- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
我有多个 JSON 文件,需要循环并从中获取某些详细信息。但是,我希望有一个一刀切的循环,因为子节点在属性方面彼此匹配。谁能建议我如何循环我的 JSON 节点?
示例:
{
"name": "Example",
"description": "Example JSON",
"properties": {
"foo": "bar",
"foo1": "bar2",
"foo3": "bar4",
},
"stages": {
"This is a stage": {
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
},
"Another Stage": {
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
}
}
}
可以有数百个阶段。但是 JSON 的模式遵循这个一般模式,阶段可以有随机名称,但它们包含相同的定义。
有什么简单的建议吗?
最佳答案
NewtonSoft JObject将为您完成繁重的工作。它将 json 文档包装为一种具有 LINQable 接口(interface)的动态对象。 .Net 项目的大多数模板都会引入对 https://www.nuget.org/packages/Newtonsoft.Json 的 NuGet 依赖项。提供它的包。
void Main()
{
var json = @"{
""name"": ""Example"",
""description"": ""Example JSON"",
""properties"": {
""foo"": ""bar"",
""foo1"": ""bar2"",
""foo3"": ""bar4"",
},
""stages"": {
""This is a stage"": {
""stageInfo1"": ""blah"",
""stageInfo2"": ""blah"",
""integration"": {
""x"": ""x"",
""y"": ""y"",
""z"": ""z""
}
},
""Another Stage"": {
""stageInfo1"": ""blah"",
""stageInfo2"": ""blah"",
""integration"": {
""x"": ""x"",
""y"": ""y"",
""z"": ""z""
}
}
}
}";
var jo = JObject.Parse(json);
Console.WriteLine("A couple of ways to access just one level of ( Path,Value(s) ) pairs --------------------------------------------");
foreach (var node in jo) { Console.WriteLine("{0} {1}", node.Key, node.Value); }
Console.WriteLine("--- Or this --------------------------------------------");
foreach (var jtoken in jo.Children()) { Console.WriteLine("{0}={1} | has {2} children", jtoken.Path, string.Join(",\n", jtoken.Values()), jtoken.Children().Count()); };
Console.WriteLine("\n\n------------------- But to walk the full tree, use recursion----------------------------------------\n");
WriteRecursively(jo);
}
void WriteRecursively(JToken topJToken)
{
foreach (var jtoken in topJToken.Children())
{
Console.WriteLine("{0}={1} | has {2} children", jtoken.Path, string.Join(",\n", jtoken.Values()), jtoken.Children().Count());
WriteRecursively(jtoken);
};
}
输出:
A couple of ways to access just one level of ( Path,Value(s) ) pairs --------------------------------------------
name Example
description Example JSON
properties {
"foo": "bar",
"foo1": "bar2",
"foo3": "bar4"
}
stages {
"This is a stage": {
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
},
"Another Stage": {
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
}
}
--- Or this --------------------------------------------
name=Example | has 1 children
description=Example JSON | has 1 children
properties="foo": "bar",
"foo1": "bar2",
"foo3": "bar4" | has 1 children
stages="This is a stage": {
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
},
"Another Stage": {
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
} | has 1 children
------------------- But to walk the full tree, use recursion----------------------------------------
name=Example | has 1 children
name= | has 0 children
description=Example JSON | has 1 children
description= | has 0 children
properties="foo": "bar",
"foo1": "bar2",
"foo3": "bar4" | has 1 children
properties=bar,
bar2,
bar4 | has 3 children
properties.foo=bar | has 1 children
properties.foo= | has 0 children
properties.foo1=bar2 | has 1 children
properties.foo1= | has 0 children
properties.foo3=bar4 | has 1 children
properties.foo3= | has 0 children
stages="This is a stage": {
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
},
"Another Stage": {
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
} | has 1 children
stages={
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
},
{
"stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
}
} | has 2 children
stages['This is a stage']="stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
} | has 1 children
stages['This is a stage']=blah,
blah,
{
"x": "x",
"y": "y",
"z": "z"
} | has 3 children
stages['This is a stage'].stageInfo1=blah | has 1 children
stages['This is a stage'].stageInfo1= | has 0 children
stages['This is a stage'].stageInfo2=blah | has 1 children
stages['This is a stage'].stageInfo2= | has 0 children
stages['This is a stage'].integration="x": "x",
"y": "y",
"z": "z" | has 1 children
stages['This is a stage'].integration=x,
y,
z | has 3 children
stages['This is a stage'].integration.x=x | has 1 children
stages['This is a stage'].integration.x= | has 0 children
stages['This is a stage'].integration.y=y | has 1 children
stages['This is a stage'].integration.y= | has 0 children
stages['This is a stage'].integration.z=z | has 1 children
stages['This is a stage'].integration.z= | has 0 children
stages['Another Stage']="stageInfo1": "blah",
"stageInfo2": "blah",
"integration": {
"x": "x",
"y": "y",
"z": "z"
} | has 1 children
stages['Another Stage']=blah,
blah,
{
"x": "x",
"y": "y",
"z": "z"
} | has 3 children
stages['Another Stage'].stageInfo1=blah | has 1 children
stages['Another Stage'].stageInfo1= | has 0 children
stages['Another Stage'].stageInfo2=blah | has 1 children
stages['Another Stage'].stageInfo2= | has 0 children
stages['Another Stage'].integration="x": "x",
"y": "y",
"z": "z" | has 1 children
stages['Another Stage'].integration=x,
y,
z | has 3 children
stages['Another Stage'].integration.x=x | has 1 children
stages['Another Stage'].integration.x= | has 0 children
stages['Another Stage'].integration.y=y | has 1 children
stages['Another Stage'].integration.y= | has 0 children
stages['Another Stage'].integration.z=z | has 1 children
stages['Another Stage'].integration.z= | has 0 children
关于c# - 循环动态 JSON 以获取所有节点 C#,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52721960/
最近开始学习MongoDB。今天老师教了我们 mongoexport 命令。在练习时,我遇到了一个典型的问题,包括教练在内的其他同学都没有遇到过。我在我的 Windows 10 机器上使用 Mongo
我是 JSON Schema 的新手,读过什么是 JSON Schema 等等。但我不知道如何将 JSON Schema 链接到 JSON 以针对该 JSON Schema 进行验证。谁能解释一下?
