Graphql 参数化查询qraphQL

我想从Apollo客户端向Apollo服务器发出参数化请求

在客户端：

const GET_VALUES = gql`
query Values($desc: String!) {      
    Values      
}    
`;

在服务器上（架构）：

结果:

[GraphQL error]: Message: Field "Values" argument "desc" of type "String!" is required, but it was not provided., Location: [object Object], Path: undefined
[Network error]: ServerError: Response not successful: Received status code 400

---

在useQuery的变量参数中，应该使用

desc

而不是

dirDesc

。试试这个：

const{loading，data，error}=useQuery（获取值{
变量：{desc:dirDesc}，
});客户端上的查询声明缺少实际变量。应该是这样的

const GET_VALUES = gql`
    query Values($desc: String!) {      
         Values(dirDesc: $desc)      
    }    
`;

然后，您可以使用提供的useQuery传递dirDesc.
dirDesc
而不是
desc
@xadm I更改了别名，但结果是samegraphqleror:Syntax Error:Expected Name，found}结果相同（您应该将查询更改为此
const GET_VALUES=gql`查询值（$desc:String！）{value（desc:$desc）}
@aOlegs
const GET_VALUES = gql`
    query Values($desc: String!) {      
      Values(desc: $desc)  
    }    
  `;

function ValueSelector({ pickValue, dirDesc }) {   
  const { loading, data, error } = useQuery(GET_VALUES, {
    variables: { desc:dirDesc},
   });

const GET_VALUES = gql`
    query Values($desc: String!) {      
         Values(dirDesc: $desc)      
    }    
`;

const GET_VALUES = gql`
    query Values($desc: String!) {      
      Values(desc: $desc)  
    }    
  `;

function ValueSelector({ pickValue, dirDesc }) {   
  const { loading, data, error } = useQuery(GET_VALUES, {
    variables: { desc:dirDesc},
   });