How can I return all individual array items as separate lines of text?

Question

Hello,

I'm working in arcade and I am trying to write an expression that gives me the value of a field and convert it to text.

What I did was create an Array that stores all non-null values in to the array. From there, I want the array to return the individual text value for each item in the array. For example....

['orange', 'apple', 'banana']

>orange

>apple

>banana

What I am getting though is just a return of the first item in the array with the code below. Is my For loop wrong?

var i = 0;

var links = [];

function addValue(feat){

if(!IsEmpty(feat)) {

links[i++] = feat;}

};

addValue($feature["PDF_LINK_01"]);

addValue($feature["PDF_LINK_02"]);

addValue($feature["PDF_LINK_03"]);

addValue($feature["PDF_LINK_04"]);

addValue($feature["PDF_LINK_05"]);

addValue($feature["PDF_LINK_06"]);

addValue($feature["PDF_LINK_07"]);

addValue($feature["PDF_LINK_08"]);

addValue($feature["PDF_LINK_09"]);

addValue($feature["PDF_LINK_10"]);

addValue($feature["PDF_LINK_11"]);

addValue($feature["PDF_LINK_12"]);

addValue($feature["PDF_LINK_13"]);

addValue($feature["PDF_LINK_14"]);

addValue($feature["PDF_LINK_15"]);

addValue($feature["PDF_LINK_16"]);

addValue($feature["PDF_LINK_17"]);

addValue($feature["PDF_LINK_18"]);

addValue($feature["PDF_LINK_19"]);

addValue($feature["PDF_LINK_20"]);

for(var k in links){

return links[k]

};

The above only returns "PDF_LINK_01".

If I take out the For loop, it returns ['PDF_LINK_01', 'PDF_LINK_02', 'PDF_LINK_03', 'PDF_LINK_04']

Any help is appreciated

blake.terhune · Accepted Answer

I'm assuming this is Arcade? The return statement will exit the function immediately, which is why it only returns the first value. You would need to build a multiline string within your for loop and then return it after the loop has finished.

However, you can use the Concatenate() function to put everything together with a NewLine and you don't even need a for loop.

return Concatenate(links, TextFormatting.NewLine)