1 回答
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TA贡献1827条经验 获得超4个赞
您的gqlSender功能不会return拒绝将被拒绝的承诺,因此不会在任何地方处理。你应该写
const gqlSender = (username, type, opt = {}) => {
return axios.post('', {
// ^^^^^^
query: gqlQuery(username, type, opt) // generates a needed graphQL query
}).then(res => {
if (res.data.errors) {
console.log('error in making axios request inside utils/index gqlSender function');
throw res.data.errors;
} else {
console.log('###',res.data);
return res.data;
}
});
};
或者
const gqlSender = async (username, type, opt = {}) => {
// ^^^^^
const res = await axios.post('', {
query: gqlQuery(username, type, opt) // generates a needed graphQL query
});
if (res.data.errors) {
console.log('error in making axios request inside utils/index gqlSender function');
throw res.data.errors;
} else {
console.log('###',res.data);
return res.data;
}
}
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