 'a' with the matched 'a'. The rest of the string isn't affected, so 'x_a_b_a_c_a' is replaced by 'a', which changes 'x_a_b_a_c_a_d' to 'a_d'

Third:

Code:

echo 'foo-bar-1.4.5-19.spkg' | sed 's/-[^-]*$//'

means "starting with a dash, take as many non-dash characters as possible until the end, and replace them with nothing", or "remove everything starting from and including the last possible dash".