使用graphql工具来模拟一个GraphQL服务器似乎打破

Question

我已经按照the documentation about using graphql-tools to mock a GraphQL server，然而这将引发错误的自定义类型，如：

Expected a value of type "JSON" but received: [object Object] 

关于明确嘲讽graphql工具文档指出他们支持自定义类型，甚至提供了使用graphql-type-json项目中的GraphQLJSON自定义类型的示例。

我提供a demo of a solution on github它采用graphql的工具来成功地嘲笑一个GraphQL服务器，但是这依赖于猴子修补内置模式：

// Here we Monkey-patch the schema, as otherwise it will fall back 
// to the default serialize which simply returns null. 
schema._typeMap.JSON._scalarConfig.serialize =() => { 
    return { result: 'mocking JSON monkey-patched' } 
} 

schema._typeMap.MyCustomScalar._scalarConfig.serialize =() => { 
    return mocks.MyCustomScalar() 
} 

也许我做错了什么在我的演示，但没有上面的猴子补丁代码，我得到上述自定义类型的错误。

有没有人有比我的演示或任何线索更好的解决方案，我可能做错了什么，以及如何改变代码，使演示工作没有猴子修补模式？

在演示index.js相关的代码如下：

/* 
** As per: 
** http://dev.apollodata.com/tools/graphql-tools/mocking.html 
** Note that there are references on the web to graphql-tools.mockServer, 
** but these seem to be out of date. 
*/ 

const { graphql, GraphQLScalarType } = require('graphql'); 
const { makeExecutableSchema, addMockFunctionsToSchema } = require('graphql-tools'); 
const GraphQLJSON = require('graphql-type-json'); 

const myCustomScalarType = new GraphQLScalarType({ 
    name: 'MyCustomScalar', 
    description: 'Description of my custom scalar type', 
    serialize(value) { 
    let result; 
    // Implement your own behavior here by setting the 'result' variable 
    result = value || "I am the results of myCustomScalarType.serialize"; 
    return result; 
    }, 
    parseValue(value) { 
    let result; 
    // Implement your own behavior here by setting the 'result' variable 
    result = value || "I am the results of myCustomScalarType.parseValue"; 
    return result; 
    }, 
    parseLiteral(ast) { 
    switch (ast.kind) { 
     // Implement your own behavior here by returning what suits your needs 
     // depending on ast.kind 
    } 
    } 
}); 

const schemaString = ` 
    scalar MyCustomScalar 
    scalar JSON 

    type Foo { 
     aField: MyCustomScalar 
     bField: JSON 
     cField: String 
    } 

    type Query { 
     foo: Foo 
    } 
`; 
const resolverFunctions = { 
    Query: { 
     foo: { 
      aField:() => { 
       return 'I am the result of resolverFunctions.Query.foo.aField' 
      }, 
      bField:() => ({ result: 'of resolverFunctions.Query.foo.bField' }), 
      cField:() => { 
       return 'I am the result of resolverFunctions.Query.foo.cField' 
      } 
     }, 
    }, 
}; 

const mocks = { 
    Foo:() => ({ 
     // aField:() => mocks.MyCustomScalar(), 
     // bField:() => ({ result: 'of mocks.foo.bField' }), 
     cField:() => { 
      return 'I am the result of mocks.foo.cField' 
     } 
    }), 

    cField:() => { 
     return 'mocking cField' 
    }, 

    MyCustomScalar:() => { 
     return 'mocking MyCustomScalar' 
    }, 

    JSON:() => { 
     return { result: 'mocking JSON'} 
    } 
} 

const query = ` 
{ 
    foo { 
     aField 
     bField 
     cField 
    } 
} 
`; 

const schema = makeExecutableSchema({ 
    typeDefs: schemaString, 
    resolvers: resolverFunctions 
}) 

addMockFunctionsToSchema({ 
    schema, 
    mocks 
}); 

// Here we Monkey-patch the schema, as otherwise it will fall back 
// to the default serialize which simply returns null. 
schema._typeMap.JSON._scalarConfig.serialize =() => { 
    return { result: 'mocking JSON monkey-patched' } 
} 

schema._typeMap.MyCustomScalar._scalarConfig.serialize =() => { 
    return mocks.MyCustomScalar() 
} 

graphql(schema, query).then((result) => console.log('Got result', JSON.stringify(result, null, 4))); 

Answer 1

我和其他几个人看到使用实时数据源的一个类似的问题（在我的情况的MongoDB /猫鼬）。我怀疑这是graphql-tools makeExecutableSchema内部的东西，以及它使用自定义类型摄取基于文本的模式的方式。

下面是在这个问题上另一篇文章：How to use graphql-type-json package with GraphQl

我还没有尝试过打造代码架构的建议，所以不能确认是否有用，或者没有。

我当前的解决方法是将JSON字段（在连接器中）提供给客户端（并在客户端进行解析）时进行字符串化，反之亦然。有点笨拙，但我没有真正使用GraphQL来查询和/或选择性地提取JSON对象中的属性。这对于我怀疑的大型JSON对象来说不是最佳的。