To search a file for a specific word in a file using shell script

Shell Programming and Scripting
1

Hi All,
I have a sql output file has below. I want to get the values 200000040 and 1055.49 .Can anyone help me to write a shell script to get this.

ACCOUNT_NO
------------------------------------------------------------
BILL_NO PAY_TYPE
------------------------------------------------------------ ----------
AMT_NOT_BILLED CASH_ON_DELIVERY
-------------- ----------------
COD_BILL_NO IS_SUSPENDED
------------------------------------------------------------ ------------
200000040
B1-57 10005
1055.49 0
0

Thanks in Advance.
Girish.

2

try..

-bash-3.2$ cat list2
ACCOUNT_NO
------------------------------------------------------------
BILL_NO PAY_TYPE
------------------------------------------------------------ ----------
AMT_NOT_BILLED CASH_ON_DELIVERY
-------------- ----------------
COD_BILL_NO IS_SUSPENDED
------------------------------------------------------------ ------------
200000040
B1-57 10005
1055.49 0
0
-bash-3.2$
-bash-3.2$ sed -n '9 p' list2
200000040
-bash-3.2$ sed -n '11 p' list2
1055.49 0

or

-bash-3.2$ sed '9 !d' list2
200000040
-bash-3.2$ sed '11 !d' list2
1055.49 0
-bash-3.2$
3

Try this one

cat inputfile | awk -F " " '/^[[:digit:]]/ { print $1 }'

4

Hi ryandegrate25,
Thanks for the reply. Your command works for my requirment.
Actually i have similar kind of records in a list2, in this case how will i use your command. I tried using it but its failing for multiple records of same type(which i posted earlier).

Thanks in advance, Giri

5

the number in

sed '11 !d' list2

refers to line number... if your input has a fixed line numbers then you can use it else you can use pattern searching like posted by oky

6

Hi oky,
In your commane "cat inputfile | awk -F " " '/^[[:digit:]]/ { print $1 }' "

Can i know what does "digit" refers to.?
Actually the input file format which i had posted will have similar kind mulitple records in a file.Considering this can you help me to get a command.

Also now i tried using your above command on multiple records, it gives me only 200000040 and not the 1055.49.. Please help.

Thanks in advance.
Giri

7

oky's solution will apply to the whole file.
it will do the same whenever the condition matches.

to remove the last line i.e "0" you can do something like below:

awk -F " " '{ if ($0 ~ /^[[:digit:]]/ && $0 != 0) { print $1 }}' inputfile

inputfile:

ACCOUNT_NO
------------------------------------------------------------
BILL_NO PAY_TYPE
------------------------------------------------------------ ----------
AMT_NOT_BILLED CASH_ON_DELIVERY
-------------- ----------------
COD_BILL_NO IS_SUSPENDED
------------------------------------------------------------ ------------
200000040
B1-57 10005
1055.49 0
0
 
ACCOUNT_NO
------------------------------------------------------------
BILL_NO PAY_TYPE
------------------------------------------------------------ ----------
AMT_NOT_BILLED CASH_ON_DELIVERY
-------------- ----------------
COD_BILL_NO IS_SUSPENDED
------------------------------------------------------------ ------------
666666040
B1-57 10005
1088.49 0
0
 
ACCOUNT_NO
------------------------------------------------------------
BILL_NO PAY_TYPE
------------------------------------------------------------ ----------
AMT_NOT_BILLED CASH_ON_DELIVERY
-------------- ----------------
COD_BILL_NO IS_SUSPENDED
------------------------------------------------------------ ------------
222220040
B1-57 10005
3355.49 0
0

output:

200000040
1055.49
666666040
1088.49
222220040
3355.49
8

[[:digit:]] - a character class keyword especially to refer the digits alone

9

Hi All,
Thanks a lot, your code snippet worked for my requirment....

Thank,

Giri