我无法弄清楚如何使这项工作：
我有：
icon-braille icon-bookmark-empty icon-blogger icon-adult icon-address-book
而且我要：
'icon-braille','icon-bookmark-empty','icon-blogger','icon-adult','icon-address-book'
问问题
83 次
4 回答
2
这是OSX友好的sed "s/ /','/g;s/^/'/;s/$/'/"：
$ echo "icon-braille icon-bookmark-empty" | sed "s/ /','/g;s/^/'/;s/$/'/"
'icon-braille','icon-bookmark-empty'
于 2013-01-03T13:13:27.410 回答
1
使用 shell 和 sed：
echo "'"$(sed "s/ /','/g")"'"
例子：
$ echo "'"$(sed "s/ /','/g")"'"
icon-braille icon-bookmark-empty icon-blogger icon-adult icon-address-book
'icon-braille','icon-bookmark-empty','icon-blogger','icon-adult','icon-address-book'
第一行被插入，第二行——产生。
于 2013-01-03T13:11:10.243 回答
0
$ echo "icon-braille icon-bookmark-empty icon-blogger icon-adult icon-address-book" | sed -e "s/ \+/','/g" -e "s/^/'/" -e "s/$/'/"
'icon-braille','icon-bookmark-empty','icon-blogger','icon-adult','icon-address-book'
于 2013-01-03T13:14:13.497 回答
0
您可以只使用外壳（假设是 bash）
$ list="icon-braille icon-bookmark-empty icon-blogger icon-adult icon-address-book"
$ result=""; sep=""
$ for word in $list; do result+=$sep\'$word\'; sep=,; done
$ echo "$result"
'icon-braille','icon-bookmark-empty','icon-blogger','icon-adult','icon-address-book'
于 2013-01-03T15:03:32.010 回答