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I am making a live search using ajax jquery therefore depending on the value typed into a textfield another php script runs that returns an array of rows in the database which is echoed and encoded with json using echo encode_json and this array is iterated in the javascript after the callback.

Why am I doing this?

  • I want to reuse the sql query where it searches for the matching rows in the database.
  • The livesearch will only display 5 matching results to the input data in the textbox and an extra "more option" tab. Theey will all be clickable eventually navigating to links within a web site but the more option tab will use the same sql query I used for the live search as it will direct into a page where it will display all the matching results.

This is my JavaScript could anyone assist me?

<script type="text/javascript">

function find(value){
    $( "#test" ).empty();
    var myExp = new RegExp(value , "i");
    $.ajax({
    url: 'searchDb.php',
    type: 'POST',
    data: 'args='+value+'&limit=2',
    success: function(data){
    var output = '<ul class="searchresults">';
    var count = 0;
    $.each(data, function(key, val) {
        if ((val.name.search(myExp) != -1)){
            output += '<li>';
            output += '<p>'+ val.name +'</p>';
            output += '</li>';
            count += 1;
            if(count == 5){
                output += '<li>';
                output += '<a href="searchResults.php">More Results</a>';
                output += '</li>';
                if(value.length == 0){             
                    output = '<ul class="searchresults">';
                }
                return false;
            }
        }
        if(value.length == 0){
            output = '<ul class="searchresults">';
        }
    });
    output += '</ul>';
    $('#test').html(output);

    }

});
}

Currently when I test this it does not work and the Google Chrome console shows an error

Uncaught TypeError: Cannot use 'in' operator to search for '22' in

["sss","sss"]

I have limit the data return to 2 just for test. It clearly it's returning it but I do not understanding cannot use 'in' operator to ........ Any help will be much appreciated! Thanks

PHP CODE:

<?php

$link = mysql_connect('localhost', 'root', '');
if (!$link) {
die('Not connected : ' . mysql_error());
 }


$db_selected = mysql_select_db('test', $link);
if (!$db_selected) {
die ('Can\'t use test : ' . mysql_error());
}
?>


<?php

function searchDb($abc, $limit = "all"){
$lists = array();
if (isset($abc)) {

$sql = "SELECT testa FROM test WHERE testa LIKE '%$abc%'";


if($limit !== "all"){
$sql .= "LIMIT ". $limit;
}

$result = mysql_query($sql) or die('Error, insert query failed');


while ( $row = mysql_fetch_assoc($result))
{

$var = $row["testa"];
array_push($lists, $var);
}
}


echo json_encode($lists);
}

$abc = $_POST['args'];
$limit = $_POST['limit'];
$lala = searchDb($abc, $limit);


?>
share|improve this question
    
post the part of php code where you build the responsed array of data –  Rami.Q Jan 16 '14 at 13:23
    
@Rami.Q thanks for the reply I just did it... –  user3104737 Jan 16 '14 at 13:30

1 Answer 1

up vote 0 down vote

you build your response Data Array this way:

while ( $row = mysql_fetch_assoc($result))
{

$var = $row["testa"];
array_push($lists, $var);
}

try to do it this way:

$RESPONSE = array();
$NAMES = array();

while ( $row = mysql_fetch_assoc($result))
    {

        $var = $row["testa"];
        array_push($NAMES,$var);
    }
$RESPONSE["NAMES"] = $NAMES;
print json_encode($RESPONSE);
exit();

in JS AJAX:

$.each(data.NAMES, function(key, val) {
        if ((val.search(myExp) != -1)){
.....

UPDATE: as requested in your comment!

the parameter of your function find(value) is undefined

so add the following at the top of your function 

function find(value){

if (typeof foo == 'undefined')
return;
share|improve this answer
    
Thank you very much!!!!!!!!! –  user3104737 Jan 16 '14 at 15:19
    
Hello there @Rami.Q I have one problem as it doesn't recgonise .length now....Uncaught TypeError: Cannot read property 'length' of undefined –  user3104737 Jan 16 '14 at 15:35
    
length of what exactly? the data array length? if yes, then do data.NAMES.length. –  Rami.Q Jan 16 '14 at 16:22
    
@user3104737, if my answer above solved your question, then mark my answer as accepted Answer please –  Rami.Q Jan 16 '14 at 16:23
    
if(value.length == 0)....the value.length i guess? –  user3104737 Jan 16 '14 at 16:34

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