Array is.Array if and else statement

Question

So I wanted to write a if else statement where if Array.isArray is false, display a message. The function I have below allows user to enter an address, click Find ZipCode and does it thing. The function works, it just that if I input an invalid address, it doesn't show the message I wanted, which is invalid zipcode

In my HTML, I have:

<label>
Address:
      <input name="address" type="text" id="address"></label>
      <input type="button" value="Find ZipCode" id="find">
<p id="output"></p>

and Javascript:

var hello = function(data) {
    var result = ""; // empty string to hold the value
    var address = $("#address").val();

    if (! Array.isArray(data))
    {
        $("#output").html("invalid zipcode");
    }
    else {
        for (var i = 0; i < data.length; i++) 
        {
            var zipcode = data[i].zipcode;
            result += address + " zipcode: " + zipcode;
        }
    $("#output").html(result);

    }
};

$("#find").click(function(){
    var address = $("#address").val();
    $("#output").html("Loading...");            
    $.getJSON("zipcodelookup.php", "address="+address , hello);  
});

prodigitalson · Accepted Answer · 2013-03-10 23:50:02Z

3

Unless you modified the Array.prototype there is no Array.isArray method. I think what youre looking for is $.isArray.

answered Mar 10, 2013 at 23:50

prodigitalson

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2 Comments

Matt Over a year ago

ES5 adds an isArray method to Array. The noticeable exception to browsers which don't support this is IE < 10, however, it can be trivially polyfilled.

Travis J Over a year ago

+1 $.isArray is definitely what the OP had in mind (or should use) since jQuery is already imported.

Mark Pieszak - Trilon.io · Accepted Answer · 2013-03-10 23:55:11Z

0

You could also test like this to see if it's an Array:

function isArray(arr) {
    return Object.prototype.toString.apply(arr) === '[object Array]';
}

answered Mar 10, 2013 at 23:55

Mark Pieszak - Trilon.io

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Comments

RobG · Accepted Answer · 2013-03-11 02:43:25Z

0

Testing for types in Javascript to infer other conditions is often a bad idea. It seems that in this case, if the zip code is found, then an array (of something) is returned.

You are therefore concluding that if you don't get an array, the zip code wasn't found. That isn't particularly robust, it would be better for the server to return a defined message or error code that was more helpful, e.g. it could advise if the zip code wasn't found, the input was invalid or the database isn't available.

You can also do tests on the returned array to see if it has the features you want (i.e. use feature detection).

answered Mar 11, 2013 at 2:43

RobG

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