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I have a a mongo db table in the following format

    [{"url":"www.example1.com", "name":"hans","age":30},
    {"url":"www.example2.com", "name":"x","age":34},
    {"url":"www.example3.com", "name":"y","age":35},
    {"url":"www.example4.com", "name":"z","age":36},
    {"url":"www.example5.com", "name":"b","age":37}]

I have two tables where i need to check for under two if conditions. I did in the following way

    val = "www.example1.com"
    if list(table1.find({"url": {"$eq": val}})):
        print("exist in table 1")
        if list(table2.find({"url": {"$eq": val}})):
            print("exist in table 2")
        else:
            print("not exist in table 2")
    else:
        print("not exist in table 1")

This gives me correct response but it seems to take more time in doing the check. Is there a better way to do this query using pymongo

2

Answers


  1. With MongoDB v4.4+, you can use $unionWith. You may notice that there are two identical $match. This is for reducing intermediate table size to improve performance.

    db.table1.aggregate([
      {
        $match: {
          "url": {
            "$eq": "www.example3.com"
          }
        }
      },
      {
        "$unionWith": {
          "coll": "table2",
          "pipeline": [
            {
              $match: {
                "url": {
                  "$eq": "www.example3.com"
                }
              }
            }
          ]
        }
      }
    ])
    

    Here is the Mongo playground for your reference.

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  2. I’d do it with findOne:

        val = "www.example1.com"
        if table1.find_one({"url": {"$eq": val}}):
            print("exist in table 1")
            if table2.find_one({"url": {"$eq": val}}):
                print("exist in table 2")
            else:
                print("not exist in table 2")
        else:
            print("not exist in table 1")
    

    Which is much simpler than find() and fine_one() returns an element if found or None and not a cursor like find()

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