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我在GraphIQL中有这个查询(注意:Graph I QL)。它给了我关于 “连接器” 的错误消息:Apollo错误:“连接器必须是功能或类”?
query ($userID: String!) {
getUserData(id: $userID) {
name_first
name_last
picture_thumbnail
id
}
}
//query variables:
{
"userID": "DsmkoaYPeAumREsqC"
}
GraphIQL将返回这样的响应:
{
"data": {
"getUserData": null
},
"errors": [
{
"message": "Connector must be a function or an class",
"locations": [
{
"line": 2,
"column": 3
}
],
"path": [
"getUserData"
]
}
]
}
这里是我的解析:
getUserData(_, args) {
debugger; <== NEVER ACTIVATES
return Promise.resolve()
.then(() => {
console.log('In getUserData');
var Users = connectors.UserData.findAll({where: args}).then((Users) => Users.map((item) => item.dataValues));
return Users;
})
.then(Users => {
return Users;
})
.catch((err)=> {
console.log(err);
});
},
的debugger
断点在分解从未激活。
我还不知道GraphIQL是什么意思的“连接器”。什么需要改变?