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确定bash中是否存在函数

scripting testing bash linux

149

terminus · 技术社区 · 16 年前

目前我正在做一些从bash执行的单元测试。单元测试是在bash脚本中初始化、执行和清理的。此脚本通常包含init()、execute()和cleanup()函数。但它们不是强制性的。我想测试一下它们是否有定义。

我以前是通过对消息源进行greping和seding来实现这一点的,但这似乎是错误的。有更优雅的方法吗?

编辑:下面的剪贴画就像一个魅力:

fn_exists()
{
    type $1 | grep -q 'shell function'
}

13 回复 | 直到 16 年前

167

David Winiecki 9 年前

$ type foo
bash: type: foo: not found

$ type ls
ls is aliased to `ls --color=auto'

$ which type

$ type type
type is a shell builtin

$ type -t rvm
function

$ if [ -n "$(type -t rvm)" ] && [ "$(type -t rvm)" = function ]; then echo rvm is a function; else echo rvm is NOT a function; fi
rvm is a function

Eliran Malka 8 年前

$ g() { return; }
$ declare -f g > /dev/null; echo $?
0
$ declare -f j > /dev/null; echo $?
1

jpaugh JotaBe 8 年前

-f

#!/bin/sh

function_exists() {
    declare -f -F $1 > /dev/null
    return $?
}

function_exists function_name && echo Exists || echo No such function

fname=`declare -f -F $1`
[ -n "$fname" ]    && echo Declare -f says $fname exists || echo Declare -f says $1 does not exist

fname=`declare -f -F $1`
errorlevel=$?
(( ! errorlevel )) && echo Errorlevel says $1 exists     || echo Errorlevel says $1 does not exist
[ -n "$fname" ]    && echo Declare -f says $fname exists || echo Declare -f says $1 does not exist

Grégory Joseph 13 年前

fn_exists() {
  # appended double quote is an ugly trick to make sure we do get a string -- if $1 is not a known command, type does not output anything
  [ `type -t $1`"" == 'function' ]
}

if ! fn_exists $FN; then
    echo "Hey, $FN does not exist ! Duh."
    exit 2
fi

jonathanserafini 14 年前

test_declare () {
    a () { echo 'a' ;}

    declare -f a > /dev/null
}

test_type () {
    a () { echo 'a' ;}
    type a | grep -q 'is a function'
}

echo 'declare'
time for i in $(seq 1 1000); do test_declare; done
echo 'type'
time for i in $(seq 1 100); do test_type; done

real    0m0.064s
user    0m0.040s
sys     0m0.020s
type

real    0m2.769s
user    0m1.620s
sys     0m1.130s

Scott 12 年前

isFunction() { [[ "$(declare -Ff "$1")" ]]; }

isFunction some_name && echo yes || echo no

isFunction() { declare -Ff "$1" >/dev/null; }

Jonas 8 年前

fn_exists()
{
   [[ $(type -t $1) == function ]] && return 0
}

isFunc () 
{ 
    [[ $(type -t $1) == function ]]
}

$ isFunc isFunc
$ echo $?
0
$ isFunc dfgjhgljhk
$ echo $?
1
$ isFunc psgrep && echo yay
yay
$

jarno 7 年前

#!/bin/bash

f () {
echo 'This is a test function.'
echo 'This has more than one command.'
return 0
}

test_declare () {
    declare -f f > /dev/null
}

test_declare2 () {
    declare -F f > /dev/null
}

test_type () {
    type -t f | grep -q 'function'
}

test_type2 () {
    local var=$(type -t f)
    [[ "${var-}" = function ]]
}

post=
for j in 1 2; do
echo
echo 'declare -f' $post
time for i in $(seq 1 1000); do test_declare; done
echo
echo 'declare -F' $post
time for i in $(seq 1 1000); do test_declare2; done
echo
echo 'type with grep' $post
time for i in $(seq 1 1000); do test_type; done
echo
echo 'type with var' $post
time for i in $(seq 1 1000); do test_type2; done
unset -f f
post='(f unset)'
done

declare -F f && echo function f exists. || echo function f does not exist.

Jason Plank IEnumerator 13 年前

fn_exists()
{
  type $1 >/dev/null 2>&1;
}

Community Lee Campbell 7 年前

function is_executable()
{
    typeset TYPE_RESULT="`type -t $1`"

    if [ "$TYPE_RESULT" == 'function' ]; then
        return 0
    else
        return 1
    fi
}

user186791 15 年前

fn_exists()
{
    type $1 2>/dev/null | grep -q 'is a function'
}

fn_exists test_function
if [ $? -eq 0 ]; then
    echo 'Function exists!'
else
    echo 'Function does not exist...'
fi

qneill Adam Burry 7 年前

$ fn_exists() { test x$(type -t $1) = xfunction; }
$ fn_exists func1 && echo yes || echo no
no
$ func1() { echo hi from func1; }
$ func1
hi from func1
$ fn_exists func1 && echo yes || echo no
yes

Noah Spurrier 14 年前

test_function () {
        ! type -f $1 >/dev/null 2>&1 && type -t $1 >/dev/null 2>&1
}