小编Dan*_*itz的帖子

MVC WebApi HttpGet具有复杂的对象

我有一个现有的WebApi动作,我想从HttpPost切换到HttpGet.它目前需要一个复杂的对象作为参数.

该模型:

public class BarRequest
{
    [JsonProperty("catid")]
    public int CategoryId { get; set; }
}
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控制器:

public class FooController : ApiController
{
    //[HttpPost]
    [HttpGet]
    [ActionName("bar")]
    public void Bar([FromUri] BarRequest request)
    {
        if (request != null)
        {
            // CategoryId should be 123, not 0
            Debug.WriteLine("Category ID :: {0}", request.CategoryId);
        }
    }
}
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现在当我发送以下请求时,一切都按预期工作.

GET /foo/bar?CategoryId=123
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旧的POST请求也按预期工作.

POST /foo/bar {"catid":123}
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但现在我需要以下请求才能工作:

GET /foo/bar?catid=123
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我怎么能做到这一点?

asp.net json.net deserialization asp.net-web-api

7
推荐指数
1
解决办法
1万
查看次数

如何获取janusgraph的组合脚本结果?

图表如下:

gremlin> a = graph.addVertex("name", "alice")
gremlin> b = graph.addVertex("name", "bobby")
gremlin> c = graph.addVertex("name", "cindy")
gremlin> d = graph.addVertex("name", "david")
gremlin> e = graph.addVertex("name", "eliza")
gremlin> a.addEdge("rates",b,"tag","ruby","value",9)
gremlin> b.addEdge("rates",c,"tag","ruby","value",8)
gremlin> c.addEdge("rates",d,"tag","ruby","value",7)
gremlin> d.addEdge("rates",e,"tag","ruby","value",6)
gremlin> e.addEdge("rates",a,"tag","java","value",10)
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我有3个脚本:

脚本#1

gremlin> g.V().has('name','alice').
           repeat(out()).
             until(has('name','alice')).
           cyclicPath().
           path().by('name')`
==>[alice,bobby,cindy,david,eliza,alice]
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脚本#2

gremlin> g.V().has('name','alice').
           repeat(outE().inV()).
             until(has('name','alice')).
           cyclicPath().
           group().
             by('name').
             by(path().unfold().has('value').values('value').fold()).
           next()
==>alice=[9, 8, 7, 6, 10]
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脚本#3

gremlin> g.V().has('name','alice').
           repeat(outE().inV()).
             until(has('name','alice')).
           cyclicPath().
           group().
             by('name').
             by(path().unfold().has('value').values('value').fold()).
           next().collect { k, v ->
             k + '=' + v.withIndex().collect { …
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gremlin tinkerpop3 janusgraph

1
推荐指数
1
解决办法
267
查看次数