编写嵌套的 GraphQL 突变

Posted 2023-03-08

【中文标题】编写嵌套的 GraphQL 突变

【英文标题】：Writing Nested GraphQL Mutations

【发布时间】：2019-04-10 01:07:13

【问题描述】：

我正在寻找编写嵌套突变的示例。我正在为配方对象进行突变，架构如下所示：

const RecipeType = new GraphQLObjectType(
  name: "Recipe",
  fields: () => (
    id:  type: GraphQLID ,
    name:  type: GraphQLString ,
    dateCreated:  type: GraphQLString ,
    authorID:  type: GraphQLID ,
    prepTime:  type: PrepTimeType ,
    cookTime:  type: CookTimeType ,
    ingredients:  type: new GraphQLList(IngredientType) ,
    steps:  type: new GraphQLList(StepType) 
  )
);

const PrepTimeType = new GraphQLObjectType(
  name: "PrepTime",
  fields: () => (
    quantity:  type: GraphQLFloat ,
    unit:  type: GraphQLString 
  )
);

const CookTimeType = new GraphQLObjectType(
  name: "CookTime",
  fields: () => (
    quantity:  type: GraphQLFloat ,
    unit:  type: GraphQLString 
  )
);

const IngredientType = new GraphQLObjectType(
  name: "Ingredients",
  fields: () => (
    name:  type: GraphQLString ,
    quantity:  type: GraphQLFloat ,
    unit:  type: GraphQLString 
  )
);

const StepType = new GraphQLObjectType(
  name: "Ingredients",
  fields: () => (
    details:  type: GraphQLString ,
    estimatedTime:  type: GraphQLFloat ,
    unit:  type: GraphQLString 
  )
);

我希望编写一个突变来为这个项目创建一个完整的对象。突变如下所示：

createRecipe: 
  type: RecipeType,
  args: 
    // Required Args
    name:  type: new GraphQLNonNull(GraphQLString) ,
    authorID:  type: new GraphQLNonNull(GraphQLID) ,
    ingredients:  type: new GraphQLList(IngredientType) ,
    steps:  type: new GraphQLList(StepType) ,
    // Not required args
    prepTime:  type: PrepTimeType ,
    cookTime:  type: CookTimeType ,
  ,
  resolve(parent, args) 
    let recipe = new Recipe(
      name: args.name,
      dateCreated: new Date().getTime(),
      authorID: args.authorID,
      ingredients: args.ingredients,
      steps: args.steps
    );

    // Check for optional args and set to recipe if they exist
    args.prepTime ? recipe.prepTime = args.prepTime : recipe.prepTime = null;
    args.cookTime ? recipe.cookTime = args.cookTime : recipe.cookTime = null;

    return recipe.save();
  

我不确定如何创建一个可以创建整个对象的突变。然后更新将是一个进一步的挑战。有没有人有任何示例或链接到支持这一点的文档？据我所知，GraphQL 并没有以有用的方式涵盖这一点。

我目前收到以下错误：

  "errors": [
    
      "message": "The type of Mutation.createRecipe(ingredients:) must be Input Type but got: [Ingredients]."
    ,
    
      "message": "The type of Mutation.createRecipe(steps:) must be Input Type but got: [Steps]."
    ,
    
      "message": "The type of Mutation.createRecipe(prepTime:) must be Input Type but got: PrepTime."
    ,
    
      "message": "The type of Mutation.createRecipe(cookTime:) must be Input Type but got: CookTime."
    
  ]

我们将不胜感激。

干杯，

【问题讨论】：

我想通了。我必须为所有子文档创建新的输入类型。从那里我将它们添加到突变中并且效果很好：

【参考方案1】：

我想通了。我需要为每个子文档创建输入类型。我已经有了对象类型，但是对于突变，我必须添加新的。从那里我将它放入突变中。

createRecipe: 
  type: RecipeType,
  args: 
    // Required Args
    name:  type: new GraphQLNonNull(GraphQLString) ,
    authorID:  type: new GraphQLNonNull(GraphQLID) ,
    ingredients:  type: new GraphQLList(IngredientInputType) ,
    steps:  type: new GraphQLList(StepInputType) ,
    // Not required args
    prepTime:  type: PrepTimeInputType ,
    cookTime:  type: CookTimeInputType ,
  ,
  resolve(parent, args) 
    let recipe = new Recipe(
      name: args.name,
      dateCreated: new Date().getTime(),
      authorID: args.authorID,
      ingredients: args.ingredients,
      steps: args.steps
    );

    // Check for optional args and set to recipe if they exist
    args.prepTime ? recipe.prepTime = args.prepTime : recipe.prepTime = null ;
    args.cookTime ? recipe.cookTime = args.cookTime : recipe.cookTime = null ;

    return recipe.save();
  
,