我有一个字符串，可以是以下任意一种：
tempString = "What? The ID: 54673 Over there"
tempString = "Jump... ID: 4561E how high"
基本上，我想从字符串中检索五个字符的 ID：。对于大多数语言，我可以通过查找 ID: 来执行此操作，将 4 添加到 ID: 的开头，然后取接下来的五个字符。
例如，在 Excel 中：
=MID("Jump... ID: 4561E how high",FIND("ID:","Jump... ID: 4561E how high")+4, 5)
将 = 4561E。
我们如何用 Swift 做到这一点？
【问题讨论】：
在“ID:”之后从 [0-9] 或 [A-F] 中查找前 5 个字符的正则表达式方法（仅当不是这种情况时才使用十六进制字母，您可以使用 A-Z）。
if let range = string.range(of: "(?<=ID: )[0-9A-F]{5}", options: .regularExpression) {
let id = string[range] //
// if you need a String instead of a substring
let stringID = String(string[range])
}
编辑/更新：
看看你的答案，看起来要求与你原来的帖子完全不同，要找到一个由 8 个十六进制字符后跟连字符然后是 4 个十六进制字符后跟连字符（3 次）的 IF 化合物，而不是你可以使用的 12 个十六进制字符以下正则表达式"([0-9a-f]{8}-)([0-9a-f]{4}-){3}([0-9a-f]{12})"：
let dataString = """
{
"@odata.context": "$metadata#GeoFences(Points())/$entity",
"ID": "2b2a2abc-5962-4290-92b4-773025ffd50b",
"Points": {
"POINT_TYPE": "F",
"POINT_NUM": 0,
"LATITUDE": 32.92197686725423,
"LONGITUDE": -117.04306932263532,
"parent_ID": "2b2a2abc-5962-4290-92b4-773025ffd50b"
},
"GEOFENCE_NAME": "New Fence",
"GEOFENCE_TYPE": "O",
"PRIVACY": "X",
"CENTER_LAT": 32.92043316309709,
"CENTER_LONG": -117.04286922250975,
"ZOOM_LAT": 0.006238797350533787,
"ZOOM_LONG": 0.005345531926053582,
"PATH_TOLERANCE": 5,
"ENTRANCE_TOLERANCE": 5
}
"""
if let range = dataString.range(of: "([0-9a-f]{8}-)([0-9a-f]{4}-){3}([0-9a-f]{12})", options: .regularExpression) {
let id = dataString[range] // 4561E
print("ID:", id)
// if you need a String instead of a substring
let stringID = String(dataString[range])
print("stringID:", stringID)
}
这将打印出来
ID：2b2a2abc-5962-4290-92b4-773025ffd50b
字符串ID：2b2a2abc-5962-4290-92b4-773025ffd50b
请注意，您的代码将导致"2b2a2abc-5962-4290-92b4-773025ffd50
编辑/更新2：
考虑到您的字符串是 JSON，您可以简单地解码您的字符串 id：
struct Root: Codable {
let id: String
enum CodingKeys: String, CodingKey {
case id = "ID"
}
}
do {
let id = try JSONDecoder().decode(Root.self, from: Data(dataString.utf8)).id
print(id) // "2b2a2abc-5962-4290-92b4-773025ffd50b"
} catch {
print(error)
}
如果您需要解码所有数据：
struct Root: Codable {
let odataContext, id: String
let points: Points
let geofenceName, geofenceType, privacy: String
let centerLat, centerLong, zoomLat, zoomLong: Double
let pathTolerance, entranceTolerance: Int
enum CodingKeys: String, CodingKey {
case odataContext = "@odata.context", id = "ID", points = "Points", geofenceName = "GEOFENCE_NAME", geofenceType = "GEOFENCE_TYPE", privacy = "PRIVACY", centerLat = "CENTER_LAT", centerLong = "CENTER_LONG", zoomLat = "ZOOM_LAT", zoomLong = "ZOOM_LONG", pathTolerance = "PATH_TOLERANCE", entranceTolerance = "ENTRANCE_TOLERANCE"
}
}
struct Points: Codable {
let pointType: String
let pointNum: Int
let latitude, longitude: Double
let parentID: String
enum CodingKeys: String, CodingKey {
