bash 目录路径作为bash中的命令行参数
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directory path as command line argument in bash
提问by user227666
The following bash script finds a .txtfile from the given directory path, then changes one word (change mountain to sea) from the .txtfile
下面的 bash 脚本.txt从给定的目录路径中找到一个文件，然后从.txt文件中更改一个单词（将山更改为海）
#!/bin/bash
FILE=`find /home/abc/Documents/2011.11.* -type f -name "abc.txt"`
sed -e 's/mountain/sea/g' $FILE
The output I am getting is ok in this case. My problem is if I want to give the directory path as command line argument then it is not working. Suppose, I modify my bash script to:
在这种情况下，我得到的输出没问题。我的问题是，如果我想将目录路径作为命令行参数提供，则它不起作用。假设，我将 bash 脚本修改为：
#!/bin/bash
FILE=`find -type f -name "abc.txt"`
sed -e 's/mountain/sea/g' $FILE
and invoke it like:
并像这样调用它：
./test.sh /home/abc/Documents/2011.11.*
Error is:
错误是：
./test.sh: line 2: /home/abc/Documents/2011.11.10/abc.txt: Permission denied
Can anybody suggest how to access directory path as command line argument?
有人可以建议如何访问目录路径作为命令行参数吗？
回答by Barmar
Your first line should be:
你的第一行应该是：
FILE=`find "$@" -type f -name "abc.txt"`
The wildcard will be expanded before calling the script, so you need to use "$@"to get all the directories that it expands to and pass these as the arguments to find.
通配符将在调用脚本之前扩展，因此您需要使用它"$@"来获取它扩展到的所有目录并将这些作为参数传递给find.
回答by anubhava
You don't need to pass .*to your script.
您不需要传递.*给您的脚本。
Have your script like this:
让你的脚本是这样的：
#!/bin/bash
# some sanity checks here
path=""
find "$path".* -type f -name "abc.txt" -exec sed -i.bak 's/mountain/sea/g' '{}' \;
And run it like:
并像这样运行它：
./test.sh "/home/abc/Documents/2011.11"
PS:See how sed can be invoked directly from find itself using -execoption.
PS：查看如何使用-exec选项直接从 find 自身调用 sed 。
