如何在基于Swift 5.2的RegEx中添加Unreedy Search Flag修饰符

Question

如何在基于Swift 5.2的RegEx中使用Ungridy标志

使用Regex101.com进行测试-

链接：https://regex101.com/r/mAvptf/1

当我分配了U标志时，它就工作了。在没有U标志的情况下，它返回单个结果块。

如何使用(?U)或.*?来获得与U标志相同的效果？

// WARNING: You included a flag that Swift doesn't support: U
//          When this flag is set, it inverts the "greediness" of the quantifiers so that they are not greedy by default, but become greedy if followed by '?'.
//          As an alternative, this effect can also be achieved by setting a (?U) modifier setting within the pattern or by a question mark behind a quantifier (e.g. .*?).

游乐场代码：

import Foundation
// WARNING: You included a flag that Swift doesn't support: U
//          When this flag is set, it inverts the "greediness" of the quantifiers so that they are not greedy by default, but become greedy if followed by '?'.
//          As an alternative, this effect can also be achieved by setting a (?U) modifier setting within the pattern or by a question mark behind a quantifier (e.g. .*?).
let pattern = #"(- examples: [{(.*[n])*).*?(?:}}]$)"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"""
/**
Gets a list of custom fields.
- GET
- OAuth:
- name:
- examples: [{contentType=application/json, example={
"s" : [ {
"name" : "name",
}, {
"name" : "name",
} ],
"fields" : [ {
"items" : [ "listItems", "listItems" ],
"name" : "name",
"message" : "message"
},
}, {
"items" : [ "listItems", "listItems" ],
"name" : "name",
} ]
}}]
- returns: Request<Fields>
*/
/**
Gets a
- examples: [{contentType=application/json, example={
"fields" : [ {
"name" : "name",
} ],
"fields" : [ {
"listItems" : [ "listItems", "listItems" ],
"name" : "name",
} ]
}}]
- returns: Request<fields>
*/
"""#
let stringRange = NSRange(location: 0, length: testString.utf16.count)
let matches = regex.matches(in: testString, range: stringRange)
var result: [[String]] = []
for match in matches {
var groups: [String] = []
for rangeIndex in 1 ..< match.numberOfRanges {
let nsRange = match.range(at: rangeIndex)
guard !NSEqualRanges(nsRange, NSMakeRange(NSNotFound, 0)) else { continue }
let string = (testString as NSString).substring(with: nsRange)
groups.append(string)
}
if !groups.isEmpty {
result.append(groups)
}
}
print(result)

我试图得到两个单独的结果：Ungreedy工作得很好，但Swift5.2 不支持它

预期结果

- examples: [{contentType=application/json, example={
"s" : [ {
"name" : "name",
}, {
"name" : "name",
} ],
"fields" : [ {
"items" : [ "listItems", "listItems" ],
"name" : "name",
"message" : "message"
},
}, {
"items" : [ "listItems", "listItems" ],
"name" : "name",
} ]
}}]

Answer 1

要获得想要的效果，可以将正则表达式更改为以下内容。

let pattern = #"(- examples: [{(.*?[n])*?).*?(?:}}]$)"#

注意，在第一个和第二个*之后，我添加了一个?。这使表达式具有与U标志相同的效果。

您现在应该可以访问这两个结果，使用上面的代码和新表达式可以得到以下输出：

[
[
"- examples: [{contentType=application/json, example={n  "s" : [ {n    "name" : "name",n  }, {n    "name" : "name",n  } ],n  "fields" : [ {n    "items" : [ "listItems", "listItems" ],n    "name" : "name",n      "message" : "message"n    },n  }, {n    "items" : [ "listItems", "listItems" ],n    "name" : "name",n  } ]n",
"  } ]n"
],
[
"- examples: [{contentType=application/json, example={n  "fields" : [ {n    "name" : "name",n  } ],n  "fields" : [ {n    "listItems" : [ "listItems", "listItems" ],n    "name" : "name",n  } ]n",
"  } ]n"
]
]

给你https://regex101.com/r/FsJ5gZ/1

---

您甚至可以去掉我添加的第一个?，只需使用以下表达式：

let pattern = #"(- examples: [{(.*[n])*?).*?(?:}}]$)"#

这似乎足以得到你的两个结果。

https://regex101.com/r/vPGRwk/1