How to check a word position in a file ?

Shell Programming and Scripting
1

Hello everybody,
I have a file like this : "window 1 truck 3 duck 2... fire 1... etc..." and I would like to print the following number of a word I am searching for. (For example here, if I search for the word "fire", I will print "1")
Thank you for your help !

2 
sed 's/.* fire \([0-9][0-9]*\) .*/\1/' file
3 
s="window 1 truck 3 duck 2... fire 1"
f=0
for i in $s;  
do  
    if [ $f -eq 1 ];then    
      echo "Found number $i"
    fi
    case $i in 
      fire) f=1 ; continue;;
    esac;  
done
4

Doesn't work here:

$ uname -a
SunOS db012a 5.8 Generic_117350-35 sun4us sparc FJSV,GPUZC-M
$ which sed
/usr/bin/sed
$ cat file
window 1 truck 3 duck 2 fire 1
$ sed 's/.* fire \([0-9][0-9]*\) .*/\1/' file
window 1 truck 3 duck 2 fire 1
5

This is because in your sample file, after the "1", there is no space.
The sed statement has a <space> after the \) bracket. It should be

# sed 's/.* fire \([0-9][0-9]*\).*/\1/' file
1

for your case of sample file

6

Then lets do this:

sed 's/ *.*fire \([0-9][0-9]*\).*/\1/' file

Works with/without leading/trailing spaces.

7

But the above pattern will also catch words like backfire, spitfire etc...

8

Thanks for pointing that out. This should lay such issues to rest:

$ cat file
 window 0 truck 3 duck 2 spitfire 8 fire 1 backfire 9
$ sed 's/.*\<fire\> \([0-9][0-9]*\).*/\1/' file
1
$ sed 's/.*\<spitfire\> \([0-9][0-9]*\).*/\1/' file
8
$ sed 's/.*\<window\> \([0-9][0-9]*\).*/\1/' file
0
$ sed 's/.*\<backfire\> \([0-9][0-9]*\).*/\1/' file
9