Unix bash 错误 - 需要二元运算符
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Unix bash error - binary operator expected
提问by pogba123
So below is the code i have in my bash script. I'm getting an error saying binary operator expected when i give the command 2 arguments (doesnt give error when i give 1 argument). It does change the file permissions when i give 2 arguments because i can see it when i do ls -l but it still gives me this error. How do i fix it?
所以下面是我在我的 bash 脚本中的代码。当我给命令 2 个参数时，我收到一个错误，提示二元运算符预期（当我给 1 个参数时不会给出错误）。当我提供 2 个参数时，它确实会更改文件权限，因为当我执行 ls -l 时我可以看到它，但它仍然给我这个错误。我如何解决它？
for file in $@
do
chmod 755 $file
done
if [ -z $@ ]
then
echo "Error. No argument."
exit $ERROR_CODE_1
fi
i have added this now
我现在已经添加了这个
if [ ! -f "$*" ]
then
echo "Error. File does not exist"
exit $ERROR_NO_FILE
fi
But now when i enter more than 1 argument it just does everything in the if statement (i.e. prints error.file does not exist) even when the file does exist.
但是现在当我输入 1 个以上的参数时，即使文件确实存在，它也会执行 if 语句中的所有操作（即打印 error.file 不存在）。
回答by Jeff Schaller
Doing it another way: just ask how many parameters were passed:
换一种方式：只需询问传递了多少参数：
...
if [ $# -eq 0 ]
...
You get the error in your code because the $@ variable expands to multiple words, which leaves the testcommand looking like this:
由于 $@ 变量扩展为多个单词，您的代码中会出现错误，这使test命令看起来像这样：
[ -z parm1 parm2 parm3 ... ]
[ -z parm1 parm2 parm3 ... ]
回答by Barmar
$@is expanding to all the arguments, with spaces between them, so it looks like:
$@扩展到所有参数，它们之间有空格，所以它看起来像：
if [ -z file1 file2 file3 ]
but -zexpects just one word after it. You need to use $*and quote it, so it expands into a single word:
但-z期望在它之后只有一个字。您需要使用$*并引用它，因此它扩展为一个单词：
if [ -z "$*" ]
This expands into:
这扩展为：
if [ -z "file1 file2 file3" ]
Or just check the number of arguments:
或者只是检查参数的数量：
if [ $# -eq 0 ]
You should also put this check before the forloop. And you should quote the argument in the forloop, so you don't have problems with filenames that have spaces:
您还应该在for循环之前进行此检查。并且您应该在for循环中引用该参数，这样您就不会遇到带有空格的文件名的问题：
for file in "$@"
回答by andlrc
Wrap your parameters in double quotes to avoid word splitting and pathname expansion:
将参数用双引号括起来以避免分词和路径名扩展：
for file in "$@"
do
chmod 755 "$file"
done
if [ -z "$*" ] # Use $* instead of $@ as "$@" expands to multiply words.
then
echo "Error. No argument."
exit "$ERROR_CODE_1"
fi
You can however change the code a little:
但是，您可以稍微更改代码：
for file # No need for in "$@" as it's the default
do
chmod 755 "$file"
done
if [ "$#" -eq 0 ] # $# Contains numbers of arguments passed
then
>&2 printf 'Error. No argument.\n'
exit "$ERROR_CODE_1" # What is this?
fi
