UNIX TOOLS - commad grep utility find match content in files

Unix Tools and Programming

Finding matching pattern using grep utility

Syntax:
grep "word-to-find" {file-name}

 
Create text file as follows:

---

$vi emo-file
hello world!
cartoons are good
especially toon like tom (cat)
what
the number one song
12221
they love us
I too

---

After saving file, issue following command,
$ grep "too" demofile
cartoons are good
especially toon like tom (cat)
I too
grep will locate all lines for the "too" pattern and print all (matched) such line on-screen. grep prints too, as well as cartoons and toon; because grep treat "too" as expression. Expression by grep is read as the letter t followed by o and so on. So if this expression is found any where on line its printed. grep don't understand words.


UNIX TOOLS - Conditional : if else fi

Unix Tools and programming :
if else fi

---

Syntax:

           if condition
           then
                       condition is zero (true - 0)
                       execute all commands up to else statement

           else
                       if condition is not true then
                       execute all commands up to fi
           fi

---

For e.g. Write Script as follows:

$ vi isnump_n #!/bin/sh # # Script to see whether argument is positive or negative # if [ $# -eq 0 ] then echo "$0 : You must give/supply one integers" exit 1 fi if test $1 -gt 0 then echo "$1 number is positive" else echo "$1 number is negative" fi

Try it as follows:
$ chmod 755 isnump_n

$ isnump_n 5
5 number is positive

$ isnump_n -45
-45 number is negative

$ isnump_n
./ispos_n : You must give/supply one integers

$ isnump_n 0
0 number is negative
Detailed explanation
First script checks whether command line argument is given or not, if not given then it print error message as "./ispos_n : You must give/supply one integers". if statement checks whether number of argument ($#) passed to script is not equal (-eq) to 0, if we passed any argument to script then this if statement is false and if no command line argument is given then this if statement is true. The echo command i.e.
echo "$0 : You must give/supply one integers"
         |              |
         |              |
        1             2
1 will print Name of script
2 will print this error message
And finally statement exit 1 causes normal program termination with exit status 1 (nonzero means script is not successfully run).
The last sample run $ isnump_n 0 , gives output as "0 number is negative", because given argument is not > 0, hence condition is false and it's taken as negative number. To avoid this replace second if statement with if test $1 -ge 0.