在 xml 中,我可以在另一个 xml 文件中包含一个文件并使用它。如果您的软件从 xml 获取配置文件但没有任何方法来分离配置,如 apache/ngnix(nginx.conf - site-av
我有一个 JSON 对象,其中包含一个本身是 JSON 对象的字符串。我如何反序列化它? 我希望能够做类似的事情: #[derive(Deserialize)] struct B { c: S
考虑以下 JSON { "a": "{\"b\": 12, \"c\": \"test\"}" } 我想定义一个泛型读取 Reads[Outer[T]]对于这种序列化的 Json import
关闭。这个问题不满足Stack Overflow guidelines .它目前不接受答案。 想改善这个问题吗?更新问题,使其成为 on-topic对于堆栈溢出。 11 个月前关闭。 Improve
我的旧项目在 MySQL 中有 Standard JSON 格式的数据。 对于我在 JS (Node.js) 和 DynamoDB 中的全新项目,关于 Standard JSON格式: 是否建议将其转
JSON 值字符串、数字、true、false、null 是否是有效的 JSON? 即,是 true 一个有效的 JSON 文档?还是必须是数组/对象? 一些验证器接受这个(例如 http://jso
我有一个 JSON 字符串,其中一个字段是文本字段。这个文本字段可以包含用户在 UI 中输入的文本,如果他们输入的文本是 JSON 文本,也许是为了说明一些编码,我需要对他们的文本进行编码,以便它不会
我正在通过 IBM MQ 调用处理数据,当由 ColdFusion 10 (10,0,11,285437) 序列化时,0 将作为 +0.0 返回,它会导致无效的 JSON并且无法反序列化。 stPol
我正在从三个数组中生成一个散列,然后尝试构建一个 json。我通过 json object has array 成功了。 require 'json' A = [['A1', 'A2', 'A3'],
我从 API 接收 JSON,响应可以是 30 种类型之一。每种类型都有一组唯一的字段,但所有响应都有一个字段 type 说明它是哪种类型。 我的方法是使用serde .我为每种响应类型创建一个结构并
我正在下载一个 JSON 文件,我已将其检查为带有“https://jsonlint.com”的有效 JSON 到文档目录。然后我打开文件并再次检查,结果显示为无效的 JSON。这怎么可能????这是
我正在尝试根据从 API 接收到的数据动态创建一个 JSON 对象。 收到的示例数据:将数据解码到下面给出的 CiItems 结构中 { "class_name": "test", "
我想从字符串转换为对象。 来自 {"key1": "{\n \"key2\": \"value2\",\n \"key3\": {\n \"key4\": \"value4\"\n }\n
目前我正在使用以下代码将嵌套的 json 转换为扁平化的 json: import ( "fmt" "github.com/nytlabs/gojsonexplode" ) func
我有一个使用来自第三方 API 的数据的应用程序。我需要将 json 解码为一个结构,这需要该结构具有“传入”json 字段的 json 标签。传出的 json 字段具有不同的命名约定,因此我需要不同
我想使用 JSON 架构来验证某些值。我有两个对象,称它们为 trackedItems 和 trackedItemGroups。 trackedItemGroups 是组名称和 trackedItem
考虑以下案例类模式, case class Y (a: String, b: String) case class X (dummy: String, b: Y) 字段b是可选的,我的一些数据集没有字
我正在存储 cat ~/path/to/file/blah | 的输出jq tojson 在一个变量中,稍后在带有 JSON 内容的 curl POST 中使用。它运作良好,但它删除了所有换行符。我知
我是一名优秀的程序员,十分优秀!