case pointType = "POINT_TYPE", pointNum = "POINT_NUM", latitude = "LATITUDE", longitude = "LONGITUDE", parentID = "parent_ID"
}
}
do {
let root = try JSONDecoder().decode(Root.self, from: Data(dataString.utf8))
print("ID:", root.id) // ID: 2b2a2abc-5962-4290-92b4-773025ffd50b
print("Root:", root) // Root: Root(odataContext: "$metadata#GeoFences(Points())/$entity", id: "2b2a2abc-5962-4290-92b4-773025ffd50b", points: __lldb_expr_111.Points(pointType: "F", pointNum: 0, latitude: 32.92197686725423, longitude: -117.04306932263532, parentID: "2b2a2abc-5962-4290-92b4-773025ffd50b"), geofenceName: "New Fence", geofenceType: "O", privacy: "X", centerLat: 32.92043316309709, centerLong: -117.04286922250975, zoomLat: 0.0062387973505337885, zoomLong: 0.005345531926053582, pathTolerance: 5, entranceTolerance: 5)
} catch {
print(error)
}
【讨论】：
您可以在字符串上使用range(of:) 找到子字符串的range，然后使用该range 的upperBound 来索引原始字符串以获取从该点开始的子字符串。最后，prefix(5) 允许您获取该子字符串的前 5 个字符，String() 将Substring.Subsequence 转换回普通的String：
if let range = tempString.range(of: "ID: ") {
let ident = String(tempString[range.upperBound...].prefix(5))
print(ident)
}
注意：range(of:)来自Foundation框架，所以需要import Foundation或import另一个使用Foundation的框架如@987654334 @ 或Cocoa。
【讨论】：
if let index = tempString.range(of: "ID: ")?.upperBound {
#"whatever"# 而不是转义你的字符串#"ID":""#
对于你的情况可以使用 NSRegularExpression：
let tempString1 = "What? The ID: 54673 Over there"
let tempString2 = "Jump... ID: 4561E how high"
func matches(for regex: String, in text: String) -> [String] {
do {
let regex = try NSRegularExpression(pattern: regex)
let results = regex.matches(in: text,
range: NSRange(text.startIndex..., in: text))
return results.map {
String(text[Range($0.range, in: text)!])
}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
var res1 = matches(for: "[0-9]+[A-Z]*", in: tempString1) // 54673
var res2 = matches(for: "[0-9]+[A-Z]*", in: tempString2) // 4561E
【讨论】：
好的，我开始这个练习是因为我懒得解码 JSON，因为我必须构建结构并让它们正确。所以，我想我会很快做一个子串。好吧，解码会更好、更快、更准确，但我将在下面展示我的代码。基本上，我想从 JSON 字符串中获取 ID：
{
"@odata.context": "$metadata#GeoFences(Points())/$entity",
"ID": "2b2a2abc-5962-4290-92b4-773025ffd50b",
"Points": {
"POINT_TYPE": "F",
"POINT_NUM": 0,
"LATITUDE": 32.92197686725423,
"LONGITUDE": -117.04306932263532,
"parent_ID": "2b2a2abc-5962-4290-92b4-773025ffd50b"
},
"GEOFENCE_NAME": "New Fence",
"GEOFENCE_TYPE": "O",
"PRIVACY": "X",
"CENTER_LAT": 32.92043316309709,
"CENTER_LONG": -117.04286922250975,
"ZOOM_LAT": 0.006238797350533787,
"ZOOM_LONG": 0.005345531926053582,
"PATH_TOLERANCE": 5,
"ENTRANCE_TOLERANCE": 5
}
我的代码如下。我将很快用解码替换它。变量 id 确实包含 id。
var foundI :Bool = false
var iIndex = 0
for (index, char) in dataString.enumerated() {
if char == "I" {
foundI = true
iIndex = index
} else if char == "D" {
if foundI == true && index == iIndex+1 {
var suffixStr = String(dataString.dropFirst(iIndex+5))
var id = String(suffixStr.prefix(36))
print("New String \(id)")
}
}
}
【讨论】：