How to Get value from PHP combobox?

I am in need of some help with my PHP and mySQL combobox. I have sucessfully created the dropdown box that is populated from a table called techs. Even that was a big achievement. The name of the column of the dropdown box is name.But now I need to post the value that is selected to "ajax.php", which then submits it to my db. I need to know how to modify the PHP code to get it to print the value to "ajax.php" when I hit the submit button. The other fields work just fine and post to the cutomers table.

I have read as much as i could online, but could not find how to accomplish this task. Some have mentioned to add a hidden field, but am unsure how to do this. I am very new to php and sql, so any help is appreciated.


It's just the PHP section that i really need some major help on.

<form id="submit" action="ajax.php" method="post">

<fieldset>

<legend>LAPTOP CHECKOUT</legend>


<label for="gid">Global ID:</label>

<input id="gid" class="text" name="gid" size="4" type="text" />

<br></br>

<label for="fname">First Name:</label>

<input id="fname" class="text" name="fname" size="20" type="text" /> 

<br></br>

<label for="lname">Last Name:</label>

<input id="lname" class="text" name="lname" size="20" type="text" />

<br></br>

<label for="ophone"> Office Phone:</label>

<input id="ophone" class="text" name="ophone" size="10" type="text" />

<br></br>

<label for="cphone"> Cell Phone:</label>

<input id="cphone" class="text" name="cphone" size="10" type="text" />

<br>

<label for="email"> E-mail:</label>

<input id="email" class="text" name="email" size="20" type="text" />

HTML:

<form>

//Write out our query

$query = "SELECT name FROM techs";


$result = mysql_query($query) or die(mysql_error());



$dropdown = "<select name='techs'>";


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

  $dropdown .= "\r\n<option value='{$row['name']}'>{$row['name']}</option>";

}


$dropdown .= "\r\n</select>";

echo $dropdown;

PHP:

<button class="button positive"> <img src="images/checkmark.png" alt="" /> Check Out Laptop </button>

</fieldset>

    </form>

HTML:

 

Actually submit your form to ajax.php... I'm not too confident that this method is safe at all.

 

ajax.php is just a normal process.php page. I was originally using ajax script to submit, but removed it to make it simple for testing.

this is the process page

$username="*removed*";

$password="*removed";

$mysql_hostname = "*removed*";

$database = "*removed*";



// command to start database connection



mysql_connect($mysql_hostname,$username,$password);



// Select the database

@mysql_select_db($database) or die( "Unable to select database");





	// CLIENT INFORMATION

	$gid          = htmlspecialchars(trim($_POST['gid']));

	$fname        = htmlspecialchars(trim($_POST['fname']));

	$lname        = htmlspecialchars(trim($_POST['lname']));

	$ophone       = htmlspecialchars(trim($_POST['ophone']));

	$cphone       = htmlspecialchars(trim($_POST['cphone']));

	$email        = htmlspecialchars(trim($_POST['email']));

	$cdate        = htmlspecialchars(trim($_POST['cdate']));

	$rdate        = htmlspecialchars(trim($_POST['rdate']));

	$ctech        = htmlspecialchars(trim($_POST['ctech']));



    $addClient  = "INSERT INTO customers (gid,fname,lname,ophone,cphone,email,cdate,rdate) VALUES ('$gid','$fname','$lname','$ophone','$cphone','$email','$cdate','$rdate')";

    mysql_query($addClient) or die(mysql_error());

PHP:

 

According to your own code the drop-down box name='techs' and the closest reference to that on your collect POST data is $_POST['ctech']. I also cant find where your INSERT INTO customers statement, references 'ctech'

So, I think you accidentally mixed names up and neglected to add 'techs' or 'ctech' to the sql statement ?

 

Do a print_r($_POST)

 

Ha ha yep you are right, I didn't have the ctech in the process.php page. I also didn't have it named right changed it to ctech. Thanks for the reply.



I still can't get it though. The current code just populates the dropdown, I can't find anything online how to go about gettiing this to the process.php page.

does this look better? I still can't get the value to submit into the mySQL table

//Write out our query

$query = "SELECT name FROM techs";





$result = mysql_query($query) or die(mysql_error());



$dropdown = "<select name='ctech'>";



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

  $dropdown .= "\r\n<option value='{$row['name']}'>{$row['name']}</option>";

}



$dropdown .= "\r\n</select>";

echo $dropdown;




//the following section needs to pass the selected variable on to process.php page


//How do i get the selected value into a variable?

PHP:

 

I GOT IT! I made your corrections and I currently have it working!! Thank you so much!

what do you mean not safe? is there a better way?

 

hi,
The php programming language basically are developed in the website
this language and this site are help to another person

thank
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