BASH: Removing last howevermany characters of a string

Is there a something I can use in a bash script that will allow me to pipe in a string and have it output that string yet with a specified number of characters from the end of it removed?

EG:

Code:

---

z="The magic string"

echo $z
echo $z $(echo $z | ultra-cool-command 3)

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output:

Code:

---

The magic string
The magic string The magic str

---

look into 'cut':

Code:

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#!/bin/sh

z="The magic string"

echo $z
echo $z $(echo $z | cut -c0-13)

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Thanks, however the length of the string in question will vary, so I'm looking for something where you don't need to specifiy a range, yet just the number of characters it needs to remove from the end of the string.

Unless there is a way to find out the length of the string as well?

echo ${z%???}

That would do it for you. Or, you could do something like this:

echo z | sed 's/.\{3\}$//'

(EDIT: had to make 2 corrections to the sed command ;))

Thanks! :cool:

echo ${#var}

gives length of var

Thanks for that as well! :D

So I guess another way of doing it would be:

Code:

---

echo $(echo $z | cut -c 0-$(expr ${#z} - 3 ))

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Also...

could someone tell me what the difference between

$()

and

${}

is?

I think I'm getting a little confused. :confused:

$(echo hi), the same as `echo hi` -- bash replaces it with the output of the command "echo hi"

${bob} is the same thing as $bob, but it clearly defines where the name begins and ends. (i.e. you could do ${bob}oo, but it wouldn't be the same as $boboo)

There are also some extensions to ${}. You can read about all this stuff in the bash manpage.

z="The magic string"

echo ${z:0:${#z}-3}
The magic str

Holy moly :D That's a lot easier on the eyes to look at than the cut one.

Although {#z} should be ${#z}.


.... so many ways to skin a cat :cool:

Quote:

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Originally posted by Gaxus

Although {#z} should be ${#z}.

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Doh! Corrected it now, thanks.:)