Efficient way to generate all combinations for an address - Javascript

Question

Our team wants to generate a concatenated address based on a user's input for street, city, state, and zip. The issue is many times a user may not fill out all four of these inputs. Is there a way to efficiently generate all combinations for the concatenated address field without writing 12 if statements?

For example, if a user has all 4 fields filled out, we want to return this:

return current.agency_street + ', ' + current.agency_city + ', ' + current.agency_state + ', ' + current.agency_zip;

If they have everything filled out except zip code, we want to return this:

return current.agency_street + ', ' + current.agency_city + ', ' + current.agency_state;

Is there a way to do this without all the if statements? Thanks!

not sure what the down vote without a comment is for, but would still love to know if this is possible — Dave, May 24 '18 at 15:52
How do you know whether your user has input something into them? — Bergi, May 24 '18 at 15:58
The implicit not-quite-normal-javascript thing going on here is that `current` is a `GlideRecord` instance. So, dot-walks into fields (eg agency_street) are going to behave a little differently since they are returning GlideElement instances. They will never be null, but concatenating to a string is going to return an empty string if the internal field value is null (otherwise it'll return the toString'd value like you'd expect). — Joey, May 24 '18 at 16:05

score 5 · Accepted Answer · answered May 24 '18 at 15:53

5

You can push all to an array and filter out the falsy values.

var values = [];
values.push(agency_street);
values.push(agency_city);
values.push(agency_state);
values.push(agency_zip);
return values.filter(x => x).join(', ');

fiddle: https://jsfiddle.net/xfqumxox/

answered May 24 '18 at 15:53

dmgig

4,400
5
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1

Thanks!! the arrow syntax doesn't work for us, but I used return values.filter(Boolean).join(', '); and it works. thanks again for this! – Dave May 24 '18 at 16:18
Great! Very good - and I learned something as well about `filter(Boolean)`. So thank you. – dmgig May 24 '18 at 16:36

score 2 · Answer 2 · answered May 24 '18 at 16:13

As always a loop helps to simplify things:

let result = "";
for(const key of ["street", "city", "state", "zip"]) {
  const value = current["agency_" + key];
  if(value) result += (result && ",") + value;
}
return result;

Which is maybe a bit more elegant with functional style:

return ["street", "city", "state", "zip"].map(key => current["agency_" + key]).filter(Boolean).join(", ");

NocNit · Answer 3 · 2018-05-24T16:05:03.467

0

You can use this if only the filled in properties of the object are set and the order of the results doesn't strictly matter:

Object.keys(current).map(key => current[key]).filter(val => val).join(", ");

edited May 24 '18 at 16:05

answered May 24 '18 at 16:00

NocNit

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1
7

This does not guarantee the order of the result – Bergi May 24 '18 at 16:02
1

You're right, if the order strictly matters this answer won't do. I'll update it to reflect that – NocNit May 24 '18 at 16:06

score 0 · Answer 4 · answered May 24 '18 at 16:03

0

You can use ternary operator as well

return (current.agency_street ? current.agency_street : '') + (current.agency_city ? ',' + current.agency_city : '') + (current.agency_state ? ',' + current.agency_state : '') + (current.agency_zip ? ',' + current.agency_zip : '');

answered May 24 '18 at 16:03

Jay

338
1
5

This might emit a leading comma – Bergi May 24 '18 at 16:19

Efficient way to generate all combinations for an address - Javascript

4 Answers4