How to define var in shell script functions

Question

I am currently using the below function

!/bin/bash

#Colour change functions

fnHotlinkG2R()
{
        sed -i 's/#hotlink {height: 200px;width: 200px;background: green;/#hotlink {height: 200px;width: 200px;background: red;/' /var/www/html/style.css
}

Rather than creating multiple difference functions I would like to enter the #hotlink as different every time I call the function from within the script.

I am fairly new to sh scripts and would like some assistance please.

nhahtdh · Accepted Answer · 2012-06-11 08:24:47Z

2

First, the first line should be hash bang #! then followed by the path to the program, not just !.

In bash, you don't declare parameter for the function. You just take the argument (and check whether it is valid/not empty) and use it. In this case, you may want to take the first argument from the function by $1 and replace #hotlink with it.

sed -i 's/'"$1"' {height: 200px; ...

In the part where the function is called, you can call it as if it is another command, and you will supply the #hotlink argument to the command.

fnHotlinkG2R '#hotlink'

edited Jun 11, 2012 at 8:24

answered Jun 11, 2012 at 8:15

nhahtdh

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2 Comments

eckes Over a year ago

I think in addition, he'll need double quotes in order to get $1 expanded inside his search/replace pattern, right?

nhahtdh Over a year ago

@eckes: Actually it will expand without quote (I tested before posting the version without quote, not that thoroughly, though). However, it may not expand correctly if the input has spaces. unix.stackexchange.com/questions/4899/…

anubhava · Accepted Answer · 2012-06-11 08:17:42Z

0

You can use it like this:

#!/bin/bash

#Colour change functions

fnHotlinkG2R()
{
    $hotlinkOld = "$1";
    $hotlinkNew = "$2";
    sed -i "s/$hotlinkOld/$hotlinkNew/i" /var/www/html/style.css
}

And call it like this:

fnHotlinkG2R "#hotlink {height: 200px;width: 200px;background: green;"\
   "#hotlink {height: 200px;width: 200px;background: red;"

answered Jun 11, 2012 at 8:17

anubhava

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1 Comment

Bernhard Over a year ago

you are missing '#' in front of the '!' for a valid hash bang.

Fredrik Pihl · Accepted Answer · 2012-06-11 08:19:43Z

0

First, your shebang is wrong. The proper one is

#!/bin/bash

Secondly, in bash you use a "different" kind of parameter passing.

$0 expands to the name of the shell or shell-script
$1 is the first argument
$2 is the second argument and so on
$@ are all arguments

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