Regex 在swift 2.2中使用正则表达式从字符串中提取精确链接
Regex 在swift 2.2中使用正则表达式从字符串中提取精确链接,regex,swift,string,extract,Regex,Swift,String,Extract,我只想按照以下代码提取网站的链接: import UIKit import Foundation func regMatchGroup(regex: String, text: String) -> [String] { do { let regex = try NSRegularExpression(pattern: regex, options: []) let nsString = text as NSString let r
我只想按照以下代码提取网站的链接:
import UIKit
import Foundation
func regMatchGroup(regex: String, text: String) -> [String] {
do {
let regex = try NSRegularExpression(pattern: regex, options: [])
let nsString = text as NSString
let results = regex.matchesInString(text,
options: [], range: NSMakeRange(0, nsString.length))
var internalString = [String]()
for result in results {
for var i = 0; i < result.numberOfRanges; ++i{
internalString.append(nsString.substringWithRange(result.rangeAtIndex(i)))
}
}
return internalString
} catch let error as NSError {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
// USAGE:
let textsearch = "mohamed amine ammach <img alt='http://fb.com' /> hhhhhhhhhhh <img alt='http://google.com' />"
let matches = regMatchGroup("alt='(.*?)'", text: textsearch)
if (matches.count > 0) // If we have matches....
{
for (var i=0;i < matches.count;i++) {
print(matches[i])
}
}
但我只是想:这里有人能帮我解决这个问题吗?我将不胜感激。您需要知道
NSTextCheckingResult.rangeAtIndex(:)
返回与索引为0的整个正则表达式模式相匹配的范围
就你而言:
rangeAtIndex(0)
->alt='(.*?)的范围http://fb.com“
范围索引(1)
->(.*)
->http://fb.com
所以,在生成匹配字符串时,需要跳过索引值0
尝试将最内部的for语句更改为:
for i in 1 ..< result.numberOfRanges { //start index from 1
internalString.append(nsString.substringWithRange(result.rangeAtIndex(i)))
}
用于1中的i..
(您还需要知道一件事,C-style for语句已被弃用。)您的结果是第一个捕获组
for i in 1 ..< result.numberOfRanges { //start index from 1
internalString.append(nsString.substringWithRange(result.rangeAtIndex(i)))
}