提问人：nb_nb_nb 提问时间：5/10/2023 更新时间：5/10/2023 访问量：51
两个函数相互调用，最终完成后发送消息
Two functions calling each other, send message when finally done
问：
我有 2 个相互调用的函数。满足所有条件时如何发送消息？
我的代码：
const func1 = async (arg) => {
let table_id = result.data.internal_id;
const response = await axios_Func({ graphQL query });
console.log(`${response.data.name} - ${table_id}`);
return func2({ query: graphQL_query, table_id });
}
func2 = async (result, table_id) => {
let dev = await getAllTables({ query: tables(123) });
let prod = await getAllTables({ query: tables(321) });
let prod_id = new Set();
result.data.data.table.table_fields.forEach((d) => {
if (d. connector != null && !(d.name in dev))
prod_id.add(prod[d.name]["id"]);
});
prod_id = [...prod_id];
prod_id.map((x) => createOne({ query: graphQL_query(x) });
});
const callback = () => console.log("all done");
现在我的函数返回：
xyz - M5WBmnWs
abc - QrqJ91Rq
qwe - VP0B0KEt
我该如何打电话？并使我的输出如下所示：callback
xyz - M5WBmnWs
abc - QrqJ91Rq
qwe - VP0B0KEt
all done
答：
0赞
jabaa
5/10/2023
#1
等待函数调用，最后调用：callback
await func1();
callback();
1赞
Roberto Yoc
5/10/2023
#2
我不明白你想做什么，但这应该有效：
const func1 = async (arg) => {
let table_id = result.data.internal_id;
const response = await axios_Func({ graphQL query });
console.log(`${response.data.name} - ${table_id}`);
await func2({ query: graphQL_query, table_id });
}
func2 = async (result, table_id) => {
let dev = await getAllTables({ query: tables(123) });
let prod = await getAllTables({ query: tables(321) });
let prod_id = new Set();
result.data.data.table.table_fields.forEach((d) => {
if (d. connector != null && !(d.name in dev))
prod_id.add(prod[d.name]["id"]);
});
prod_id = [...prod_id];
prod_id.map((x) => createOne({ query: graphQL_query(x) });
// if createOne is a promise you need:
// return Promise.all(prod_id.map((x) => createOne({ query: graphQL_query(x) }));
});
const callback = () => console.log("all done");
const execute = async () => {
await func1()
callback()
}
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