我有一个JSON:
{
"scbs_currentstatus": "",
"scbs_primaryissue": "",
"_umb_id": "Test",
"_umb_creator": "Admin",
"_umb_createdate": "0001-01-01 00:00:00",
"_umb_updatedate": "0001-01-01 00:00:00",
"_umb_doctype": "Test",
"_umb_login": "Test",
"_umb_email": "Test",
"_umb_password": {
"newPassword": "Test",
"oldPassword": null,
"reset": null,
"answer": null
},
"_umb_membergroup": {
" User": false,
"Line User": true,
"Callback User": false,
"Su User": false,
},
"umbracoMemberComments": "Test",
"umbracoMemberFailedPasswordAttempts": ""
}
Iam尝试删除所有属性,首先是" umb _" 。这可能在json.net中吗?
并输出如下:
{
"scbs_currentstatus": "",
"scbs_primaryissue": "",
"umbracoMemberComments": "Test",
"umbracoMemberFailedPasswordAttempts": ""
}
使用删除我能够一次完成但不是全部。
result.Property("_umb_id").Remove();
有什么建议吗?
答案 0 :(得分:9)
您可以先解析字符串:
var temp = JArray.Parse(json);
temp.Descendants()
.OfType<JProperty>()
.Where(attr => attr.Name.StartsWith("_umb_"))
.ToList() // you should call ToList because you're about to changing the result, which is not possible if it is IEnumerable
.ForEach(attr => attr.Remove()); // removing unwanted attributes
json = temp.ToString(); // backing result to json
更新或:
result.Properties()
.Where(attr => attr.Name.StartsWith("_umb_"))
.ToList()
.ForEach(attr => attr.Remove());
更新#2
您可以在where
子句中指定更多条件:
.Where(attr => attr.Name.StartsWith("_umb_") && some_other_condition)
OR
.Where(attr => attr.Name.StartsWith("_umb_") || some_other_condition)
或者你需要什么。
答案 1 :(得分:-1)
我最终反序列化为JObject并递归循环以删除不需要的字段。这是感兴趣的人的功能。
private void removeFields(JToken token, string[] fields)
{
JContainer container = token as JContainer;
if (container == null) return;
List<JToken> removeList = new List<JToken>();
foreach (JToken el in container.Children())
{
JProperty p = el as JProperty;
string propertyName = p.hasOwnProperty(key);
if (p != null && fields.Contains(p.propertyName) && p.propertyName.substring(0,4) == "_umb" )
{
removeList.Add(el);
}
removeFields(el, fields);
}
foreach (JToken el in removeList)
{
el.Remove();
}
}