Passing Parameter Variable To Outside

Question

CategoryModel.find, gathers all records inside a db. Error if something goes wrong and categories is my output array.

categoryModel.find(function(error, categories) {
  if(error){
    console.log(error)
  }
});

I need to reach that categories outside of the function.

var result = categoryModel.find(function(error, categories) {
  if(error){
    console.log(error)
  }
  console.log(categories)
});

console.log(result.categories)

This type is not working. Im beginner at javascript :( Thanks

SOLVED

var result = {}
categoryModel.find(function(err, categories) {
  console.log(categories)
  result = categories
  callback(result)
});

console.log(result)

Hello `async` my old friend.... – briosheje Mar 03 '18 at 21:50 — briosheje, Mar 03 '18 at 21:50
I guess i have to use callback function? – Berkin Mar 03 '18 at 21:57 — Berkin, Mar 03 '18 at 21:57
what does `categoryModel` actually return? – oniondomes Mar 03 '18 at 22:03 — oniondomes, Mar 03 '18 at 22:03
A json array which contains output from db – Berkin Mar 03 '18 at 22:04 — Berkin, Mar 03 '18 at 22:04
I mean what do you get if you output `result`? – oniondomes Mar 03 '18 at 22:05 — oniondomes, Mar 03 '18 at 22:05
i ll make some changes in another function – Berkin Mar 03 '18 at 22:05 — Berkin, Mar 03 '18 at 22:05

score 0 · Answer 1 · answered Mar 03 '18 at 22:03

Like @briosheje said in the comment, you should use async / await.

For your case, it would look like this:

async function someFunction() {
  var result = await categoryModel.find();
  if (result.error) {
    console.log(error);
  }
  console.log(result.categories);
}

You can read more about async / await there.

Passing Parameter Variable To Outside

SOLVED

1 Answers1