Question

来自输入文件，其中包含多行'name = value'字段（前20行）



Job=C201A005 include=PC201
Job=C201A005 proc=RUNTRACN
Job=C201A005 proc=RUNTRACN utilpgm=IEFBR14
Job=C201A005 proc=DD0005
Job=C201A005 proc=DD0005 utilpgm=SORT
Job=C201A005 proc=DD0005 include=SORT
Job=C201A005 proc=DD0005 file=PC201.AD.SRVCPT.INPUT
Job=C201A005 proc=DD0005 pdsmem=ULDH.REHOST3.PARM/SA005
Job=C201A005 proc=DD0005 include=BATC2P
Job=C201A005 proc=DD0005 include=BATC2P file=ULDH.REHOST3.LOADBTCH
Job=C201A005 proc=DD0005 include=BATC2P file=SYS2.DB2DBP1.SDSNLOAD
Job=C201A005 proc=DD0005 include=BATC2P file=SYS2.DB2DBP1.RUNLIB.LOAD
Job=C201A005 proc=DD0005 include=BATC2P file=SYS2.S99I0062.PROD.SCSQLOAD
Job=C201A005 proc=DD0005 include=BATC2P file=ULDH.REHOST3.REXX
Job=C201A005 proc=DD0005 include=BATC2P file=ULDH.REHOST3.REXX
Job=C201A005 proc=ABENDJOB
Job=C201A005 proc=ABENDJOB utilpgm=IEFBR14
Job=C201A005 proc=ABENDJOB file=PROD.NET.MSGS
Job=C201A018 include=PC201
Job=C201A018 proc=RUNTRACN

Job=C201A005 include=PC201 Job=C201A005 proc=RUNTRACN Job=C201A005 proc=RUNTRACN utilpgm=IEFBR14 Job=C201A005 proc=DD0005 Job=C201A005 proc=DD0005 utilpgm=SORT Job=C201A005 proc=DD0005 include=SORT Job=C201A005 proc=DD0005 file=PC201.AD.SRVCPT.INPUT Job=C201A005 proc=DD0005 pdsmem=ULDH.REHOST3.PARM/SA005 Job=C201A005 proc=DD0005 include=BATC2P Job=C201A005 proc=DD0005 include=BATC2P file=ULDH.REHOST3.LOADBTCH Job=C201A005 proc=DD0005 include=BATC2P file=SYS2.DB2DBP1.SDSNLOAD Job=C201A005 proc=DD0005 include=BATC2P file=SYS2.DB2DBP1.RUNLIB.LOAD Job=C201A005 proc=DD0005 include=BATC2P file=SYS2.S99I0062.PROD.SCSQLOAD Job=C201A005 proc=DD0005 include=BATC2P file=ULDH.REHOST3.REXX Job=C201A005 proc=DD0005 include=BATC2P file=ULDH.REHOST3.REXX Job=C201A005 proc=ABENDJOB Job=C201A005 proc=ABENDJOB utilpgm=IEFBR14 Job=C201A005 proc=ABENDJOB file=PROD.NET.MSGS Job=C201A018 include=PC201 Job=C201A018 proc=RUNTRACN

我正在尝试构建一个数据结构来为每个“Job”保存一个每种类型的数组结构（proc，include，file，program，utilpgm ..）' 这是我写的代码。我无法在适当的数组上推送'值'。

 #!/usr/bin/ruby

jobs={} #holds everything
currentjob=""
files=[], pdsmems=[], includs=[], procs=[], programs=[], utilpgms=[]

ARGF.each do |line|
    # chop the line into fields,
    # where every field is a 'name=value' pair.
    line.chomp!
    fields = line.split(' ')

    print "Currentjob = >#{currentjob}<\n"

    for field in fields do
        fieldname, value = field.split('=')        # split into name, value
        #print ">> #{fieldname}=#{value}\n"
        if fieldname =~ /[jJ]ob/
            if currentjob != value
                print "NEW JOB #{fields[0].split('=')[1]}\n"
                # this is a new job, so build the hash for old one
                files=[], pdsmems=[], includs=[], procs=[], programs=[], utilpgms=[]
                jobelements = {'file' => files, 'pdsmem' => pdsmems,
                               'include' => includs, 'proc' => procs,
                               'program' => programs, 'utilpgm' => utilpgms}
                jobs[value] = jobelements
                #print "Hash assigned. job = #{jobs} \n"
                currentjob = value
            end
            next
        end
        print "fieldname = #{fieldname}\n"

        # push 'value' on the appro. array in the 'jobelements' hash
        # which is value of the 'currentjob' key.
        #jobs[currentjob][fieldname].push(value)   # <== HELP!
        #jobs[currentjob][fieldname] << value      # <== HELP!

    end
end

 #print the data structure
print jobs

我似乎无法正确地引用正确的数组并将elemnt推入其中。帮助赞赏。干杯! P.S。

预期输出如下：

 #!/usr/bin/ruby

jobs={} #holds everything
currentjob=""
files=[], pdsmems=[], includs=[], procs=[], programs=[], utilpgms=[]

ARGF.each do |line|
    # chop the line into fields,
    # where every field is a 'name=value' pair.
    line.chomp!
    fields = line.split(' ')

    print "Currentjob = >#{currentjob}<\n"

    for field in fields do
        fieldname, value = field.split('=')        # split into name, value
        #print ">> #{fieldname}=#{value}\n"
        if fieldname =~ /[jJ]ob/
            if currentjob != value
                print "NEW JOB #{fields[0].split('=')[1]}\n"
                # this is a new job, so build the hash for old one
                files=[], pdsmems=[], includs=[], procs=[], programs=[], utilpgms=[]
                jobelements = {'file' => files, 'pdsmem' => pdsmems,
                               'include' => includs, 'proc' => procs,
                               'program' => programs, 'utilpgm' => utilpgms}
                jobs[value] = jobelements
                #print "Hash assigned. job = #{jobs} \n"
                currentjob = value
            end
            next
        end
        print "fieldname = #{fieldname}\n"

        # push 'value' on the appro. array in the 'jobelements' hash
        # which is value of the 'currentjob' key.
        #jobs[currentjob][fieldname].push(value)   # <== HELP!
        #jobs[currentjob][fieldname] << value      # <== HELP!

    end
end

 #print the data structure
print jobs

Answer 1

以下是一些如何创建数组和哈希的基础知识：

#create empty array called jobs
jobs = []
#or use jobs = Array.new
#create a hash  from one of your lines:
line = {"Job" => "C201A018", "proc" => "RUNTRACN" }
#add the line to your jobs
jobs.push(line)
#or do it this way:
#jobs << line

Answer 2

初始化数组变量的方式可能无法按照您希望的方式工作。

files=[], pdsmems=[], includs=[], procs=[], programs=[], utilpgms=[]

不会创建一个名为files的新空数组，但文件将如下所示：

[[], [], [], [], [], []]

这是因为,不是语句的结尾，而是会在其后面和之前的所有其他值上创建一个数组。

示例：

a = [1], b = [2], c = [3]
puts a.to_s # returns '[[1], [2], [3]]' and not '[1]'

您可以通过将,替换为;或新行来解决此问题。 ;终止一个语句，就像换行符一样;

要想方便地测试短红宝石片段，请查看pry或irb（可能已经安装了红宝石）。

Ruby：在Hash中的数组上推送元素

2 个答案: