如何在CSV中查找和替换元素

debugcn 发表于 Dev

用户3839012

如何查找和替换"West"以下CSV文件示例中的模式（例如引号）？

"LastName","FirstName","","","890","","6G","","S "West" AVENUE","","CITY","ZIP"

锡人

您不能使用CSV类读取此内容，因为它是格式错误的CSV字符串。有时发生这种情况是因为无论生成它的人都不知道他们在做什么：

require 'csv'
foo = '"LastName","FirstName","","","890","","6G","","S "West" AVENUE","","CITY","ZIP"'
arr_of_arrs = CSV.parse(foo)

然后导致异常：

Missing or stray quote in line 1 (CSV::MalformedCSVError)

相反，要解决此问题，您必须先修复数据，然后解析。这是一个起点：

/(?<=\s)("[^"]+")(?=\s)/

http://rubular.com/r/sWEkx07Zyo

模式在引号之间寻找东西，并用前导和尾随空格括起来。不会捕获空格。

以下是适用于此特定示例的一些代码：

foo = '"LastName","FirstName","","","890","","6G","","S "West" AVENUE","","CITY","ZIP"'

REGEX = /(?<=\s)("[^"]+")(?=\s)/

word = foo[REGEX]
foo[REGEX] = word[1..-2]
puts foo
# >> "LastName","FirstName","","","890","","6G","","S West AVENUE","","CITY","ZIP"

此时可以使用CSV：

require 'csv'
arr_of_arrs = CSV.parse(foo)
# => [["LastName",
#      "FirstName",
#      "",
#      "",
#      "890",
#      "",
#      "6G",
#      "",
#      "S West AVENUE",
#      "",
#      "CITY",
#      "ZIP"]]

这些东西可能会令人困惑：

word = foo[REGEX]
foo[REGEX] = word[1..-2]

foo\[...\]是String类的一部分，并且是查找和替换字符串中字符的一种很好的便捷方法。

可以使CSV解析器对嵌入的引号感到满意，因此，如果扔掉它们太繁琐，则可以执行以下操作：

word = foo[REGEX]
foo[REGEX] = '"%s"' % word

require 'csv'
arr_of_arrs = CSV.parse(foo)
# => [["LastName",
#      "FirstName",
#      "",
#      "",
#      "890",
#      "",
#      "6G",
#      "",
#      "S \"West\" AVENUE",
#      "",
#      "CITY",
#      "ZIP"]]