bash 嵌套 if/else 的 zsh 脚本解析器错误
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zsh script parser error for nested if/else
提问by George H
I have the following humble zsh function:
我有以下不起眼的 zsh 功能：
function remember()
{
if [ "" != "" ]
then
if[ "" != "clean" ]
then
echo "Why";
#echo >> {~/.remember_info};
else
rm -r ~/.remember_info;
touch ~/.remember_info;
fi
else
cat .remember_info;
fi
}
When I try to source it I get:
当我尝试获取它时，我得到：
parse error near `echo' (The echo being the line with echo "Why";)
`echo' 附近的解析错误（回声是带有回声“为什么”的那一行；）
The error is quite non descriptive and I assume its related to part of the loop's logic (since no matter what instruction I give after then it error out there).
该错误是非常非描述性的，我认为它与循环逻辑的一部分有关（因为无论我在此之后给出什么指令，它都会出错）。
Is there any way to "debug" this kind of thing ? zsh -n doesn't help much (at all)
有没有办法“调试”这种事情？zsh -n 没有太大帮助（根本）
回答by chepner
You forgot the space between ifand [when comparing to clean.
if与[相比时，您忘记了和之间的空格clean。
This is case, though, where your function can be made simpler by handling the =case first.
但是，在这种情况下，您可以通过=首先处理案例来简化您的功能。
function remember()
{
if [ "" = "" ]; then
cat ~/.remember_info
elif [ "" = clean ]; then
rm -r ~/.remember_info
touch ~/.remember_info
else
echo "" >> ~/.remember_info;
fi
}
Or, use a casestatement.
或者，使用case语句。
remember () {
f=~/.remember_info
case of
"")
cat "$f"
;;
clean)
rm -r "$f"
touch "$f"
;;
*)
print "" >> "$f"
;;
esac
}
回答by Arkadiusz Drabczyk
You are missing a whitespace after [. It should be:
之后缺少空格[。它应该是：
function remember()
{
if [ "" != "" ]
then
if [ "" != "clean" ]
then
echo "Why";
#echo >> {~/.remember_info};
else
rm -r ~/.remember_info;
touch ~/.remember_info;
fi
else
cat .remember_info;
fi
}
[is the same as test. It is a separate command, described in man test:
[与 相同test。它是一个单独的命令，描述man test如下：
TEST(1)
NAME
test - check file types and compare values
SYNOPSIS
test EXPRESSION
test
[ EXPRESSION ]
[ ]
[ OPTION
