shell - create a path from lines in a file on the unix command line

Question

I want to build a basic Java class path in a bash script.

This command works out all the jar files to go in my path:

find ~/jars | grep \.jar$

This lists all the jar files I want on my path, one per line.

How can I join together these lines so they are on one line, separated by : and put that into a variable I can use in my script when I invoke Java?

E.g.:

JARS=`find ~/jars | grep \.jar$`
#### somehow make JARS into one line with : between entries
java -cp $JARS ...

Answer 1

如果您已经使用过find，则不需要管道。

查找有-regex和-name选项，也有printf可用的：

您可以根据您的要求尝试此操作：

find ~/jars -name "*.jar" -printf"%p "

我没有测试，但它会输出所有带有完整路径的罐子，空格分隔。

Answer 2

jars=($(find ~/jars -type f -name \*.jar))   # save paths into array
classpath=$(IFS=:; echo "${jars[*]}")   # convert to colon-separated
java -cp "$classpath" ...               # and use it

Answer 3

您可以通过管道传输它并使用 sed 更改新行,：

sed ':a;N;$!ba;s/\n/:/g'

全部一起，

find ~/jars | grep \.jar$ | sed ':a;N;$!ba;s/\n/:/g'

您也可以使用 来执行此操作tr，尽管它会留下一个尾随冒号（格伦杰克曼评论）：

find ~/jars | grep \.jar$ | tr '\n' ':'

基于SED：如何替换换行符 (\n)？