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[英]how would I write linq query that would display user name instead of userID in a gridview
[英]GraphQL How would I write a query to find by first name
我正在嘗試了解 Graph QL 並有一個基本的示例工作。 例如,如果我傳入此查詢,我將取回與 id 的匹配。
query {
person(id:"4090D8F6-EFC4-42CD-B55C-E2203537380C")
{
firstname
surname
}
}
我的數據只是一組靜態的測試數據,我現在想做的是返回所有名字與我提供的名字相匹配的用戶。 我很困惑如何寫這個,因為 id null 檢查似乎阻止了我!?
我的 PersonQuery 看起來像這樣:
public class PersonQuery : ObjectGraphType<Person>
{
public PersonQuery(ShoppingData data)
{
Field<PersonType>(
"person",
description: "A Person",
arguments: new QueryArguments(
new QueryArgument<NonNullGraphType<IdGraphType>>
{
Name = "id",
Description = "The id of the person"
}),
resolve: ctx =>
{
return data.GetById(ctx.GetArgument<Guid>("id"));
});
}
}
我將如何做到這一點,以便我可以按名字返回人員列表,不知道這是否是下面的有效查詢,但想要一些關於如何執行此操作以及工作 id 示例的幫助。
query {
person
{
firstname: ("Andrew")
surname
}
}
答案更新 - 由 DavidG 提供
我按照上面提到的做了,所以我的 PersonQuery 現在看起來像這樣
public class PersonQuery : ObjectGraphType<Person>
{
public PersonQuery(ShoppingData data)
{
Field<PersonType>(
name: "person",
description: "A Person",
arguments: new QueryArguments(
new QueryArgument<IdGraphType>
{
Name = "id",
Description = "The id of the person"
}),
resolve: ctx =>
{
return data.GetById(ctx.GetArgument<Guid>("id"));
});
Field<ListGraphType<PersonType>>(
name : "persons",
description: "Persons",
arguments: new QueryArguments(
new QueryArgument<StringGraphType>
{
Name = "firstname",
Description = "The firstname of the person"
},
new QueryArgument<StringGraphType>
{
Name = "surname",
Description = "The surname of the person"
}),
resolve: ctx =>
{
var firstName = ctx.GetArgument<String>("firstname");
var surname = ctx.GetArgument<String>("surname");
return data.Filter(firstName, surname);
});
}
}
然后我可以按如下方式運行 graphql 查詢:
query {
persons(firstname: "Andrew", surname: "P")
{
firstname
surname
}
}
您將需要要么改變你在這里現場制作的id
參數可選,或創建一個新的領域(也許叫persons
或people
),並添加解析為您的數據存儲庫的新參數。 就我個人而言,我更喜歡做后者並創造一個新領域。 例如:
public PersonQuery(ShoppingData data)
{
Field<PersonType>( /* snip */ );
//Note this is now returning a list of persons
Field<ListGraphType<PersonType>>(
"people", //The new field name
description: "A list of people",
arguments: new QueryArguments(
new QueryArgument<NonNullGraphType<StringGraphType>>
{
Name = "firstName", //The parameter to filter on first name
Description = "The first name of the person"
}),
resolve: ctx =>
{
//You will need to write this new method
return data.GetByFirstName(ctx.GetArgument<string>("firstName"));
});
}
現在您只需要自己編寫GetByFirstName
方法。 查詢現在看起來像這樣:
query {
people(firstName:"Andrew")
{
firstname
surname
}
}
現在您可能會發現GetByFirstName
是不夠的,您還需要一個姓氏參數,並且它們是可選的,因此您可以執行以下操作:
Field<ListGraphType<PersonType>>(
"people",
description: "A list of people",
arguments: new QueryArguments(
new QueryArgument<StringGraphType>
{
Name = "firstName", //The parameter to filter on first name
Description = "The first name of the person"
},
new QueryArgument<StringGraphType>
{
Name = "surname",
Description = "The surname of the person"
}),
resolve: ctx =>
{
//You will need to write this new method
return data.SearchPeople(
ctx.GetArgument<string>("firstName"),
ctx.GetArgument<string>("surame"));
});
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