Why is this array.length return 1 for an empty array? [closed]

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Using:

<input id="selectedRecords" name="selectedRecords" type="hidden" />

Executing this code:

var keyValues = $('#selectedRecords').val().split(",").map(item => item.trim());

console.log('$(#selectedRecords).val(): [' + $('#selectedRecords').val() + ']');
console.log('keyValues: [' + keyValues + ']');
console.log('keyValues.length: [' + keyValues.length + ']');

Gives this console readout (copy pasted from console window in Brave browser):

$(#selectedRecords).val(): []
keyValues: []
keyValues.length: [1]

Why is keyValues.length returning a count of 1 ?

var keyValues = $('#selectedRecords').val().split(",").map(item => item.trim());

console.log('$(#selectedRecords).val(): [' + $('#selectedRecords').val() + ']');
console.log('keyValues: [' + keyValues + ']');
console.log('keyValues.length: [' + keyValues.length + ']');

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input id="selectedRecords" name="selectedRecords" type="hidden" />

keyValues contains an array with one element, the empty string ''. — user16540390
– user16540390, Commented Nov 23, 2023 at 17:03
Your debugging approach is flawed and based on false assumptions. Instead of concatenating keyValues into strings in a variety of ways, log keyValues itself to the console and observe what it actually is. — David
– David, Commented Nov 23, 2023 at 17:05
.split(',') can't return an empty array. The only way for String.prototype.split to return an empty array is ''.split(''): developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — user16540390
– user16540390, Commented Nov 23, 2023 at 17:09
It's closed because it's not useful for future visitors. It may be reproducible but so what? The question is "if I don't debug properly, why do I reach wrong conclusions" which is hardly something for the ages. Even if we don't go with the current close reason, this is a duplicate. The main dupe target would be Javascript console.log(object) vs. concatenating string then maybe a few others can be added. — VLAZ
– VLAZ, Commented Nov 24, 2023 at 10:38
@PaulZahra I was under the impression you did not understand the closure, so I explained the rationale. In the future, I shall endeavour to not do that, since you appear to consider the explanation "unkind". — VLAZ
– VLAZ, Commented Nov 27, 2023 at 11:04

moonwave99 · Accepted Answer · 2023-11-23 17:08:44Z

2

You are concatenating things while logging them:

console.log('$(#selectedRecords).val(): [' + $('#selectedRecords').val() + ']');
console.log('keyValues: [' + keyValues + ']');
console.log('keyValues.length: [' + keyValues.length + ']');

keyValues is [""], but if you concatenate it with a string, its string representation will be used - the comma separated list of its values, in this case just "".

Solution: log things right!

console.log('$(#selectedRecords).val()', $('#selectedRecords').val());
console.log('keyValues', keyValues);
console.log('keyValues.length', keyValues.length);

answered Nov 23, 2023 at 17:08

moonwave99

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