CODEWITHSUNDEEP
phpjquerymysqlajax
TogrulZade
TogrulZade

Reputation: 83

How to get loop array from php to AJAX

I want to get all of array value with ajax which that is coming from MySQL. I can't get all of result. I can get only 1 result.

My JQuery codes are:

$("input.input-search").keyup(function(){
    var name = $(this).val();
    if(name !='')
    {
        $.ajax({
            type: 'post',
            url: 'ajax.php?bol=search',
            data: {'name':name},
            dataType: 'json',
                success: function(val)
                {    
                     x = val.length;
                     for (i = 1; i<=x; i++){
                        $(".search-result").html(val[i].user+' * '+x);
                    }
                },
                error: function(name){
                $(".search-result").html("Nəticə yoxdur...");
                }
            });
    }
});

PHP Codes are:

case "search":
$name = trim($_POST['name']);
$q = mysql_query("SELECT * FROM `users` WHERE `user` LIKE '%".$name."%' ORDER by id;");
if(mysql_affected_rows() > 0){
    while($arr = mysql_fetch_array($q)){
        $array[] = $arr;
}
    echo json_encode($array);
}
break;

Upvotes: 1

Views: 99

Answers (2)

Torsten Barthel
Torsten Barthel

Reputation: 3476

It's simple. You are overwriting your elements HTML content every time your loop runs. With the jQuery

.html("...")

method you set a new content for your selected element every time your loop runs.

So use

.append('...')

instead.

Of course at the very beginning of your success - method empty your element with

.html("");

Upvotes: 5

Jeffwa
Jeffwa

Reputation: 1143

If your query is only returning 2 rows, the problem lies in your Javascript for loop. You're starting at 1 when the array index starts at 0. Try this:

for (i = 0; i <= x; i++) {...}

Upvotes: 4

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