Question

您好我有以下python递归函数，它汇总了所有子节点的值，我想在NodeJS中移植，但我对异步调用有一些问题。

def getTree(parent_id, level=1):
    c.execute('select * from users where parent_id=?', (parent_id,))
    rows = c.fetchall()
    total = 0
    for child in children:
        total += getAsyncValue(child.id)
        total += getTree(child.id, level+1)
    return total

我试图这样做，但我可能需要用promises链接它，因为当我从异步函数中获取它时循环时总计数不可用

getTree = function(parent_id, level=1) {
  c.all("select * from users where parent_id="+parent_id, function(err, children) {
    var total = 0;
    children.forEach(function(child) {
      total += getAsyncValue(child.id)
      total += getTree(child.id, level+1)
    });
    return total;
  });
}

Answer 1

没有看到getAsyncValue我无法提供完整的答案 - 但是

var getAsyncValue = function(id) {
    return new Promise((resolve, reject) => {
       // resolve some value some how
    });
};
// helper to make the getTree function "nicer"
c.allAsync = function(str) {
    return new Promise((resolve, reject) => 
        this.all(str, (err, children) => {
            if (err) {
                return reject(err);
            }
            resolve(children);
        })
    );
};
var getTree = function(parent_id, level=1) {
    return c.allAsync("select * from users where parent_id="+parent_id).then(children => 
        Promise.all(children.map(child => 
            Promise.all([getAsyncValue(child.id), getTree(child.id, level+1)])
            .then(([a, b]) => a + b)
        )).then(results => results.reduce((a, b) => a + b))
    );
};

我认为使用async / await，代码可以写成：

var getAsyncValue = async function(id) {
    return new Promise((resolve, reject) => {
       // resolve some value some how
    });
};
// helper to make the getTree function "nicer"
c.allAsync = async function(str) {
    return new Promise((resolve, reject) => 
        this.all(str, (err, children) => {
            if (err) {
                return reject(err);
            }
            resolve(children);
        })
    );
};
var getTree = async function(parent_id, level=1) {
    let children = await c.allAsync("select * from users where parent_id="+parent_id);
    let total = 0;
    for (let i = 0; i < children.length; i++) {
        let child = children[i];
        total += await getAsyncValue(child.id);
        total += await getTree(child.id, level + 1);
    }
    return total;
};

NodeJS - sum async导致递归函数

1 个答案: