Question

环境

本地节点驱动程序
MongoDB 4.0.2

期望的行为

根据对象的id从对象数组中删除对象，并在适用的情况下减小其他对象的position值。

具体地说，我正在尝试：

遍历MongoDB中的对象数组并更新其中的属性每个对象

也许它可能涉及使用$inc和-1，但是我仍然不确定如何进行迭代。

我尝试过的事情

我有一种适用于JavaScript场景的逻辑（在浏览器控制台中有效）：

function remove_and_update_position(id) {

    // faux array in collection document
    var statements = [{
            position: 0,  // < i want to update 'position' of all objects after an object is removed  
            id: "a"
        },
        {
            position: 1,
            id: "b"
        },
        {
            position: 2,
            id: "c"
        },
        {
            position: 3,
            id: "d"
        },
        {
            position: 4,
            id: "e"
        }

    ];

    // delete an object from the 'statements' array by its id
    statements.splice(statements.indexOf(statements.find(item => item.id == id)), 1);

    // iterate over the modified array
    for (var i = 0; i < statements.length; i++) {

        // for all objects in the modified array,  
        // with position numbers greater than the  
        // position removed ( minus 1),  
        // decrement their position numbers
        if (i > position - 1) {
            var new_position = statements[i].position - 1;
            statements[i].position = new_position;
        }

    }

    // verify adjustments were correct
    console.log(statements);

}

// call the function
remove_and_update_position("c");

我只需要将其转换为MongoDB/Node方案-到目前为止，我只拥有对象删除部分：

const remove_and_update_position = (req, res) => {

    var request_body = req.body;

    var topic_id = request_body.topic_id;

    var o_id = new ObjectID(topic_id);

    var collection = mongo_client.db("topics").collection("topics");

    var statement_id = request_body.statement_id; 

    var position = Number(request_body.position); 

    // query for the relevant document *and* target nested statement by id
    var filter = { _id: o_id, "statements.id": statement_id };

    // remove from statements array
    var update = { $pull: { "statements": { id: statement_id } } };

    // perform the update
    collection.updateOne(filter, update, function(err, doc) {
        if (err) {
            res.send(err);
        } else {
            res.json({ response_type: "success" });
        }
    });

}

Answer 1

已根据answer here更新。

const remove_and_update_position = (req, res) => {

    var request_body = req.body;

    var topic_id = request_body.topic_id;

    var o_id = new ObjectID(topic_id);

    var collection = mongo_client.db("topics").collection("topics");

    var statement_id = request_body.statement_id;

    var position = Number(request_body.position);

    // query for the relevant document *and* target nested statement by id
    var filter = { _id: o_id, "statements.id": statement_id };

    // remove from statements array
    var update = { $pull: { "statements": { id: statement_id } } };

    // perform the update
    collection.updateOne(filter, update, function(err, doc) {
        if (err) {
            res.send(err);
        } else {

            // BEGIN decrement others objects position
            var position_indicator = position - 1;

            console.log("deleted object position: " + position);
            console.log("will decrement objects with position greater than: " + position_indicator);

            var new_filter = { _id: o_id, "statements.position": { $gt: position_indicator } };

            var new_update = { $inc: { "statements.$[elem].position": -1 } };

            var array_filters = { "arrayFilters": [{ "elem.position": { $gt: position_indicator } }], "multi": true }; 

            collection.updateOne(new_filter, new_update, array_filters, function(err, doc) {
                if (err) {
                    res.send(err);
                } else {
                    res.json({ response_type: "success" });
                }
            });
            // END decrement others objects position
        }
    });

}

如何从数组中删除对象并更新所有其他对象的位置值？

1 个答案: