Jquery $.get() passing a variable issue

Question

I can't pass different values of variable through $.get() function please check this code to learn more about my problem

var addressFieldValues = ['address1', 'address2', 'address3'];

for(i=0; i<addressFieldValues.length; i++) {
 var address = addressFieldValues[i];
 $.get('function.php', address, function(data){
  alert(address); // alerts address1 all time
 });
}

Why is it alerting "address1" these 3 times? since it should alert 3 different addresses at all.

To resolve this, you can either use let address in support browsers or a closure function. The reason: var address only exists once for the entire loop and, because $.get() behaves asynchronously, alert(address); isn't evaluated until after the for loop is done and has reassigned var address multiple times. — Jonathan Lonowski
– Jonathan Lonowski, Commented Apr 30, 2016 at 19:49
Using a .forEach() instead of a loop would solve the problem too. — Pointy
– Pointy, Commented Apr 30, 2016 at 19:54

dYale · Accepted Answer · 2016-04-30 21:44:21Z

2

One option would be to wrap your get request in a function, and pass in address as an argument. This way, you avoid the asynchronous issues.

function get(address) {
    $.get('', address, function(data) {
        alert(address);
    });
}

var addressFieldValues = ['address1', 'address2', 'address3'];
for (i = 0; i < addressFieldValues.length; i++) {
    var address = addressFieldValues[i];
    get(address);
}

edited Apr 30, 2016 at 21:44

answered Apr 30, 2016 at 19:51

dYale

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Comments

József Cserkó · Accepted Answer · 2016-04-30 19:51:39Z

0

Set request in object format, not string:

$.get('function.php', {'address': address}, function(data){
  alert(data);
});

Second problem: alert server data instead.

answered Apr 30, 2016 at 19:51

József Cserkó

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