bash SH linux：语法错误：单词意外
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SH linux: Syntax error: word unexpected
提问by pedro.olimpio
I want know what am I doing wrong in this code:
我想知道我在这段代码中做错了什么：
#!/bin/sh
SERVICE_NAME=neocloud
PATH_TO_JAR=/etc/neocloud/cloud.jar
PID_PATH_NAME=/tmp/neocloud-pid
case in
start)
echo "Starting $SERVICE_NAME ..."
if [ ! -f $PID_PATH_NAME ]; then
nohup java -jar $PATH_TO_JAR /tmp 2>> /dev/null >> /dev/null &
echo $! > $PID_PATH_NAME
echo "$SERVICE_NAME started ..."
else
echo "$SERVICE_NAME is already running ..."
fi
;;
stop)
if [ -f $PID_PATH_NAME ]; then
PID=$(cat $PID_PATH_NAME);
echo "$SERVICE_NAME stoping ..."
kill $PID;
echo "$SERVICE_NAME stopped ..."
rm $PID_PATH_NAME
else
echo "$SERVICE_NAME is not running ..."
fi
;;
restart)
if [ -f $PID_PATH_NAME ]; then
PID=$(cat $PID_PATH_NAME);
echo "$SERVICE_NAME stopping ...";
kill $PID;
echo "$SERVICE_NAME stopped ...";
rm $PID_PATH_NAME
echo "$SERVICE_NAME starting ..."
nohup java -jar $PATH_TO_JAR /tmp 2>> /dev/null >> /dev/null &
echo $! > $PID_PATH_NAME
echo "$SERVICE_NAME started ..."
else
echo "$SERVICE_NAME is not running ..."
fi
;;
esac
executing via shI get the following error message:
通过执行sh我收到以下错误消息：
Syntax error: word unexpected (expecting "in")
语法错误：单词意外（期望“in”）
But in the casecommand I've the inword.
但是在case命令中我有这个in词。
Anyone know how to fix this bug?
有谁知道如何修复这个错误？
Thanks a lot!
非常感谢！
回答by mklement0
As posted here, your code is syntactically valid POSIX-like shell code, which you can verify at shellcheck.net(which will, however, warn you about potentially unwanted side effects due to not double-quoting your variable references (e.g., echo $PID_PATH_NAMErather than echo "$PID_PATH_NAME"), which however does NOT apply to $1in the casestatement[1]).
正如此处发布的那样，您的代码是语法上有效的类似 POSIX 的 shell 代码，您可以在shellcheck.net 上对其进行验证（但是，由于没有双引号您的变量引用（例如，echo $PID_PATH_NAME而不是echo "$PID_PATH_NAME"），然而这并不适用于$1在case声明中[1] ）。
Similarly, copying the code in your question and pasting it into a new file and using shon Ubuntu (which is Dash) to execute it, works fine too.
同样，复制问题中的代码并将其粘贴到一个新文件中，然后sh在 Ubuntu（即 Dash）上使用来执行它，也可以正常工作。
Thus - unless your shis not what it should be - I suspect that you have "weird" characters in your shell file, such as Unicode whitespace outside the standard ASCII range, which may looklike normal whitespace, but isn't; the Unicode no-break space character (U_00A0, UTF8-encoded as 0xC2 0xA0) is an example.
因此 - 除非你sh不是它应该的样子 - 我怀疑你的 shell 文件中有“奇怪”的字符，例如标准 ASCII 范围之外的 Unicode 空格，它可能看起来像正常的空格，但不是；Unicode 不间断空格字符（U_00A0, UTF8 编码为0xC2 0xA0）就是一个例子。
To look for such characters, run the following (where scriptrepresents your script):
要查找此类字符，请运行以下命令（其中script代表您的脚本）：
LC_ALL=C cat -e script
and look for M-and ^<letter>sequences in the output; for instance, the aforementioned no-break space shows up as M-BM-.
并在输出中查找M-和^<letter>序列；例如，前面提到的不间断空间显示为M-BM-。
[1] Double-quoting the argument given to the casestatement doesn't hurt, but is not necessary.
[1] 用双引号引用case声明中的论点没有坏处，但不是必须的。
While unquoted parameter/variable references are word-split and pathname-expanded in mostplaces in POSIX-like shells, caseis a curious exception.
虽然未加引号的参数/变量引用在 POSIX 类 shell中的大多数地方都是分词和路径名扩展的，但这case是一个奇怪的例外。
The following demonstrates this, and works with all major POSIX-like shells (dash, bash, ksh, zsh):
下面演示了这一点，并适用于所有主要的 POSIX 类 shell ( dash, bash, ksh, zsh)：
$ sh -c 'case in "foo *") echo "match";; *) echo "nomatch"; esac' - 'foo *'
match
Literal argument foo *matches the casebranch, even though $1is unquoted.
(Contrast this with the typical situation (e.g., echo $1), where the value of $1would be subject to both word-splitting and pathname expansion (globbing).)
文字参数foo *匹配case分支，即使$1是不带引号的。
（将此与典型情况（例如，echo $1）进行对比，其中 的值$1将受到分词和路径名扩展（通配）的影响。）
