我如何使用变量X的值

Question

docRef.update({

            sequencenumber: {
                x: false
              }

         })

In this case consider the value

x=3

I want the field value to be

3:false

But in my database the field value is

x:false

Answer 1

In ES6 use like [variable] .

 var x=3; var a = { sequencenumber: { [x]: false } } console.log(a) 

Answer 2

You can use javascript's bracket notation for objects however you will have to initialise your data object sequencenumber before you can edit it. See bellow:

sequencenumber = {};
sequencenumber[x] = true;
docRef.update(sequencenumber);

Answer 3

Converting my comment to answer:

1. use ES2015+ Computed property names : [x]:false

2. Use bracket notation: obj.sequencenumber[x]=false;

 var x = 3 // ES2015+: obj = { sequencenumber: { [x]: false // } } console.log(obj) // previous versions: var obj = { sequencenumber: {} } obj.sequencenumber[x] = false; // bracket notation console.log(obj) // To read it: console.log(obj.sequencenumber[3]) // or console.log(obj.sequencenumber[x])