With first 9 arguments being referred from
With first 9 arguments being referred from $1-$9, $10 gets interpreted as $1 followed by a 0. How do I account for this and access arguments to functions greater than 10?
前9个参数从$ 1- $ 9引用，$ 10被解释为$ 1后跟0.我如何解释这个并访问大于10的函数的参数？
Thanks.
谢谢。
3 个解决方案
#1
26
Use :
使用 ：
#!/bin/bash
echo ${10}
To test the difference with $10, code in foo.sh :
要测试10美元的差异，foo.sh中的代码：
#!/bin/bash
echo $10
echo ${10}
Then :
然后 ：
$ ./foo.sh first 2 3 4 5 6 7 8 9 10
first0
10
the same thing is true if you have :
如果您有以下情况，情况也是如此：
foobar=42
foo=FOO
echo $foobar # echoes 42
echo ${foo}bar # echoes FOObar
Use {} when you want to remove ambiguities ...
当您想要消除含糊不清时使用{}
my2c
MY2C
#2
4
If you are using bash, then you can use ${10}.
如果您使用的是bash，则可以使用$ {10}。
${...} syntax seems to be POSIX-compliant in this particular case, but it might be preferable to use the command shift like that :
在这种特殊情况下，$ {...}语法似乎与POSIX兼容，但最好使用命令shift，如下所示：
while [ "$*" != "" ]; do echo "Arg: $1" shift done
EDIT: I noticed I didn't explain what shift does. It just shift the arguments of the script (or function). Example:
编辑：我注意到我没有解释什么转变。它只是移动脚本（或函数）的参数。例：
> cat script.sh echo "$1" shift echo "$1" > ./script.sh "first arg" "second arg" first arg second arg
In case it can help, here is an example with getopt/shift :
如果它可以提供帮助，这里有一个getopt / shift示例：
while getopts a:bc OPT; do case "$OPT" in 'a') ADD=1 ADD_OPT="$OPTARG" ;; 'b') BULK=1 ;; 'c') CHECK=1 ;; esac done shift $( expr $OPTIND - 1 ) FILE="$1"
#3
2
In general, to be safe that the whole of a given string is used for the variable name when Bash is interpreting the code, you need to enclose it in braces: ${10}
通常，为了确保在Bash解释代码时将整个给定字符串用于变量名称，您需要将其括在大括号中：$ {10}
-, gets interpreted as
With first 9 arguments being referred from $1-$9, $10 gets interpreted as $1 followed by a 0. How do I account for this and access arguments to functions greater than 10?
前9个参数从$ 1- $ 9引用，$ 10被解释为$ 1后跟0.我如何解释这个并访问大于10的函数的参数？
Thanks.
谢谢。
3 个解决方案
#1
26
Use :
使用 ：
#!/bin/bash
echo ${10}
To test the difference with $10, code in foo.sh :
要测试10美元的差异，foo.sh中的代码：
#!/bin/bash
echo $10
echo ${10}
Then :
然后 ：
$ ./foo.sh first 2 3 4 5 6 7 8 9 10
first0
10
the same thing is true if you have :
如果您有以下情况，情况也是如此：
foobar=42
foo=FOO
echo $foobar # echoes 42
echo ${foo}bar # echoes FOObar
Use {} when you want to remove ambiguities ...
当您想要消除含糊不清时使用{}
my2c
MY2C
#2
4
If you are using bash, then you can use ${10}.
如果您使用的是bash，则可以使用$ {10}。
${...} syntax seems to be POSIX-compliant in this particular case, but it might be preferable to use the command shift like that :
在这种特殊情况下，$ {...}语法似乎与POSIX兼容，但最好使用命令shift，如下所示：
while [ "$*" != "" ]; do echo "Arg: $1" shift done
EDIT: I noticed I didn't explain what shift does. It just shift the arguments of the script (or function). Example:
编辑：我注意到我没有解释什么转变。它只是移动脚本（或函数）的参数。例：
> cat script.sh echo "$1" shift echo "$1" > ./script.sh "first arg" "second arg" first arg second arg
In case it can help, here is an example with getopt/shift :
如果它可以提供帮助，这里有一个getopt / shift示例：
while getopts a:bc OPT; do case "$OPT" in 'a') ADD=1 ADD_OPT="$OPTARG" ;; 'b') BULK=1 ;; 'c') CHECK=1 ;; esac done shift $( expr $OPTIND - 1 ) FILE="$1"
#3
2
In general, to be safe that the whole of a given string is used for the variable name when Bash is interpreting the code, you need to enclose it in braces: ${10}
通常，为了确保在Bash解释代码时将整个给定字符串用于变量名称，您需要将其括在大括号中：$ {10}
followed by a 0. How do I account for this and access arguments to functions greater than 10? With first 9 arguments being referred from
With first 9 arguments being referred from $1-$9, $10 gets interpreted as $1 followed by a 0. How do I account for this and access arguments to functions greater than 10?
前9个参数从$ 1- $ 9引用，$ 10被解释为$ 1后跟0.我如何解释这个并访问大于10的函数的参数？
Thanks.
谢谢。
3 个解决方案
#1
26
Use :
使用 ：
#!/bin/bash
echo ${10}
To test the difference with $10, code in foo.sh :
要测试10美元的差异，foo.sh中的代码：
#!/bin/bash
echo $10
echo ${10}
Then :
然后 ：
$ ./foo.sh first 2 3 4 5 6 7 8 9 10
first0
10
the same thing is true if you have :
如果您有以下情况，情况也是如此：
foobar=42
foo=FOO
echo $foobar # echoes 42
echo ${foo}bar # echoes FOObar
Use {} when you want to remove ambiguities ...
当您想要消除含糊不清时使用{}
my2c
MY2C
#2
4
If you are using bash, then you can use ${10}.
如果您使用的是bash，则可以使用$ {10}。
${...} syntax seems to be POSIX-compliant in this particular case, but it might be preferable to use the command shift like that :
在这种特殊情况下，$ {...}语法似乎与POSIX兼容，但最好使用命令shift，如下所示：
while [ "$*" != "" ]; do echo "Arg: $1" shift done
EDIT: I noticed I didn't explain what shift does. It just shift the arguments of the script (or function). Example:
编辑：我注意到我没有解释什么转变。它只是移动脚本（或函数）的参数。例：
> cat script.sh echo "$1" shift echo "$1" > ./script.sh "first arg" "second arg" first arg second arg
In case it can help, here is an example with getopt/shift :
如果它可以提供帮助，这里有一个getopt / shift示例：
while getopts a:bc OPT; do case "$OPT" in 'a') ADD=1 ADD_OPT="$OPTARG" ;; 'b') BULK=1 ;; 'c') CHECK=1 ;; esac done shift $( expr $OPTIND - 1 ) FILE="$1"
#3
2
In general, to be safe that the whole of a given string is used for the variable name when Bash is interpreting the code, you need to enclose it in braces: ${10}
通常，为了确保在Bash解释代码时将整个给定字符串用于变量名称，您需要将其括在大括号中：$ {10}
-$
