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我正在尝试将文件从我的前端应用程序上传到我的服务器，但我不断收到这个奇怪的错误，不知道为什么会这样。

错误是前端：

Variable "$file" got invalid value {}; Expected type Upload. Upload value invalid.

后端错误：

GraphQLError: Upload value invalid.
at GraphQLScalarType.parseValue (/app/backend/node_modules/graphql-upload/lib/GraphQLUpload.js:66:11)

这就是我添加上传标量的方式


// scalars/upload.ts
import { scalarType } from "nexus/dist";
import { GraphQLUpload } from "graphql-upload";

export const Upload = scalarType({
  name: "Upload",
  asNexusMethod: "upload", // We set this to be used as a method later as `t.upload()` if needed
  description: GraphQLUpload.description,
  serialize: GraphQLUpload.serialize,
  parseValue: GraphQLUpload.parseValue,
  parseLiteral: GraphQLUpload.parseLiteral,
});

export default Upload;

// scalar/index.ts
export { default as Upload } from "./Upload";

// index.ts
import * as allTypes from "./resolvers";
import * as scalars from "./scalars";
const schema = makePrismaSchema({
  types: [scalars, allTypes],
  prisma: {
    datamodelInfo,
    client: prisma,
  },
...

// my mutation 
t.string("uploadFile", {
  args: {
    file: arg({ type: "Upload", nullable: false }),
  },
  resolve: async (_: any, { file }, {  }: any) => {
    console.log(file);
    return "ok";
  },
});

这就是我在我的 React/NextJS 应用程序中制作 uplaod 客户端的方式

// the link
const link = () => {
  return createUploadLink({
    uri: process.env.backend,
    credentials: "include",
    fetch,
  });
};

// the mutation ( I use graphql codegen )

mutation TestUpload($file: Upload!) {
  uploadFile(file: $file)
}

const [upload] = useTestUploadMutation();

<input
  type="file"
  onChange={(e) => {
    const { files, validity } = e.target;
    if (validity.valid && files && files[0]) {
      upload({
        variables: {
          file: files[0],
        },
      });
    }
  }}
/>

即使尝试使用我的前端应用程序以外的其他东西来运行突变，我也会遇到同样的错误，所以我想问题出在我的后端设置上，即使我搜索了很多方法来做到这一点，但似乎没有一个工作。

4

1 回答 1

4

我只是遇到了完全相同的问题。这解决了它。所以基本上手动定义上传而不是使用 GraphQLUpload

import { objectType, scalarType } from "@nexus/schema";
import { GraphQLError } from "graphql";
import * as FileType from "file-type";

export const Upload = scalarType({
  name: "Upload",
  asNexusMethod: "upload", // We set this to be used as a method later as `t.upload()` if needed
  description: "desc",
  serialize: () => {
    throw new GraphQLError("Upload serialization unsupported.");
  },
  parseValue: async (value) => {
    const upload = await value;
    const stream = upload.createReadStream();
    const fileType = await FileType.fromStream(stream);

    if (fileType?.mime !== upload.mimetype)
      throw new GraphQLError("Mime type does not match file content.");

    return upload;
  },
  parseLiteral: (ast) => {
    throw new GraphQLError("Upload literal unsupported.", ast);
  },
});

export const File = objectType({
  name: "File",
  definition(t) {
    t.id("id");
    t.string("path");
    t.string("filename");
    t.string("mimetype");
    t.string("encoding");
  },
});
于 2020-05-25T21:10:33.290 回答