PHP / AJAX pass a form input and select value to AJAX

Question

I have an HTML form and it contains basically of one input text box and one select...

Input box

echo '<input type="input" id="category_typex" name="category_typex" value="" />';

Select

echo '<select name="primary_cat" id="primary_cat" onChange="getSecondaryCat(this.value);">';

The Ajax code I'm using to pass the select is as follows:

function getSecondaryCat(val, cat_var) {
$.ajax({
    type: "POST",
    url: "category-get-secondary.php",
    data:'secondary_cat='+val,
    success: function(data){
        $("#secondary_cat").html(data);
        $("#tertiary_cat").html('<option value="">Select specific category</option>')
    }
});
}

I'm not being able to pass both the select and input values onto the next page....

I'm only able to pass the select value and not the input value

Use $("#primary_cat").val() in the function to get the value of the other input. — Barmar
– Barmar, Commented Apr 8, 2016 at 23:06
Use .ajax() method and pass the argument in json format via POST http method, like this question stackoverflow.com/a/15637150/4386714 — user4386714
– user4386714, Commented Apr 8, 2016 at 23:20
@Barmar sir what is cat_var in function getSecondaryCat(val, cat_var) {...} — KUMAR
– KUMAR, Commented Jul 11, 2020 at 16:33

Rafael Cuevas · Accepted Answer · 2016-04-08 23:13:36Z

1

Try this:

function getSecondaryCat(val, cat_var) {
    var category_typex = $('#category_typex').val();
    $.ajax({
        type: "POST",
        url: "category-get-secondary.php",
        data:{secondary_cat:val, category_typex:category_typex},
        success: function(data){
            $("#secondary_cat").html(data);
            $("#tertiary_cat").html('<option value="">Select specific category</option>');
        }
    });
}

answered Apr 8, 2016 at 23:13

Rafael Cuevas

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