4
我试图让击到正确执行以下最小化例子:我想获得在引用的字符串为命令(或没有)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)
总结,并没有任何的命令,包括通过反引号的嵌套shell命令,计算值等,在run()被调用之前评估。这可能吗?我怎样才能做到这一点?
请参阅[BashFAQ/050](http://mywiki.wooledge.org/BashFAQ/050)。 – 2011-01-12 15:38:27