4
我試圖讓擊到正確執行以下最小化例子:我想獲得在引用的字符串爲命令(或沒有)的能力
# Runs a command, possibly quoted (i.e. single argument)
function run()
{
$*
}
run ls # works fine
run "ls" # also works
run "ls `pwd`" # also works, but pwd is eagerly evaluated (I want it to evaluate inside run)
run "ls \\\`pwd\\\`" # doesn't work (tried other variants as well)
總結,並沒有任何的命令,包括通過反引號的嵌套shell命令,計算值等,在run()被調用之前評估。這可能嗎?我怎樣才能做到這一點?
請參閱[BashFAQ/050](http://mywiki.wooledge.org/BashFAQ/050)。 – 2011-01-12 15:38:27