Ajax Live Search - Get 2 Fields instead of 1

Question

I have a small problem, I want to a live search that returns me a POST_TITLE and a POST_ID. the title is for the people to see but my main reason is that I want the POST_ID to work with it.

Can someone help me, I posted the code below...

   <script>
//Gets the browser specific XmlHttpRequest Object
function getXmlHttpRequestObject() {
    if (window.XMLHttpRequest) {
        return new XMLHttpRequest();
    } else if(window.ActiveXObject) {
        return new ActiveXObject("Microsoft.XMLHTTP");
    } else {
        alert("Your Browser Sucks!\nIt's about time to upgrade don't you think?");
    }
}

//Our XmlHttpRequest object to get the auto suggest
var searchReq = getXmlHttpRequestObject();

//Called from keyup on the search textbox.
//Starts the AJAX request.
function searchSuggest() {
    if (searchReq.readyState == 4 || searchReq.readyState == 0) {
        var str = escape(document.getElementById('txtSearch').value);
        searchReq.open("GET", '/wp-content/themes/twentyten/livesearch.php?search=' + str, true);
        searchReq.onreadystatechange = handleSearchSuggest;
        searchReq.send(null);
    }        
}

//Called when the AJAX response is returned.
function handleSearchSuggest() {
    if (searchReq.readyState == 4) {


           var sx = document.getElementById('restaurantid')
        sx.innerHTML = '';

        var ss = document.getElementById('search_suggest')
        ss.innerHTML = '';
        var str = searchReq.responseText.split("\n");
        for(i=0; i < str.length - 1; i++) {
            //Build our element string.  This is cleaner using the DOM, but
            //IE doesn't support dynamically added attributes.
            var suggest = '<div onmouseover="javascript:suggestOver(this);" ';
            suggest += 'onmouseout="javascript:suggestOut(this);" ';
            suggest += 'onclick="javascript:setSearch(this.innerHTML);" ';
            suggest += 'class="suggest_link">' + str[i] + '</div>';
            ss.innerHTML += suggest;
            ss
        }
    }
}

//Mouse over function
function suggestOver(div_value) {
    div_value.className = 'suggest_link_over';
}
//Mouse out function
function suggestOut(div_value) {
    div_value.className = 'suggest_link';
}
//Click function
function setSearch(value) {
    document.getElementById('txtSearch').value = value;
    document.getElementById('restaurantid').value = value;
    document.getElementById('search_suggest').innerHTML = '';
}
</script>


<form id="frmSearch" action="">
    <input type="text" id="restaurantid" name="restaurantid" style="display: none;" />
                    <input type="text" id="txtSearch" name="txtSearch" alt="Search Criteria" onkeyup="searchSuggest();" autocomplete="off" />
                    <input type="submit" id="cmdSearch" name="cmdSearch" value="Search" alt="Run Search" />
                    <div id="search_suggest"></div>
                </form>
</code>

livesearch.php - THE AJAX PAGE

<code>
<?php

$con = mysql_connect('x', 'x', 'x);
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("xx", $con);
if (isset($_GET['search']) && $_GET['search'] != '') {
    //Add slashes to any quotes to avoid SQL problems.
    $search = addslashes($_GET['search']);
    //Get every page title for the site.
    $suggest_query = mysql_query('SELECT * FROM `mrr_posts` WHERE `post_title` LIKE \'%'.$search.'%\' AND `post_status` LIKE \'publish\' LIMIT 0, 30') or trigger_error("Query: $suggest_query\n<br />MySQL Error: " .mysql_error());            
    while ($suggest = mysql_fetch_array($suggest_query, MYSQL_ASSOC)) {    
    //while($suggest = db_fetch_array($suggest_query)) {
        //Return each page title seperated by a newline.
        echo $suggest['post_title'] . "\n";
    }
}





mysql_close($con);
?>

Answer 1

I noticed in the discussion above you're returning JSON now, and parsing it from the client side. And I noticed you tagged your question with jQuery, so I guess you're using that. This isn't an answer to your question, but here are some tips for javascript coding with jQuery that will help simplify your code a ton.

• instead of doing your ajax calls using the XMLHttpRequest object directly, just use $.get(url, successFunction)
• instead of using getElementById('some-id'), use $('#some-id'), then to do things like empty out the inner html, you can do $('#some-id').html(''). Using the jQuery element instead of HTMLElement, you can also manipulate the DOM in a cross-browser compatible way: http://api.jquery.com/category/manipulation/
• instead of building your javascript into your HTML (all those onmouseover and onmouseout handlers), use $('div.suggest_link') to select all divs on the page that have a class of "suggest_link". Then, attach a live event handler which will work on dynamically generated html, like this: $('div.suggest_link').live('mouseover', handleMouseOverForSuggestLink). You can read more about this on jQuery's page: http://api.jquery.com/live/

All these suggestions will work in modern browsers, and will help cut down a lot of code. Good luck!

Answer 2

You should return data from server in JSON (or XML, but JSON is easier), and then parse it in Javascript. Show titles to user, id keep for yourself.

Answer 3

In general xajax might simplify things quite much. Have a look at my answer here:

how to assign a javascript variable to a smarty variable