如何将引用的参数传递给另一个程序

如何将引用的参数传递给另一个程序

本文介绍了如何将引用的参数传递给另一个程序的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

在bash脚本中,我需要将参数传递给另一个程序.参数中有空格,因此必须用引号引起来.在这种简单情况1中,一切正常:

In a bash script, I need to pass a parameter to another program. The parameter has spaces in so must be quoted. Everything works fine in this simple case 1:

/bin/echo /some/command --param="abc def ghi"

输出:

/some/command --param=abc def ghi

如果我想使bash脚本更加复杂,那么问题就开始了,在情况2和3中,参数值有所改变.

The problem begins if I want to make the bash script more sophisticated, somehow the parameter value has changed in case 2 and 3:

FOO="ghi"
DEFAULTS="--param=\"abc def $FOO\""
/bin/echo /some/command $DEFAULTS

DEFAULTS='--param="abc def '$FOO'"'
/bin/echo /some/command $DEFAULTS

输出:

/some/command --param="abc def ghi"
/some/command --param="abc def ghi"

abc def ghi周围的双引号在2和3中显示,而在1中没有显示.

The double quotes surrounding abc def ghi are shown in 2 and 3, whereas they are not shown for 1.

通过/some/command实际打印它作为第一个参数收到的内容,可以更好地说明这一点.在这里,整个字符串"abc def ghi"作为param参数的值被接收:

This is better illustrated by having /some/command actually print what it receives as the first parameter. Here the whole string "abc def ghi" is received as the value of the param parameter:

perl -le'print $ARGV[0]' param="abc def ghi"

输出(右):

param=abc def ghi

但是这里只收到"abc,并带有一个引号:

But here only "abc is received, with one leading quote:

DEFAULTS='param="abc def ghi"'
perl -le'print $ARGV[0]' $DEFAULTS

DEFAULTS="param=\"abc def ghi\""
perl -le'print $ARGV[0]' $DEFAULTS

输出(错误):

param="abc
param="abc

如何使用bash动态构造类似param="abc def ghi"的字符串并将其原封不动地传递给另一个命令?

How do I use bash to dynamically construct a string like param="abc def ghi" and be able to pass it intact to another command?

推荐答案

使用数组;这就是为什么引入它们的原因.

Use an array; this is why they were introduced.

FOO="ghi"
# Contains one element, but not the literal quotes
DEFAULTS=(--param="abc def $FOO")
# command gets 1 argument, the single element of the array
/some/command "${DEFAULTS[@]}"

# Two elements
DEFAULTS=(--param="abc def $FOO" --other="foo bar")
# command gets 2 arguments
/some/command "${DEFAULTS[@]}"

这篇关于如何将引用的参数传递给另一个程序的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

09-04 23:09