在 bash 中的 if 语句中将参数传递给函数
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Passing arguments to a function within an if statement in bash
提问by user1117603
I am trying to pass arguments to a function within an if statement and then evaluating what the function returns (in bash). The result I am getting is incorrect. How do you properly do this?
我试图在 if 语句中将参数传递给函数，然后评估函数返回的内容（在 bash 中）。我得到的结果不正确。你如何正确地做到这一点？
#!/bin/bash
foo() {
if [ = "zero" ]; then
echo "0"
else
echo "1"
fi
}
arg="zero"
if foo $arg -eq 0 ; then
echo "entered 0"
else
echo "entered something else"
fi
arg="blah"
if foo $arg -eq 0 ; then
echo "entered 0"
else
echo "entered something else"
fi
Not the desired results:
不是想要的结果：
cknight@macbook:~/work$ ./test.sh
0
entered 0
1
entered 0
Thank you in advance. ~Chris
先感谢您。~克里斯
回答by cnicutar
You can either returna result from your function or use $()to capture its output:
您可以return从您的函数$()中获取结果或用于捕获其输出：
if [ $(foo $arg) -eq 0 ] ; then
回答by chepner
Are you sure you need your function to echo 0 or 1? It looks more like you want something like
你确定你需要你的函数来回显 0 或 1 吗？看起来更像是你想要的东西
foo() {
if [ = "zero" ]; then
return 0
else
echo 1
fi
}
# Or even just simply foo () { [ = "zero ]; }
arg="zero"
if foo $arg; then
echo "entered 0"
else
echo "entered something else"
fi
arg="blah"
if foo $arg; then
echo "entered 0"
else
echo "entered something else"
fi
回答by Karoly Horvath
Not sure what you want to do...
不知道你想做什么...
Your foo()function always returns with success, use return.
您的foo()函数总是成功返回，请使用return.
You pass to your foo()function a second and a third parameter -eq 0but they aren't used. Do you want to check the exit status? That's simply if foo $arg; then ..; fi.
您将foo()第二个和第三个参数传递给您的函数-eq 0，但未使用它们。是否要检查退出状态？那简直了if foo $arg; then ..; fi。
