Populated Dropdown needs results on selection

Hi all, I have looked everywhere (obviously) to find the snipet of code required to show results from a dropdown list. I just need something simple that works. Here is my code so far:

<?php

    $sql="SELECT * FROM licensing ORDER BY id DESC"; 
    $result=mysql_query($sql); 

    $options=""; 

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

    $id=$row["id"]; 
    $url=$row["l_url"]; 
    $phone=$row["l_phone"]; 
    $state=$row["l_state"];
    $sullstate=$row["l_statefull"];  
    $options.="<option value=\"$id\">".$state;
    } 
?> 

    <select name="listing"><option value=0>Choose<?=$options?></option></select>
    <input type="submit" value="Search"/>

If someone could complete the echo results from the selected state, showing the variable results.

I appreciate any help.

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nuttyspice

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Not really sure what you're looking for here. Could you elaborate?
Every <option> tag needs a corresponding closing </option> tag, but you only have one.
Change line 15 to $options.="<option value=\"$id\">".$state."</option>;
Then change line 19 to

<select name="listing">
    <option value="0">Choose</option>
    <?php echo $options; ?>
</select>

EvolutionFallen

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OK, so... I have a table with state, url, etc.

The dropdown menu consists of the state abbreviations NM, CA.

I need the "search" button to then print the corresponding state information from the database onto the screen.

Results:
URL
Statefull
etc....

nuttyspice

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Do you need this to go to the server every time you chose state or can the data be held in a javascript variable (json format) on page load and the output updated via js on searchbutton click?

diafol

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Server

nuttyspice

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Where are your form tags? Not present in the code supplied.
Are you sending to the same page or are you sending to a form handler and redirecting to the form page?
Do you want to send by get or post?

diafol

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Same page as stated in a reply, forgot the form tags, post

nuttyspice

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You're going to want to use AJAX to call a php script from your page and then use the php script to query your database and to echo the results back to the page.

<html>
    <head>
        <script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js"></script>
        <script type="text/javascript">
            function searchOccupation () {
                $.ajax({
                    url: "searchOccupation.php?search=" + $('#searchTxt').attr('value'),
                    success: function (data) {
                        alert(data);
                    }
                });
            }
    </script>
    </head>
    <body>
        <input type="text" id="searchTxt">
        <input type="button" value="Search" id="searchBtn" onclick="searchOccupation()">
    </body>
    </html>

Then your php script (whose name should match that in the "url" field of the ajax call (in this case it should be named "searchOccupation.php") will look like this:

<?php
    $searchTxt = $_GET['search'];
    $con = new mysqli('server', 'user', 'password', 'database');
    if (!$con) {die("failed to connect: " . $con->connect_error;)}
    $sql = "SELECT * FROM tableName WHERE occupation = '" . $searchTxt . "'";
    $result = $con->query($sql);
    if (!$result) {die("No result set");}
    while($row = $result->fetch_assoc()) {
        echo $row['firstName'];  //This sends data back to the page 
    } 
?>

jacksparrow0109

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Sorry Jack, but as requested I just need php help. I have a dropdown menu, which searches my database for the state (this works just fine). I just need the data result to be shown on the page, after hitting search. Just something simple. I already have a connection include file to my database in my .php page.

nuttyspice

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So what's the problem? Jack's solution should work fine for you, just drop the JavaScript part, it's not necessary, just prevents a page reload.
The only thing that's missing is your <form> tag from what I see.

<form id="[your form id here]" action="[page filename here]" method="get">
    <!-- your SELECT element goes here -->
</form>
<?php
if(!empty($_GET['listing'])) {
    $searchTxt = mysql_real_escape_string($_GET['listing']);
    $con = new mysqli('server', 'user', 'password', 'database');
    if (!$con) {die("failed to connect: " . $con->connect_error;)}
    $sql = "SELECT * FROM tableName WHERE fieldname = '" . $searchTxt . "'"; // edit tablename and fieldname as needed
    $result = $con->query($sql);
    if (!$result) {die("No result set");}
    while($row = $result->fetch_assoc()) {
        echo $row['firstName'];  // change firstName as needed, use more echo statements for all data you need to output 
    }
}

EvolutionFallen

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Ok, never mind... I have a drop down list, I just needed a little extra code added on to show results on the same page, I don't have a search box, I already have an include file, was just looking for print/echo results from a drop down list selection.

Thanks for the somewhat off the subject help.

nuttyspice

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Your terminology is really confusing. It seems we've covered all that. My last script (which is just a modification of jacksparrow0109's) along with the <select> element I showed in my first post shows exactly how to use a "dropdown" (aka SELECT tag/element) which print results to the same page based on the user's selection.

EvolutionFallen

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Populated Dropdown needs results on selection

I need the "search" button to then print the corresponding state information from the database onto the screen.

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