bash shell 脚本中的非法数字
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Illegal number in shell script
提问by whla
I am new to writing scripts and am trying to start out with a simple one. I am stumped as to why I am receiving the error: [: 13: Illegal number: countfrom the code below. Line 13 is the last fi.
我是编写脚本的新手，正在尝试从一个简单的脚本开始。我很困惑为什么我会收到错误：[: 13: Illegal number: count来自下面的代码。第 13 行是最后一个fi。
count=grep "^$(date -d -30minute +'%Y-%m-%d %H:%M')" /var/log/****/zlsapp.log | wc -l
if [ count -ge 50 ]
then
if [ count -lt 100 ]
then
exit 1
fi
if [ count -ge 100 ]
then
exit 2
fi
exit 0
fi
Also is there anyway to do compound if statements like if(count >= 50 && count < 100)?
还有无论如何要做复合 if 语句if(count >= 50 && count < 100)吗？
采纳答案by Jonathan Leffler
The line:
线路：
count=grep "^$(date -d -30minute +'%Y-%m-%d %H:%M')" /var/log/zumigo/zlsapp.log | wc -l
probably does not do what you think it does (or you've not accurately copied and pasted your actual code into the question). If run in the middle of 2014-08-01, it runs the command "2014-08-01 12:00" with the log file as an argument and the environment variable countset to the value grep, and pipes the output from the probably non-existent command to wc -l.
可能没有按照您的想法去做（或者您没有准确地将实际代码复制并粘贴到问题中）。如果在 2014-08-01 中间运行，它会运行命令“2014-08-01 12:00”，以日志文件为参数，将环境变量count设置为 value grep，并将输出从可能非- 存在的命令wc -l。
When you subsequently go to test $countin the test statements, it is an empty string, which doesn't convert properly to a number.
当您随后$count在测试语句中进行测试时，它是一个空字符串，无法正确转换为数字。
You probably meant:
你可能的意思是：
count=$(grep "^$(date -d -30minute +'%Y-%m-%d %H:%M')" /var/log/zumigo/zlsapp.log | wc -l)
This captures the output of running grepon the log file and piping the output to wc -l.
这会捕获在grep日志文件上运行的输出并将输出通过管道传输到wc -l.
If you run your script with bash -xor equivalent (the -xoption usually shows the execution of the script), you should see this.
如果你运行你的脚本bash -x或等价物（该-x选项通常显示脚本的执行），你应该看到这个。
回答by David C. Rankin
Two reasons. (1)in bash variables are untyped (could be int, could be char). In order to remove ambiguity, you can specify the type with:
两个原因。(1)在 bash 中的变量是无类型的（可以是 int，也可以是 char）。为了消除歧义，您可以使用以下命令指定类型：
declare -i count
To tell bash it should be an int. (2)you need to dereference your variables with $to get the number back. I.E.
告诉 bash 它应该是一个整数。(2)您需要取消引用您的变量$以取回数字。IE
[ $count -lt 100 ]
(it is also good practice to quote your variables - not required, but good practice: [ "$count" -lt 100 ]. Drop a comment if you still have trouble.
（引用您的变量也是一种很好的做法 - 不是必需的，但很好的做法：[ "$count" -lt 100 ]如果您仍然遇到问题，请发表评论。
回答by Keith Thompson
The problem is that countdoes not refer to the variable count; it's simply the string "count".
问题是count没有引用变量count；它只是字符串"count"。
Change:
改变：
if [ count -ge 50 ]
to
到
if [ $count -ge 50 ]
and make the corresponding change elsewhere (but not in the initial assignment).
并在其他地方进行相应的更改（但不在初始分配中）。
You should also use double quotes:
您还应该使用双引号：
if [ "$count" -ge 50 ]
Also is there anyway to do compound if statements like if(count >= 50 && count < 100)?
还有无论如何要执行复合 if 语句，例如 if(count >= 50 && count < 100)？
Yes:
是的：
if [ "$count" -ge 50 -a "$count" -lt 100 ]
if [ "$count" -ge 50 -a "$count" -lt 100 ]
is likely to work, but the -a(logical and) operator is marked as obsolescent by POSIX. Instead write
可能有效，但-a（逻辑与）运算符被POSIX 标记为过时。而是写
if [ "$count" -ge 50 ] && [ "$count" -lt 100 ]
If you're using bash, info bashand search for the "test" command ([is an alias for test).
如果您使用 bash，info bash并搜索“test”命令（[是 的别名test）。
And if you're using bash, you should consider using [[ ... ]]rather than [ ... ]-- or you can use (( ... ))for arithmetic expressions. See the bash documentationfor more information (follow the iink or type info bash).
如果您使用 bash，您应该考虑使用[[ ... ]]而不是[ ... ]-- 或者您可以使用(( ... ))算术表达式。有关更多信息，请参阅bash 文档（按照 iink 或 type info bash）。
In addition to the missing $signs, the first line of your script:
除了缺少的$符号，脚本的第一行：
count=grep "^$(date -d -30minute +'%Y-%m-%d %H:%M')" /var/log/****/zlsapp.log | wc -l
doesn't set the countvariable, since you're not capturing the output of the grep ... | wc -lcommand. To do so:
不设置count变量，因为您没有捕获grep ... | wc -l命令的输出。这样做：
count=$(grep "^$(date -d -30minute +'%Y-%m-%d %H:%M')" /var/log/****/zlsapp.log | wc -l)
(Yes, $(...)can be nested.)
（是的，$(...)可以嵌套。）
