Ruby 如何读取文件并从特定行中删除字符串
我有以下文件:Ruby 如何读取文件并从特定行中删除字符串,ruby,Ruby,我有以下文件: ids_to_remove.txt ID File_Name Row Name id\a_po87y GMT_dealer auto_dev id\ruio66 dao_wells auto_dev id\rzd766 123_Smart_option cia_read ... ... etc #This configuration file was written by: org.eclipse.equinox.internal [group
ids_to_remove.txt
ID File_Name Row Name
id\a_po87y GMT_dealer auto_dev
id\ruio66 dao_wells auto_dev
id\rzd766 123_Smart_option cia_read
...
...
etc
#This configuration file was written by: org.eclipse.equinox.internal
[groups]
auto_dev = id\a_po87y, id\rt7890, id\sdfs09, id\rzdo9k
qa_check = id\op9iu, id\guijm0, id\a_po87y
AD_read = id\a_po87y
#Content may be structured and packaged into modules to facilitate delivering, extending, and upgrading the Content
[groups]
AD_read = id\a_po87y
auto_dev = id\guijm0, id\oikju8, id\ruio66
CSI = id\kiopl, id\ruio66, id\o9i8u7
GMT\u dealer-details.txt
ID File_Name Row Name
id\a_po87y GMT_dealer auto_dev
id\ruio66 dao_wells auto_dev
id\rzd766 123_Smart_option cia_read
...
...
etc
#This configuration file was written by: org.eclipse.equinox.internal
[groups]
auto_dev = id\a_po87y, id\rt7890, id\sdfs09, id\rzdo9k
qa_check = id\op9iu, id\guijm0, id\a_po87y
AD_read = id\a_po87y
#Content may be structured and packaged into modules to facilitate delivering, extending, and upgrading the Content
[groups]
AD_read = id\a_po87y
auto_dev = id\guijm0, id\oikju8, id\ruio66
CSI = id\kiopl, id\ruio66, id\o9i8u7
dao_wells-details.txt
ID File_Name Row Name
id\a_po87y GMT_dealer auto_dev
id\ruio66 dao_wells auto_dev
id\rzd766 123_Smart_option cia_read
...
...
etc
#This configuration file was written by: org.eclipse.equinox.internal
[groups]
auto_dev = id\a_po87y, id\rt7890, id\sdfs09, id\rzdo9k
qa_check = id\op9iu, id\guijm0, id\a_po87y
AD_read = id\a_po87y
#Content may be structured and packaged into modules to facilitate delivering, extending, and upgrading the Content
[groups]
AD_read = id\a_po87y
auto_dev = id\guijm0, id\oikju8, id\ruio66
CSI = id\kiopl, id\ruio66, id\o9i8u7
在
ids_to_remove.txt
中,有近500个条目,一行中的项目以制表符分隔。在其他文件中,组值以空格分隔
File\u Name
的值表示E:/my\u files/*-details.txt
上的文件夹或文件,其中*
是File\u Name
值,如GMT\u dealer
、dao\u wells
或123\u Smart\u选项
对于每一行,我想删除由file\u Name
表示的文件中行ID为row Name
的行中出现的ID
值。例如,我想从文件GMT\u dealer
中id为auto\u dev
的行中删除字符串id\a\u po87y
<代码>id\a_po87y只应从组自动开发
中删除,组qa_检查
和AD_读取
中存在的相同id应保持原样。同样,它必须在E:/my_files
下的所有文件上执行
我写了下面的代码:
file_dir = 'E:/my_files'
file = File.open("E:/ids_to_remove.txt", "r")
contents = file.each_line.map { |line| line.split("\t") }.transpose
id, file_name, group = contents
id.each do |ids|
puts "For id: #{ids}"
file_name.each do |name|
value = File.open("#{file_dir}/#{name}-details.txt")
text = File.read(value)
text.each_line do |el|
group.each do |gr|
if el.match(/#{gr}/) then
print "group row #{gr}\n"
replace = text.gsub( /#{Regexp.escape(ids)}\,\s/, '').gsub( /#{Regexp.escape(ids)}/, '' ).gsub /,\s*$/, ''
end
group.shift
end
end
file_name.shift
end
end
id.shift
它不能满足我的需要。寻找任何建议
为了调试,我添加了一些put和输出
For ID: id\a_po87y
group row auto_dev
group row auto_dev
For ID: id\ruio66
For ID: id\rzd766
我会这样做:
file_dir = 'E:/my_files'
file = File.open("E:/ids_to_remove.txt", "r")
file.each_line.map do |line|
id, file_name, group = line.split
old_text = File.read("#{file_dir}/#{file_name}-details.txt")
new_text = []
old_text.each_line do |line|
if line =~ /=/
line_group, line_ids = line.split("=")
if line_group.strip == group.strip
line_ids = line_ids.split(",").reject { |l_id| l_id.strip == id }.join(",")
end
new_text << "#{line_group}=#{line_ids.chomp("\n")}"
else
new_text << line.chomp("\n")
end
end
File.write("#{file_dir}/#{file_name}-details.txt", new_text.join("\n"))
end
file\u dir='E:/my\u files'
file=file.open(“E:/ids\u to\u remove.txt”、“r”)
file.each_line.map do|line|
id,文件名,组=line.split
old_text=File.read(“#{File_dir}/#{File_name}-details.txt”)
新文本=[]
旧文本。每行都有|
如果行=~/=/
行组,行ID=line.split(“=”)
如果行_group.strip==group.strip
line_id=line_id.split(“,”).reject{l_id | l_id.strip==id}.join(“,”)
结束
新的文字萨瓦-感谢编辑,我是新到这个论坛,正在调整。你能帮我解决这个问题吗?谢谢“它不能满足我的需要”——它能做什么呢?@SergioTulentsev,谢谢你的调查。它不会替换目标文件中的字符串。也没有错误。不确定我应该在哪里更改。group.shift和file\u name.shift是怎么回事?你需要它们做什么?@SergioTulentsev,我正在尝试删除所有3个数组中的第一个元素-id、group和file_name,一旦第一个id或字符串在对应行的第一个文件中被删除。我的逻辑可能是错的。请寻找你的专业建议。太棒了,为什么我没有像你一样思考:)谢谢。