Question

脚本非常简单：

#!/bin/bash
if [[ 0 ]]; then
   echo foo
fi

错误表现为：

$ source ./sample.sh
./sample.sh:2: parse error near `]]'

但请注意bash能够很好地执行脚本：

$ /bin/bash ./sample.sh
foo

$ /bin/bash --version
GNU bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu)
Copyright (C) 2009 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

来自bash docs：

[[ expression ]]
   Return a status of 0 or 1 depending on  the  evaluation  of  the
   conditional  expression expression.  Expressions are composed of
   the primaries described  below  under  CONDITIONAL  EXPRESSIONS.

   ...

CONDITIONAL EXPRESSIONS
   Conditional  expressions  are  used  by the [[ compound command and the
   test and [ builtin commands to test file attributes and perform  string
   and  arithmetic comparisons.  Expressions are formed from the following
   unary or binary primaries.

   ...

   string
   -n string
          True if the length of string is non-zero.

请注意，语法错误会以0或"0"作为表达式显示。

添加一个运算符（例如-n）可以解决解析错误，但它似乎不是必须从文档中解释，也不能解释为什么bash评估它就好了。

Answer 1

你的shell不是bash。当你运行source ./sample.sh时，它会在当前shell的上下文中运行脚本，无论是什么，忽略#!/bin/bash hash-bang行。

顺便问一下，if [[ 0 ]]打算做什么？这有点荒谬。 [[ 0 ]]相当于[[ -n 0 ]]，它会检查0是否为非空字符串。这保证是真的。

要写一个简单的真或假检查，我会写：

if true; then

或

if false; then

当bash执行代码时，为什么源代码会出现语法错误？

1 个答案: