mongodb : use variable instead of property name in find

Question

I need to add condition to find depends to variable value : if it is null don't add to condition if not null - add. I am new in mongodb and for example I gen get it from SQL

DECLARE Variable int = null
SELECT * FROM Table
WHERE field = isnull(Variable ,field)

In mongodb I try to do find

let Variable = "aa";
db.Table.find({
  
         $or: [{field:Variable },{Variable : null }]
         
})

But its don't work as expected Please assist

Tushar Gupta - curioustushar · Accepted Answer · 2021-04-23 10:07:36Z

1

Use { [Variable] : null }

db.Table.find({
   $or: [{field:Variable },{ [Variable] : null }]       
})

let Variable = "aa";

const obj = { [Variable] : null };

console.log(obj);

answered Apr 23, 2021 at 10:07

Tushar Gupta - curioustushar

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5 Comments

Michael Over a year ago

Could you please write whole script ?

Tushar Gupta - curioustushar Over a year ago

@Michael docs.mongodb.com/manual/reference/operator/aggregation/ifNull

Michael Over a year ago

I need second condition in OR true when value of Variable is null

Tushar Gupta - curioustushar Over a year ago

docs.mongodb.com/manual/reference/operator/query/exists

Tushar Gupta - curioustushar Over a year ago

docs.mongodb.com/manual/reference/operator/query/or

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