How to get loop array from php to AJAX

Question

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;

Please stop using mysql_* functions. These extensions have been removed in PHP 7. Learn about prepared statements for PDO and MySQLi and consider using PDO, it's really pretty easy. — Jay Blanchard
– Jay Blanchard, Commented Feb 9, 2016 at 13:17
You should use mysql_num_rows() instead of mysql_affected_rows(). What do you see when you look in your console? Do you see JSON? — Jay Blanchard
– Jay Blanchard, Commented Feb 9, 2016 at 13:19
I know. But i thing that if i filter the name value with mysql_real_escape string and stripslahs and close mysql connection it will ok. Is there other mistake about injection? — TogrulZade
– TogrulZade, Commented Feb 9, 2016 at 13:20

Torsten Barthel · Accepted Answer · 2016-02-09 13:48:13Z

5

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("");

edited Feb 9, 2016 at 13:48

answered Feb 9, 2016 at 13:25

Torsten Barthel

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5 Comments

TogrulZade Over a year ago

Ok i got all of result. But when i keyup i get new result under the other.

TogrulZade Over a year ago

Yes. Thanks. it was ok about getting all result. But now when i keyup then the result is written until count result. Example: i want to search users Like Togrul name. and i have 2 users like togrul. 1. Togrul. 2 Togrul2. When i write "T" it is written under to other Togrul and Togrul2. so i write "To" and it writes the results again under the result.

Torsten Barthel Over a year ago

Maybe just place the call to .html('') at the very beginning of your .keyup() - method.

TogrulZade Over a year ago

.html(""); It did not happen.

TogrulZade Over a year ago

Yes. yes. first i used in after $(".search-result").keyup(... so i used in success: like $(".search-result").html('');

Jeffwa · Accepted Answer · 2016-02-09 13:24:04Z

4

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++) {...}

answered Feb 9, 2016 at 13:24

Jeffwa

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1 Comment

TogrulZade Over a year ago

first of all i wrote i=0 but when i can't get result i changed that maybe the ptoblem is from this.

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