PHP and MySQL Live Search

Are you getting an error? The PHP file doesn't have anything to do with putting the results in the text box. The javascript in the top code box does that. Technically there isn't a text box in this example, but I think you're referring to the div that displays the results.

The ajax request is sent to the PHP file which in this example searches an XML file. In your case that PHP file would query a database. You've expressed your ability to handle that part.

When the ajax receives it's response, that response is inserted into the div's innerHTML using the javascript showResults function. Is that where you're having trouble?

I don't know how to take results from sql and return it to the textbox that would allow the user to select to populate text box.   There is a text box in the example above unless I really have issues :)  Maybe the results are going to the div but I don't know how to populate the div with a dataset that would be selectable like example.

<form>
<input type="text" size="30" onkeyup="showResult(this.value)">
<div id="livesearch"></div>
</form>

This is a three step process.

1. AJaX request

Javascript, which is in the top code window in the example, sends a request to a php file using AJaX. The request is based on the value of the text field and is requested each time a key is released in that field.

2. PHP processes request

The PHP file collects the results of that request and returns it. In this case that is done with a simple "return" of a string which contains a series of html tags and content. If you have ever generated a dynamic web page using PHP and MySQL, it's the same principle. Instead of echoing the dynamic html to the browser, you echo it to the AJaX request. No part of the PHP file is any different except that in this case you're only echoing a small portion of html, not the entire page.

3. Javascript processes AJaX response

The html which is echoed by the PHP file is then received as the AJaX response and inserted in the <div id="livesearch"><div> by javascript.

Well maybe I don't know what I'm doing.  Below is what I have and normally would echo the results to an html table or div or something where I have  $hint=$row['LightNumber']; and close the db connection after.  

$result = mysql_query($sql,$con);

while($row = mysql_fetch_array($result))
{
     
        $hint=$row['LightNumber'];
       
}

if ($hint=="")
  {
  $response="no suggestion";
  }
else
  {
  $response=$hint;
  }

//output the response
echo $response;
?>