Ajax多重回应

I need to fetch the names using mysql query for which I'm trying to send some values via ajax to php.

Js file:

var dataselect = 'catg='+ $('#catg_list').val() + '&brand='+ $('#brand_list').val(); // get data in the form manual

$.ajax({
    type="POST",
    url:"check.php"
    data: dataselect,
    success: function(data) {
        alert (data);
    }
});

check.php file

<?php
    include(database connection);
    $catg_list= $_POST['catg'];
    $brand_list= $_POST['brand'];

    if ($catg_list!="") {
     $catg_query = mysql_query("SELECT name FROM categories WHERE id='$catg_list'");
     if ($catg_query) {
        while ($row_catg=mysql_fetch_assoc($catg_query)) {
            echo $row_catg['name'];
        }
     }
    }

    if ($brand_list!="") {
     $brand_list = mysql_query("SELECT name FROM brand WHERE id='$brand_list'");
     if ($brand_list) {
        while ($row_brand=mysql_fetch_assoc($brand_list)) {
            echo $row_brand['name'];
        }
     }
    }

?>

Problem is I need to display both the above names separately in . Is it possible? I am very much new to ajax. Any help would be great.

Thanks

You should be sending your data back in JSON format. You can send back a JSON object, which is like an associative array in PHP. You could use this php code:

<?php
    include(database connection);
    $catg_list= $_POST['catg'];
    $brand_list= $_POST['brand'];
    $results = array('categories' => array(), 'brands' => array());

    if ($catg_list!="") {
     $catg_query = mysql_query("SELECT name FROM categories WHERE id='$catg_list'");
     if ($catg_query) {
        while ($row_catg=mysql_fetch_assoc($catg_query)) {
            $results['categories'][] = $row_catg['name'];
        }
     }
    }

    if ($brand_list!="") {
     $brand_list = mysql_query("SELECT name FROM brand WHERE id='$brand_list'");
     if ($brand_list) {
        while ($row_brand=mysql_fetch_assoc($brand_list)) {
            $results['brands'][] = $row_brand['name'];
        }
     }
    }

    echo json_encode($results);
?>

Then, in your javascript success function, your data variable will be an object with two fields, each containing an array.

{
    categories: [],
    brands: []
}

You can access them by iterating over data.categories and data.brands.

Lastly, do not use SQL statements with data straight from your $_POST array. You have to sanitize that.

js

var dataselect = 'catg='+ $('#catg_list').val() + '&brand='+ $('#brand_list').val(); // get data in the form manual

var x, y;

$.ajax({
    type="POST",
    url:"check.php"
    data: dataselect,
    success: function(data) {
        var x = data.categories[0];
        var y = data.brand[0];
    }
});