PHP save dropdown value to mysql

Question

First of all let me say, I am new to this and my code is probably laughable to those of you who know what you're doing...

I am trying to retrieve data from my mysql database and display the information in a matching dropdown list. I have this part working with some ugly basic code.

$url = $_SERVER["REQUEST_URI"];
$url_path = parse_url($url, PHP_URL_PATH);
$devid = pathinfo($url_path, PATHINFO_BASENAME);

$con = mysqli_connect("localhost","xxxxxx","xxxxxx","xxxxxxxx");
  //Run a query
  $result = mysqli_query($con,"SELECT type FROM `xxxxx`.`xxxxxx` WHERE device_id='$devid'");


$pulldown1 = '<select name="extension_type">';

while($row = mysqli_fetch_array($result))
{
if($row['type'] == "Unlimited")
{
$pulldown1 .= "<option selected value=\"Unlimited\">Unlimited Extension</option>
<option value=\"Metered\">Metered Extension</option>
<option value=\"Virtual\">Virtual Extension</option>
\n";
}
if($row['type'] == "Metered"){
$pulldown1 .= "<option selected value=\"Metered\">Metered Extension</option>
<option value=\"Unlimited\">Unlimited Extension</option>
<option value=\"Virtual\">Virtual Extension</option>
\n";
}
if($row['type'] == "Virtual"){
$pulldown1 .= "<option selected value=\"Virtual\">Virtual Extension</option>
<option value=\"Unlimited\">Unlimited Extension</option>
<option value=\"Metered\">Metered Extension</option>
\n";
}

}
$pulldown1 .= '</select>';

echo $pulldown1;

mysqli_close($con);

Now I'd like to be able to CHANGE the value and have it save the corresponding value and update that column in the database. This is where I am stuck. Can someone guide me in the right direction, please?

Specify what value you want to change, and also you might wanna look into protecting your SQL queries from SQL-injection. — Simon Carlson
– Simon Carlson, Commented Apr 22, 2013 at 20:11
this is how to save dropdownlist stackoverflow.com/questions/16125893/… . To update a list , you need to populate in a form. — Wayne Tun
– Wayne Tun, Commented Apr 22, 2013 at 20:11
Change the dropdown from it's current selected option to a different option and update the value in the db. — Epsilon
– Epsilon, Commented Apr 22, 2013 at 20:12

JasonMortonNZ · Accepted Answer · 2013-04-22 20:24:17Z

3

You could use jQuery to send an AJAX post request when the select box has it's value changed. In turn this could update your mysql database entry. If this is the kind of thing your wanting let me know and I'll expand on my answer further with a code example.

JS Code example

$("select").change(function () {
    // Send ajax request
    $.ajax
    ({
       type: "POST",
       url: "/change-value.php",
       dataType: 'json',
       data: { location: $(this).val() },
       cache: false,
       success: function(data)
       {
       }
    });

});

// PHP change-value.php example

<?php

// Get the changed value
$value = $_POST['value'];

// Code to update you mysql database entry would go here...

edited Apr 22, 2013 at 20:24

answered Apr 22, 2013 at 20:14

JasonMortonNZ

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4 Comments

Epsilon Over a year ago

That would be helpful, please elaborate.

JasonMortonNZ Over a year ago

I'll need more to go on than that?

JasonMortonNZ Over a year ago

If you want $_POST['extype'] to be accessible from within your change-value.php, you'll have to send it through in you ajax request. So where you have data { location: $(this).val() } you'll want to add the applicable code to grab the value of extype. E.g. data{location: $(this).val(), extype: $('#extype').val() }

JasonMortonNZ Over a year ago

Are you able to select if any of these answers provided to you were correct :)

Revent · Accepted Answer · 2013-04-22 21:33:42Z

0

Modified from the PHP documentation on php.net:

<?php
if ( !empty($_POST['extension_type']) )
{
   // Your MySQL connection code should be at the top of the script.
   $query = "UPDATE `xxxxx` SET `extension_type` = ? WHERE `[your_id_field_here]` = ?";
   /* Prepared statement, stage 1: prepare */
   if ( !($stmt = $mysqli->prepare($query) ) {
       echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
   }
   /* Prepared statement, stage 2: bind and execute */
   if ( !$stmt->bind_param("s", $_POST['extension_type']) || !$stmt->bind_param("i", intval($id)) ) {
       echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
   }
   if ( !$stmt->execute() ) {
       echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
   }
}
?>

As noted by @Simon Carlson, you also need to account for SQL injection. If you have a set list of extension types, you could improve on the code above by checking the $_POST value against known valid values.

edited Apr 22, 2013 at 21:33

answered Apr 22, 2013 at 20:25

Revent

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3 Comments

Epsilon Over a year ago

I tried this exact method and I get "Unexpected T_VARIABLE on "$query = "UPDATE xxxxx SET extension_type = ? WHERE ID = ?";" "

Revent Over a year ago

Right, you have to fill in your information for the MySQL connection, table, etc. You can't just copy and paste this and expect it to work. This is just an example to show you in general terms how to do it. I can be more specific if you give us more specifics.

Epsilon Over a year ago

I have already specified the connection, table, info etc, see above. So long as I do this before the "mysqli_close($con);" it should inherit what I have already defined when I execute mysqli_query($con,$query), right?

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