Error: invalid variable initialization

Question

I have to call a function only if ajax call success.

Code:

<script>
    function removeFromCart(id) {
        var value = "";
        for(var i=0; i<id.length; i++)
            if(!isNaN(id[i]))
                value += id[i];
        $.ajax({
            type: "POST",
            url: "http://url.com/removefromcart.php",
            data: "id="+value,
            success: function(response){
                afterRemove(response);
            }
        });
    }

    function afterRemove(response) {
        if(response[0] == "r") {
            var prezzo = "";
            for(var i = 1; i<respose.length; i++)
                var prezzo += response[i];
            document.getElementByid("prezzo").innerHTML = prezzo+".00 &euro;";
            $("#segnale"+value).fadeOut();
        }
    }
</script>

The problem is that Firebug gives me this error on page load:

invalid variable initialization: var prezzo += response[i];

This because the afterRemove function is called on page load and not only if ajax success. How can fix this?

You wrote respose instead of response in your for initialization. — Blazemonger
– Blazemonger, Commented Nov 11, 2011 at 15:28
Just FYI, be consistent; don't mix jQuery ($("#segnale"+value)) lookups and non-jQuery lookups (document.getElementByid("prezzo")). $("#prezzo").html(prezzo+".00 €"); — gen_Eric
– gen_Eric, Commented Nov 11, 2011 at 15:35

Manuel van Rijn · Accepted Answer · 2011-11-11 15:28:24Z

6

you define/var the variable prezzo twice

var prezzo = ""; // <-- here
for(var i = 1; i<respose.length; i++)
    var prezzo += response[i]; // <-- and here

answered Nov 11, 2011 at 15:28

Manuel van Rijn

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Comments

James Montagne · Accepted Answer · 2011-11-11 15:29:54Z

2

var prezzo += response[i]

This is invalid because var prezzo means you are creating a new variable named prezzo. You are then trying to add to that variable. You can't add to a variable that is brand new because it doesn't have a value yet. Just drop the var.

answered Nov 11, 2011 at 15:29

James Montagne

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Comments

gen_Eric · Accepted Answer · 2011-11-11 15:30:12Z

1

var prezzo += response[i];

This line makes no sense. a += b means a = a+b.

In your case, you're doing var prezzo += response[i]; which is var prezzo = prezzo + response[i];. prezzo doesn't exist yet (because you are declaring it right there), so you can't add to its value, because it doesn't have a value (yet).

(Also, it's invalid syntax)

EDIT: I noticed you declared var prezzo = ""; earlier, so just remove the var:

prezzo += response[i];

It should look like:

var prezzo = "";
   for(var i = 1; i<response.length; i++)
      prezzo += response[i];

answered Nov 11, 2011 at 15:30

gen_Eric

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Comments

balaphp · Accepted Answer · 2011-11-11 15:30:21Z

1

if you initialize the variable when you do ++. you will always get error...

function afterRemove(response) {
    if(response[0] == "r") {
        var prezzo = "";
        for(var i = 1; i<respose.length; i++)
             prezzo += response[i];
        document.getElementByid("prezzo").innerHTML = prezzo+".00 &euro;";
        $("#segnale"+value).fadeOut();
    }
}

answered Nov 11, 2011 at 15:30

balaphp

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