英文:
How can I pass parameter correctly?
问题
I want to call the arrow function in root.user, but I think I can't pass the parameter correctly, so I tried this: let user = root.user('101'), and on the console, I got this:
returning object undefined
[{"firstName":"Gokhan","lastName":"Coskun","login":"gcoskun","id":101}]
{"firstName":"George","lastName":"Clooney","login":"gclooney"}
[{"firstName":"Gokhan","lastName":"Coskun","login":"gcoskun","id":101}]
I wanted the user with the id 101 to get returned, but instead, I got all of the users returned.
英文:
const root = {
user: (id) => {
console.log("returning object " + JSON.stringify(id.id) + " " + JSON.stringify(storage.select("users", id.id)))
return storage.select("users", id.id)
}
}
I want to call the arrow function in root.user but I think I can't pass the parameter correctly, so I tried this --> let user = root.user('101')
and on the console I got this -->
returning object undefined
[{"firstName":"Gokhan","lastName":"Coskun","login":"gcoskun","id":101}]
{"firstName":"George","lastName":"Clooney","login":"gclooney"}
[{"firstName":"Gokhan","lastName":"Coskun","login":"gcoskun","id":101}]
I wanted the user with the id 101 get returned and got instead all of the users returned.
答案1
得分: 2
以下是翻译的内容：
为什么你在做 id.id，但传递的是一个 string？你可以要么传递一个带有 id 属性的对象（root.user({ id: '101' })），要么将 id.id 替换为简单的 id。
另外，看起来你的用户对象中的 id 字段是 number 类型，但你传递的是一个 string，所以根据 storage.select 内部的逻辑，你可能需要进行更改。
传递一个 number 类型的 id：
// 仅供示例使用的模拟对象：
const storage = {
select(key, id) {
return [
{ firstName: 'Gokhan', lastName: 'Coskun', login: 'gcoskun', id: 101 },
{ firstName: 'George', lastName: 'Clooney', login: 'gclooney' },
{ firstName: 'Gokhan', lastName: 'Coskun', login: 'gcoskun', id: 101 },
// 根据此处的逻辑，这些类型需要匹配。
// 使用 == 而不是 === 以便在这里不是必需的。
].filter(user => user.id == id)
},
};
const root = {
user: (id) => {
console.log(`ID = ${ id }`);
// 确保我们只返回一个用户或者如果没有用户则返回 null：
return storage.select('users', id)[0] || null;
},
};
const user = root.user('101');
console.log(user);
传递一个带有 number 类型的 id 属性的对象：
// 仅供示例使用的模拟对象：
const storage = {
select(key, id) {
return [
{ firstName: 'Gokhan', lastName: 'Coskun', login: 'gcoskun', id: 101 },
{ firstName: 'George', lastName: 'Clooney', login: 'gclooney' },
{ firstName: 'Gokhan', lastName: 'Coskun', login: 'gcoskun', id: 101 },
// 根据此处的逻辑，这些类型需要匹配。
// 使用 == 而不是 === 以便在这里不是必需的。
].filter(user => user.id == id);
},
};
const root = {
user: (query) => {
console.log(`ID = ${ query.id }`);
// 确保我们只返回一个用户或者如果没有用户则返回 null：
return storage.select('users', query.id)[0] || null;
},
};
const user = root.user({ id: '101' });
console.log(user);
英文:
Why are you doing id.id but passing a string? You either pass an object with an id prop (root.user({ id: '101' })) or replace id.id with simply id.
Also, it looks like the id fields in your user objects are of type number, while you are passing a string, so depending on the logic inside storage.select you might have to change that.
Passing a number id:
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
// Just mocking it for the example:
const storage = {
select(key, id) {
return [
{ firstName: 'Gokhan', lastName: 'Coskun', login: 'gcoskun', id: 101 },
{ firstName: 'George', lastName: 'Clooney', login: 'gclooney' },
{ firstName: 'Gokhan', lastName: 'Coskun', login: 'gcoskun', id: 101 },
// Depending on the logic here, these types need to match.
// Using == instead of === so that it's not required here.
].filter(user => user.id == id)
},
};
const root = {
user: (id) => {
console.log(`ID = ${ id }`);
// We make sure we only return a single user or null if there isn't one:
return storage.select('users', id)[0] || null;
},
};
const user = root.user('101');
console.log(user);
<!-- end snippet -->
Passing an object with an id prop of type number:
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
// Just mocking it for the example:
const storage = {
select(key, id) {
return [
{ firstName: 'Gokhan', lastName: 'Coskun', login: 'gcoskun', id: 101 },
{ firstName: 'George', lastName: 'Clooney', login: 'gclooney' },
{ firstName: 'Gokhan', lastName: 'Coskun', login: 'gcoskun', id: 101 },
// Depending on the logic here, these types need to match.
// Using == instead of === so that it's not required here.
].filter(user => user.id == id);
},
};
const root = {
user: (query) => {
console.log(`ID = ${ query.id }`);
// We make sure we only return a single user or null if there isn't one:
return storage.select('users', query.id)[0] || null;
},
};
const user = root.user({ id: '101' });
console.log(user);
<!-- end snippet -->
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