egrep in an archieve
by ektoren from LinuxQuestions.org on (#53TZF)
I tried to solve an exercise I find in a website but I am wrong.. in every answer..I have wrote the commands using the egrep but I am wrong.What I am doing wrong ?
In every question you must read from the archieve /etc/catalog. In the surname and name it can be any character except of space or :
Iris Ksimis,:3553:asop:asop@yahoocom,asop@hotmail.com:22494012
Smenos Fomis:1242:smenos:smenos@hotmail.comm:391310495,5301-203
Aminiou Emos:2356:aminiou:aminiou@yahoo.com:3012349104852
So as you can see there is name-surname,code,username,emails,phones. The fields are split by ":". The emails or phones(if existed more than 1) are split by ","
1)Give the command that counts users whose names ends with y(Y). what I did and it is wrong .
egrep --color -wc -i "y$" /etc/catalog
I use -wc to count -i so that it doesn't play role the y or Y. y$ end of the line.
it's wrong and I have no idea why.
2)Give the command that counts the uses whose (at least ) one of the phones starts with 1
egrep --color -wc "^([^:]*:){3}:1.*$" /etc/catalog
^([^:]*:){3} I did that so those : for 3 times will not be appeared.
3)Give the command that counts the users whose they don't have the digit 1 in their phones.
egrep --color -wc "^([^:]*:){3}:[^1]" /etc/catalog
4)Give the command that appears the users (all the line ) whose usernames are not have those characters l until m(L until M)and they have email in hotmail.
egrep --color -i ' [lm]' /etc/catalog
5)Give the command that counts the users whose usernames ends with u(U) and they have email yahoo.
egrep --color -i "^([^:]*:){2}:u$:.*yahoo.com.*" /etc/catalog
6)Give the commands that appears all the users (all the line ) whose username(whole username) are similar with the whole surname I have no idea how to do that..
In every question you must read from the archieve /etc/catalog. In the surname and name it can be any character except of space or :
Iris Ksimis,:3553:asop:asop@yahoocom,asop@hotmail.com:22494012
Smenos Fomis:1242:smenos:smenos@hotmail.comm:391310495,5301-203
Aminiou Emos:2356:aminiou:aminiou@yahoo.com:3012349104852
So as you can see there is name-surname,code,username,emails,phones. The fields are split by ":". The emails or phones(if existed more than 1) are split by ","
1)Give the command that counts users whose names ends with y(Y). what I did and it is wrong .
egrep --color -wc -i "y$" /etc/catalog
I use -wc to count -i so that it doesn't play role the y or Y. y$ end of the line.
it's wrong and I have no idea why.
2)Give the command that counts the uses whose (at least ) one of the phones starts with 1
egrep --color -wc "^([^:]*:){3}:1.*$" /etc/catalog
^([^:]*:){3} I did that so those : for 3 times will not be appeared.
3)Give the command that counts the users whose they don't have the digit 1 in their phones.
egrep --color -wc "^([^:]*:){3}:[^1]" /etc/catalog
4)Give the command that appears the users (all the line ) whose usernames are not have those characters l until m(L until M)and they have email in hotmail.
egrep --color -i ' [lm]' /etc/catalog
5)Give the command that counts the users whose usernames ends with u(U) and they have email yahoo.
egrep --color -i "^([^:]*:){2}:u$:.*yahoo.com.*" /etc/catalog
6)Give the commands that appears all the users (all the line ) whose username(whole username) are similar with the whole surname I have no idea how to do that..