Question

我有一个mongo架构，如下所示：

{
"_id" : ObjectId("58e4222497b2735ba3cd9ec4"),
"place" : "",
"plant" : "Test1",
"eventDate" : ISODate("2017-04-05T00:00:00Z"),
"event" : "Test123",
"toBeTested" : [
    {
        "_id" : ObjectId("58e453a07c9f94702ebac93d"),
        "thingsTested" : [
            "A1",
            "A2",
            "A3"
        ]
    }
]}

我正在使用mongoose删除thingsTested的单个元素。我在mongoose中的代码是：

Layout
    .update(
        {_id: req.params.parentid}, 
        {$pull: {toBeTested: {thingsTested: 'A3'}}},
        function (err, docs){
        if (err) {
            res.send(err);
        }
            res.json(docs);    
        }
    );

正如您所看到的，我已经硬编码我要从A3集中移除thingsTested。但是，展示的行为是所有thingsTested都被删除。

作为后续问题，如何确保mongoose命令仅使用A3 thingsTested（子ID）删除_id中的58e453a07c9f94702ebac93d？< / p>

谢谢！

Answer 1

您将不得不在查询中选择外部数组元素，然后可以使用#include <iostream> #include <fstream> #include <string> #include <algorithm> #include <cctype> #include <cstdlib> #include <ctime> using namespace std; struct information{ string first, middle, last, contribution; int birthyear; }; int main() { int i; information employee[1000]; ifstream fin; ofstream fout; string input; int comma; //stores the location of the comma int space, space2, space3; //store the location of the spaces. int dollarsign; //checks where the dollar sign is and then the numbers after. fin.open("oldretirement.txt"); //opening input.. if (fin.fail()){ fout<<"\nError! failed to open file!"; exit(1); } fout.open("newretirement.txt"); if (fout.fail()){ fout<<"\nError! failed to open file!"; exit(1); } while(!fin.eof()){ //loops till file failure getline(fin, input); comma = input.find(","); if (isalpha(input[comma-1])){ space = input.find(" "); //finds first space employee[i].last = input.substr(0, space-2); space2 = input.find(" ", space + 1); //finds the next space employee[i].first = input.substr(space+1, space2-space); //gets the substring of the area between spaces space3 = input.find(" ", space2+1); //you get the idea employee[i].middle = input.substr(space2+1, space3-space2); employee[i].birthyear = atoi((input.substr(space3+1, 4)).c_str()); employee[i].contribution = input.substr(space3+5); } else{ space = input.find(" "); employee[i].first = input.substr(0, space-1); space2 = input.find(" ", space+1); employee[i].middle = input.substr(space+1, space2-space); space3 = input.find(" ", space2+1); employee[i].last = input.substr(space2+1, space3-space2); employee[i].birthyear = atoi((input.substr(space3, 4)).c_str()); dollarsign = input.find("$"); employee[i].contribution = input.substr(dollarsign+2); } i++; } i--; //makes sure to make the number of indexes not too many. int money_made; //something to save how much money they made the company other than a string, so that it is easier to compare. string conversion; //something to store a string into to make things easier. for (int j= 0;j<i;j++){ //looping for printing stuff fout<<employee[j].last<<", "<<employee[j].first<<" "<<employee[j].last<<" "<<employee[j].birthyear<<employee[j].contribution; conversion = employee[j].contribution; conversion.erase((conversion.find(","), 1)); //erases the comma to make conversion to int easier. money_made = atoi(conversion.c_str()); //converts string to integer int age = (2017 - employee[j].birthyear); if( (money_made > 300000)&&(age >= 60)) fout<<" Ready\n"; // else fout<<endl; } fin.close(); fout.close(); }引用匹配的元素，如下所示：

Mongoose $ pull正在删除嵌套的整个数组，而不是数组中的单个值

1 个答案: