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Home/ Questions/Q 7777441
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T18:08:33+00:00 2026-06-01T18:08:33+00:00

In the head section i have: script src=https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js type=text/javascript></script> script type=text/javascript src=https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.18/jquery-ui.min.js ></script> $(document).ready(function()

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In the head section i have:

  script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js" type="text/javascript"></script>
  script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.18/jquery-ui.min.js" ></script>

$(document).ready(function() 
{
    $("#auto_input").autocomplete({
        source:"../autocomplete.php",
        minLength: 5 
});
});

and the input where i would like to show suggestions:

<form action=\"../add_name.php\" method=\"get\">
b>".$NAME_INPUT.": </b>
<input name=\"auto_input\" id=\"auto_input\" class=\"input\" type=\"text\" value=\"";
                    if(isset($_GET[auto_input]))
                    {
                        $page.=$_GET['auto_input'];
                    }$page.="/>";

</form>

The php file which returns results contains:

  $string = trim(strip_tags($_GET['term']));
  $query = "SELECT `id`,`name` FROM `table` WHERE `name` LIKE '%".$string."%'";
  $table=$database->getTable($database->execute($query));

 for ($i = 1; $i<=$table[0];$i++){
 $row['value'] = $table[$i]['name'];
 $row['id'] = $table[$i]['id'];
 $final[] = $row;
 }
 echo json_encode($final);

The suggestions are not displayed, after typing 5 characters. What am i doing wrong ?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T18:08:34+00:00Added an answer on June 1, 2026 at 6:08 pm

    Regarding your last comment i think that the path of source is incorrect.
    The line:

    source:"../autocomplete.php"
    
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