Autosuggest box not working correctly using json and php?


I am using the following autosuggest code:

Have been able to successfully tap into my database using the JSON method, however, when I type something into the autosuggest field, the auto suggestions do not change with the different key presses.

For example: When I type in any letter to start the auto-suggest, I always get the first 4 results in my database.  No matter what I type, I always get the same results.  If my letter matches a letter in the results, it changes color as expected. If I make another key press, the auto suggest box does not pair down the results, it just continues to show my first 4 db entries, and colors matching letters.

Can anyone help?  This is my code so far:


<form method="get" action="" class="asholder">
	<small style="float:right">Hidden ID Field: <input type="text" id="testid" value="" style="font-size: 10px; width: 20px;" disabled="disabled" /></small>
	<label for="testinput">Person</label>
	<input style="width: 200px" type="text" id="testinput" value="" /> 
	<input type="submit" value="submit" />
<script type="text/javascript">
	var options = {
		callback: function (obj) { document.getElementById('testid').value =; }
	var as_json = new bsn.AutoSuggest('testinput', options);




header ("Expires: Mon, 26 Jul 1997 05:00:00 GMT"); // Date in the past
	header ("Last-Modified: " . gmdate("D, d M Y H:i:s") . " GMT"); // always modified
	header ("Cache-Control: no-cache, must-revalidate"); // HTTP/1.1
	header ("Pragma: no-cache"); // HTTP/1.0
	if (isset($_REQUEST['json']))
		header("Content-Type: application/json");
			$sql="SELECT * FROM members";
			$data = mysql_query($sql);
			echo "{\"results\": [";
			while($results = mysql_fetch_array($data)){
			echo "{\"id\": \"".$results['id']."\", \"value\": \"".$results['LastName'].", ".$results['FirstName']."\", \"info\": \"".$results['NUID']."\"},";
			echo'{ id: "", value: "", info: "" }';
		echo "]}";


the bsn.AutoSuggest_2.1.3.js file has not been modified.

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evibesmusicAuthor Commented:

I found out that if I limit the number of query results the function works properly but, I don't want to limit the number of query results, I want to be able to select from all entries in the db.
then try putting a large number in the options
Richard DavisSenior Web DeveloperCommented:
Looking at your SELECT statement, I noticed that you are never actually creating a WHERE clause by which to filter the select on. I do something similar on one of my custom admin systems I developed, only not with AJAX, but same principle.

I would modify your code so that the ajax call is sending the contents of the input box as each character is typed in and then in the dynamic_content.php, I would scrub the value passed in, then create a WHERE clause in the SELECT that uses a LIKE condition with %$value% as the like qualifier.

This will reduce the query's dataset to a much smaller number of returned records while also giving you back the data that specifically applies to the search criteria that you're wanting to have returned.

Hope this helped.

if (isset($_REQUEST['json']))

if (isset($_REQUEST['input']))

AND also:
 $sql="SELECT * FROM members";

 $sql="SELECT * FROM members WHERE `lastName` LIKE '" . mysql_real_escape_string($_REQUEST['input']) . "%' ";


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