打印封装的依赖关系树

本文介绍了打印封装的依赖关系树的处理方法，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

使用这个文件，我想打印一个软件包的依赖关系树，给定一个单一的基础包。例如，拿bash包

@庆典＃几行删除要求：coreutils的libintl8 libncursesw10 libreadline7 _Update-INFO-DIR cygwin的

我想找到般所需的输出的包，部分例如

庆典庆典的coreutils庆典的coreutils libattr1庆典的coreutils libattr1 libintl8庆典的coreutils libattr1 libintl8 libiconv2庆典的coreutils libattr1 libintl8 _autorebase庆典的coreutils libattr1 libintl8 _autorebase底垫庆典的coreutils libattr1 libintl8 _autorebase变基冲刺庆典的coreutils libattr1 libintl8 _autorebase变基冲刺的cygwin庆典的coreutils libattr1 libintl8 _autorebase变基冲刺cygwin的基础cygwin的

我有这样的命令，但它不会递归

＃！AWK -f$ 1 ==@{  PKG = $ 16}$ 1 ==要求：{  对于（i = 2; I＆LT; = NF;我++）    请求数[PKG] [I-1] = $我}结束 {  查询=庆典  对（在请求数[查询] PKG）{    打印请求数[查询] [PKG]  }}

解决方案

＃！的/ usr /斌/的awk -f@include加盟$ 1 ==@{  助攻= $ 16}$ 1 ==要求：{  对于（Z = 2; z，其中= NF; Z ++）    请求数[助攻] [Z-1] = $ž}结束 {  PRPG（庆典）}功能smartmatch（小，大，价值）{  对于（每个大）    值[大[每]  在值返回小}功能PRPG（FPG）{  如果（smartmatch（FPG，SPATH））回报  SPATH [长度（SPATH）+1] = FPG  打印已加入（SPATH，1，长度（SPATH））  如果（IsArray的（请求数[FPG））    对于（每个请求数[FPG]）      PRPG（请求数[FPG] [每]）  删除SPATH [长度（SPATH）}

源

Using this file, I would like to print a tree of package dependencies, given a single base package. For example, take the Bash package

@ bash # few lines removed requires: coreutils libintl8 libncursesw10 libreadline7 _update-info-dir cygwin

I would like find-like output of the required packages, partial example

bash bash coreutils bash coreutils libattr1 bash coreutils libattr1 libintl8 bash coreutils libattr1 libintl8 libiconv2 bash coreutils libattr1 libintl8 _autorebase bash coreutils libattr1 libintl8 _autorebase rebase bash coreutils libattr1 libintl8 _autorebase rebase dash bash coreutils libattr1 libintl8 _autorebase rebase dash cygwin bash coreutils libattr1 libintl8 _autorebase rebase dash cygwin base-cygwin

I have this command but it does not recurse

#!awk -f $1 == "@" { pkg = $2 } $1 == "requires:" { for (i=2; i<=NF; i++) reqs[pkg][i-1] = $i } END { query = "bash" for (pkg in reqs[query]) { print reqs[query][pkg] } }

解决方案

#!/usr/bin/awk -f @include "join" $1 == "@" { apg = $2 } $1 == "requires:" { for (z=2; z<=NF; z++) reqs[apg][z-1] = $z } END { prpg("bash") } function smartmatch(small, large, values) { for (each in large) values[large[each]] return small in values } function prpg(fpg) { if (smartmatch(fpg, spath)) return spath[length(spath)+1] = fpg print join(spath, 1, length(spath)) if (isarray(reqs[fpg])) for (each in reqs[fpg]) prpg(reqs[fpg][each]) delete spath[length(spath)] }

Source