为什么`读线; 声明$ line`和`find ... +`？

Question

以下脚本有效,但有什么用途declare $line？如果我删除它,它不起作用.

又什么的差异{} \;,并{} +在find命令？

awk '{print "old="$1" new="$2}' list.txt |\
while IFS= read line; do
    declare $line
    find /path -name '*.ext' -exec sed -i "s/\b$old\b/$new/" {} +
done

Answer 1

该declare是设置变量:您的awk命令发出形式的内容old=foo new=bar.运行declare old=foo new=bar设置这两个变量.

也就是说,这是一种错误和草率的方式.相反,用于read直接从输入文件中读取所需的字段并分配给变量(在BashFAQ#1中有更多内容):

while read -u 3 -r old new _; do
    find /path -name '*.ext' -exec sed -i "s/\b$old\b/$new/" {} +
done 3<list.txt

为了使这更安全一点,人们也可以逃避文字内容而不被视为正则表达式:

requote() { sed 's/[^^]/[&]/g; s/\^/\\^/g' <<< "$1"; };
substquote() { sed 's/[&/\]/\\&/g' <<< "$1"; }
while read -u 3 -r old new _; do
    find /path -name '*.ext' -exec \
        sed -i "s/\b$(requote "$old")\b/$(substquote "$new")/" {} +
done 3<list.txt

请注意,我没有改变使用\b,这是许多实现sed不支持的扩展.有关进行文字字符串替换的替代方法,请参阅BashFAQ#21.

为了完整性(虽然这个不相关的主题确实应该作为一个单独的问题提出 - 并且在这种情况下,可能已经被关闭为重复,因为之前已经被要求和回答过),请允许从find手册页中引用:

  -exec command {} +
         This  variant  of the -exec action runs the specified command on
         the selected files, but the command line is built  by  appending
         each  selected file name at the end; the total number of invoca?
         tions of the command will  be  much  less  than  the  number  of
         matched  files.   The command line is built in much the same way
         that xargs builds its command lines.  Only one instance of  `{}'
         is  allowed  within the command.  The command is executed in the
         starting directory.