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Check if argument passed is an integers


 
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# 8  
Old 11-04-2002
They do not work.They returned 1.

Thanks again.
# 9  
Old 11-04-2002
Ooops!! Thank you, Perderabo for correcting my typo!!...

grep -E should work for elchalateco, he should have tried it on the command line...

Cheers!
Vishnu.
# 10  
Old 11-05-2002
MySQL

Vishnu,
YOU'R THE MAN!

Thanks a million Smilie
# 11  
Old 11-06-2002
--------------------------------------------------------------------------------checkinput()
{

echo $1 | grep -Eq '^(\+|-)?[0-9]+$'

return $?

}

The above function does not allow the use of + sign in front of a number. even thoug positive numbers don't need the + sing to indicate that they are positives. I need this for this function.

I have tryed couple of alternatives with no luck.


Thanks
# 12  
Old 11-06-2002
That was a wrong claim... try this on command line...
echo +2345 | grep -E '^(\+|-)?[0-9]+$'

I hope you understood what '^(\+|-)?[0-9]+$' means...

Since you seem concerned with selective and intelligent matches, I would suggest you to go through some good book on regular expressions...

here is one worth it's bucks...

http://www.oreilly.com/catalog/regex2/

Cheers!
Vishnu.
# 13  
Old 11-07-2002
i couldn't resist

Although I think its great passing around all these regexes, I thought I might break it down anyway, it might help someone. Please correct me if I am wrong.


'^(\+|-)?[0-9]+$' :

^ = start of the line

(\+|-) = either positive or a negative

? = preceeding character, zero or once; therefore either + or - occurs or it doesn't

[0-9] =any numeric character

+ = preceeding character, once or more times therefore [0-9]+ is any number between 0-9 more than once

$ = end of line
Shakey21
# 14  
Old 11-07-2002
I want to thank every one who contributed .

Guys I offer an apology. The scritp does not work on my box at home; however it run perfectly at work.

Thanks again.

Frank
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