BASH：if 语句执行命令和函数
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BASH: if statement execute command and function
提问by drewrockshard
I've run into an issue in which I think should be easy to resolve, but for the life of me, I can't figure it out. Could be that it's really late; not sure.
我遇到了一个我认为应该很容易解决的问题，但对于我的生活，我无法弄清楚。可能是真的晚了；没有把握。
So I have a shell script and I have an if statement that I need to run. The problem is that I have a function inside this bash script that I am using to actually build part of this find command inside the if statement. I want to know how I can do both, without receiving an error [: too many arguments.
所以我有一个 shell 脚本，我有一个需要运行的 if 语句。问题是我在这个 bash 脚本中有一个函数，我用来在 if 语句中实际构建这个 find 命令的一部分。我想知道如何在不收到错误的情况下做到这两点[: too many arguments。
Here's the current code:
这是当前的代码：
if [ -n `find ./ `build_ext_names`` ];then
That's all I really need to post. I need to figure out how to run that build_ext_namesinside that find command, which in-turn is inside the ifstatement
这就是我真正需要发布的内容。我需要弄清楚如何build_ext_names在该 find 命令中运行该命令，而该命令又位于 if 语句中
回答by l0b0
Michael Aaron Safyanhas the right idea, but to fix the immediate problem you can just use the simpler $(command)construct instead of ```command` `` for command substitution. It allows for much simpler nesting:
Michael Aaron Safyan的想法是正确的，但要解决眼前的问题，您可以使用更简单的$(command)构造而不是 ```command``` 进行命令替换。它允许更简单的嵌套：
if [ -n "$(find ./ "$(build_ext_names)")" ]; then
回答by Michael Aaron Safyan
This is easier if you split it up:
如果你把它分开，这会更容易：
function whateverItIsYouAreTryingToDo() {
local ext_names=$(build_ext_names)
local find_result=$(find ./ $ext_names)
if [ -n "$find_result" ] ; then
# Contents of if...
fi
}
