Pass multiple values using ajax and echo in another page

Question

function getData(des, poid, divid){
            $.ajax({
                type: 'GET',
                url: 'loaddata.php?des=' + des + '&poid='+poid, //call storeemdata.php to store form data

                success: function(data) {
                    var ajaxDisplay = document.getElementById(divid);
                    ajaxDisplay.innerHTML = data;
                }
            });
        }

//Function call
<select name="descr" id="descr" onchange="getData(this.value,this.value,   'displaydata')">

This is the code and I passed multiple values in url. when I try to echo those values in next page it echo only one value (des) two times.Any help would be appreciated.

$slno = $_GET['des'];
$poid = $_GET['poid'];
echo $slno;
echo $poid;

I agree that it seems like the error is in the actual calling of getData can you see the url being called in the request? — Jesse Hockenbury
– Jesse Hockenbury, Commented Jan 24, 2017 at 6:00
<select name="descr" id="descr" onchange="getData(this.value,this.value, 'displaydata')"> This is the code @pAsh — vino
– vino, Commented Jan 24, 2017 at 6:00
your answer shows your passing the same variable (this.value) twice — Jesse Hockenbury
– Jesse Hockenbury, Commented Jan 24, 2017 at 6:01

Jesse Hockenbury · Accepted Answer · 2017-01-24 06:03:54Z

1

<select name="descr" id="descr" onchange="getData(this.value,this.value, 'displaydata')">

The issue is that in the display data function you are setting the des and poid url variables to the same value as shown in your code above.

answered Jan 24, 2017 at 6:03

Jesse Hockenbury

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3 Comments

vino Over a year ago

Thanks for feedback @jesse. Can you please show me how to?

Jesse Hockenbury Over a year ago

for the second parameter it would be a long line becase you would need to get the value of the other select with document.getElementbyid

vino Over a year ago

Thanks @jesse. I'll try that and let you know if it worked.

Arkowsky · Accepted Answer · 2017-01-24 06:24:01Z

1

You should pass ass second param inner text from option tag instead of the same value. To do this you can just use:

$(this).find('option:selected').text();

answered Jan 24, 2017 at 6:24

Arkowsky

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3 Comments

vino Over a year ago

Thanks @Arkowsky. I'm trying to pass another value from another dropdown as well. I don't this this will work.

Arkowsky Over a year ago

Can you show us more code? I mean how did you try implement this passing text from option in html

vino Over a year ago

First I choose poid from a drop down and by matching poid I fetched des in another drop down. On selecting des these values should be passed by ajax. poid <select name="poid" onchange="javascript: submit()" > It reloads page and fetxh data in des.. des <select name="descr" id="descr" onchange="getData(this.value,this.value, 'displaydata')">

vino · Accepted Answer · 2017-01-27 07:18:36Z

I found a way to do it. Thanks to everyone who looked into it to help me. Here is the code and it works like a charm.

function getData(des, divid){
    var poid = $("select[name='poid']").val();//this line pass second value "poid" in my case
            $.ajax({
                type: 'GET',
                url: 'loaddata.php?',
                data: {des : des, poid : poid},
                success: function(data) {
                    var ajaxDisplay = document.getElementById(divid);
                    ajaxDisplay.innerHTML = data;
                }
            });
        }

function call

<select name="des" id="des" onchange="getData(this.value, 'displaydata')">

ajax.php file

$slno = $_GET['des'];
$pid1 = $_GET['poid'];
echo $slno;
echo $pid1;

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