CODEWITHSUNDEEP
javascriptjquery
user838437
user838437

Reputation: 1501

jQuery ajax post success - saving data to variable

I have this function:

function getReport(name,x) {
    var xArr = [];
    var yArr = [];
    $.ajax({
        async: false,
        type: "POST",
        //async: false,
        //dataType: "json",
        url: "reportAjax.php",
        data: "name="+ name,
        success: function(data){
            var json = $.parseJSON(data);
            var chartDesc = json.INFO.DESC;
            $.each(json.RESULT, function(i, object) {
                $.each(object, function(property, value) {
                    //alert(property + "=" + value);
                    if (property == x) {
                        xArr.push(value);
                    }
                    else {
                        yArr.push(parseInt(value));
                    }

                });
            });                
        }
    }); 
    console.log(xArr);
    console.log(yArr);
    console.log(chartDesc);
    drawChart(xArr,yArr,chartDesc);
}

For some reason, I can see in the console.log the values of xArr and yArr, but I get chartDesc is not defined for chartDesc.

If I move the console.log(chartDesc) line to be under this line var chartDesc = json.INFO.DESC I can see it properly.

Why is that?

Upvotes: 1

Views: 1288

Answers (3)

Nemoden
Nemoden

Reputation: 9056

chartDesc variable scope is tied to the success function, that's why outside it, the variable is undefined. in Javascript functions create scopes for variables defined with var

Upvotes: 1

thecodeparadox
thecodeparadox

Reputation: 87073

make your

console.log(xArr);
console.log(yArr);

within success function. and declire chartDesc variable outside of success function,

console log outside of success function execute immediately when function called, but it takes some time to insert value of xArr and yArr due to success function execution. so you get nothing in you log.

Upvotes: 0

Rory McCrossan
Rory McCrossan

Reputation: 337560

You are declaring the chartDesc variable inside the AJAX callback function, so it is outside the scope of your later reference to it.

To fix this, remove the var from the start of the line, and instead declare it at the top of the function with the xArr and yArr variables:

var xArr = [];
var yArr = [];
var chartDesc = "";

Upvotes: 3

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