在 Bash 别名/功能中使用参数

Question

I am trying to create a helpful bash alias/function. I can't seem to get it to properly interpret the string parameters.

alias phpr="php -r '${*}'"
alias decode="php -r 'echo base64_decode($1);'"

function phpr ()
{
    php -r "${*}"
}

function decode ()
{
    php -r "echo base64_decode($1);"
}

an example of how Im trying to use decode and it failing:

decode "MT0xO2JhbGFuY2VfaXNfc2hvd249MDtiZXRfYW1vdW50PTE7YmV0X3R5cGU9MTE7ZmF2b3JpdGVfdHJhY2sxPUZSRTtmYXZvcml0ZV90cmFjazI9TU1BO2Zhdm9yaXRlX3RyYWNrMz1UQU07ZmF2b3JpdGVfdHJhY2s0PU1FRDtmYXZvcml0ZV90cmFjazU9QVFVO3dhZ2VyX3N0YXR1cz0wO2N1cnJlbnRUcmFjaz1GUkU7Y3VycmVudFJhY2U9NjtyYWNlRGF0ZT0yMDExLTA0LTI4O3JhY2V1cGRhdGVkPTs%3D"

PHP Warning:  base64_decode() expects at least 1 parameter, 0 given in Command line code on line 1
PHP Stack trace:
PHP   1. {main}() Command line code:0
PHP   2. base64_decode() Command line code:1

I'm not sure when I'd use a function vs an alias for cases like these.

Answer 1

For your alias - alias doesn't take parameters. Have you tried alias phpr="php -r" ?

For the function - it could be something being parsed in your string. Double-quotes allow parsing, Single-quotes take the string verbatim. You want the $1 to be parsed, but include any whitespace, wo you want double-quotes around that. But you can do single quotes for the rest of it:

function decode ()
{
    php -r 'echo base64_decode('"$1"');'
}

Note that this includes 4 single-quotes and 2 double-quotes.

EDIT Updated with Glenn's change. If php needs the parameter to be in quotes, you may need to do:

function decode ()
{
    php -r 'echo base64_decode("'"$1"'");'
}

Note that this includes 4 single-quotes and 4 double-quotes.

Ah, shell scripting...

Answer 2

I can answer one part of your question straight away: Bash aliases cannot take arguments.