从c＃中的字符串中提取列表中的文件名字符串＃

-4

我有以下大字符串，我希望所有的.jpg文件名都带有字符串中的扩展名列表，但我不知道如何获取它。从c＃中的字符串中提取列表中的文件名字符串＃

[{\"url\":\"https:\\/\\/\\/data\\/menus\\/799\\/799\\/dd2d49c78b1fe3a9cea5761d90132ff1.jpg\",\"href\":\"https:\\/\\/\\/data\\/menus\\/799\\/799\\/dd2d49c78b1fe3a9cea5761d90132ff1.jpg\",\"filename\":\"dd2d49c78b1fe3a9cea5761d90132ff1.jpg\",\"url_master\":\"menus_original\\/799\\/799\\/dd2d49c78b1fe3a9cea5761d90132ff1.jpg\",\"path_master\":\"\\/home\\/foodie\\/zomato_data\\/menus_original\\/799\\/799\\/dd2d49c78b1fe3a9cea5761d90132ff1.jpg\",\"data_center\":\"\",\"menu_type\":\"FOOD\",\"title\":\"FOOD\",\"menu_type_class\":\"FOOD\",\"real_menu_type\":\"FOOD\",\"is_salt_special_menu\":0,\"start_date\":\"\",\"consumer_upload\":0,\"start_date_formatted\":\"\",\"end_date\":\"\",\"end_date_formatted\":\"\",\"id\":131698200},{\"url\":\"https:\\/\\/\\/data\\/menus\\/799\\/799\\/f9f923c43b6b2d2a87ad8ce22b9995da.jpg\",\"href\":\"https:\\/\\/\\/data\\/menus\\/799\\/799\\/f9f923c43b6b2d2a87ad8ce22b9995da.jpg\",\"filename\":\"f9f923c43b6b2d2a87ad8ce22b9995da.jpg\",\"url_master\":\"menus_original\\/799\\/799\\/f9f923c43b6b2d2a87ad8ce22b9995da.jpg\",\"path_master\":\"\\/home\\/foodie\\/zomato_data\\/menus_original\\/799\\/799\\/f9f923c43b6b2d2a87ad8ce22b9995da.jpg\",\"data_center\":\"\",\"menu_type\":\"FOOD\",\"title\":\"FOOD\",\"menu_type_class\":\"FOOD\",\"real_menu_type\":\"FOOD\",\"is_salt_special_menu\":0,\"start_date\":\"\",\"consumer_upload\":0,\"start_date_formatted\":\"\",\"end_date\":\"\",\"end_date_formatted\":\"\",\"id\":131698203},{\"url\":\"https:\\/\\/\\/data\\/menus\\/799\\/799\\/ea3117de65882f14723480841940b5b1.jpg\",\"href\":\"https:\\/\\/\\/data\\/menus\\/799\\/799\\/ea3117de65882f14723480841940b5b1.jpg\",\"filename\":\"ea3117de65882f14723480841940b5b1.jpg\",\"url_master\":\"menus_original\\/799\\/799\\/ea3117de65882f14723480841940b5b1.jpg\",\"path_master\":\"\\/home\\/foodie\\/zomato_data\\/menus_original\\/799\\/799\\/ea3117de65882f14723480841940b5b1.jpg\",\"data_center\":\"\",\"menu_type\":\"FOOD\",\"title\":\"FOOD\",\"menu_type_class\":\"FOOD\",\"real_menu_type\":\"FOOD\",\"is_salt_special_menu\":0,\"start_date\":\"\",\"consumer_upload\":0,\"start_date_formatted\":\"\",\"end_date\":\"\",\"end_date_formatted\":\"\",\"id\":131698204}]

我想文章从刺：

dd2d49c78b1fe3a9cea5761d90132ff1.jpg 
dd2d49c78b1fe3a9cea5761d90132ff1.jpg 
dd2d49c78b1fe3a9cea5761d90132ff1.jpg 
f9f923c43b6b2d2a87ad8ce22b9995da.jpg 
f9f923c43b6b2d2a87ad8ce22b9995da.jpg

谢谢...提前

来源

2016-11-14 RAJNIK PATEL

对象创建一个类更容易和convienent方法是什么你有没有尝试过？ –

我曾尝试从此：http://stackoverflow.com/questions/13780654/extract-all-strings-between-two-strings –

是像json或什么的格式的字符串？ –

如果你只需要.JPG文件名，你可以试试这个简单的正则表达式：

\w+\.jpg

演示：https://regex101.com/r/KMWtZY/1

可以按如下方式使用C＃中使用它：

var regex = new Regex(@"\w+\.jpg"); 
return regex.Matches(strInput);

来源：1

来源

2016-11-14 05:16:57 Ibrahim

是的，我想这个，但我怎么可以在C＃中使用？ –

你应该逃脱“。”字符。你的正则表达式也匹配2ff1_jpg，asdasd，jpg –

@RAJNIKPATEL我更新了我的答案。 – Ibrahim

你的字符串好像它是一个JSON字符串。因而这将是desealize的JSON到一个对象，并获取像

价值为您的JSON字符串

public class YourData 
{ 
    public string url { get; set; } 
    public string href { get; set; } 
    public string filename { get; set; } 
    public string url_master { get; set; } 
    public string path_master { get; set; } 
    public string data_center { get; set; } 
    public string menu_type { get; set; } 
    public string title { get; set; } 
    public string menu_type_class { get; set; } 
    public string real_menu_type { get; set; } 
    public int is_salt_special_menu { get; set; } 
    public string start_date { get; set; } 
    public int consumer_upload { get; set; } 
    public string start_date_formatted { get; set; } 
    public string end_date { get; set; } 
    public string end_date_formatted { get; set; } 
    public int id { get; set; } 
}

比

//From wherever you are reading it. 
string jsonstr = "Your Json String"; 
//I Removed all the/and \ from the string 
jsonstr = jsonstr.Replace("/", ""); 
jsonstr = jsonstr.Replace("\\", ""); 
//at last of the string you have something like this 
//;\n  .menuTypes = [\"DEFAULT\",\"FOOD\",\"BAR\",\"DELIVERY\",\"SPECIAL\",\"TAKEAWAY\",\"INTERNAL\"];\n 
//which is not the part of the JSON string. So I removed that part as well to make it a valid JSON 
jsonstr = jsonstr.Remove(jsonstr.IndexOf(";n")); 
//Console.WriteLine(jsonstr); //You can uncomment it to see how JSON looks after cleaning. 
//Just Deserialize the JSON 
List<YourData> yd = JsonConvert.DeserializeObject<List<YourData>>(jsonstr); 
//Loop to get all the filenames or any other fields you want 
foreach(YourData ydd in yd) 
{ 
    Console.WriteLine(ydd.filename); 
}

来源

2016-11-14 06:32:34

嘿！愤怒的downvoter。请说明downvote的原因。 –

从c＃中的字符串中提取列表中的文件名字符串＃

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