[SOLVED] dropdown send the selected value

[phpnewbie112]

Hello, how can I send only a selected value in a dropdown bearing in mind that I cannot use the $_POST['total']

 

I am using a payment gateway, I have to submit in hidden form for example:

 

<input type="hidden" name="txtAmount" value="">

 

where  the value should be extracted from the "selected" value of the dropdown. thanks a million

[redarrow]

you mean this .....

<?php

if($txtAmount){

//do somethink

}else{

//dont do nothink

}

?>

[phpnewbie112]

I mean the dropdown is called txtAmount, what do I have to add inside value="" to submit the selected value to my payment gateway

[redarrow]

try this you mean this mate..........

 

<?php

if($_POST['submit']){

$names=$_POST['names'];

if($names=="john"){

	echo $names;

}else{

	echo "sorry that was not the correct selected item";
}


}

?>

<form method="POST" action="<?php $_SERVER['PHP_SELF']?>">

<select name="names">

<option value="john">john</option>

<option value="terry">terry</option>

<option value="chris">chris</option>

</select>

<br><br>

<input type="submit" name="submit" value="send">

</form>

[ashishag67]

What method is paypal asking inorder to process the transaction. You can also use 2 forms instead. i.e. the first form with dropdown will pass the value to the second form wherein you can post the value from the first to the second page where in you can also declare a form variable with hidden field with respective drop down value..

 

//page1.php

<form method=post action=page2.php>
<select name=txtAmt>
<option value="1000">1000</option>
<option value="2000">2000</option>
<input type=submit value="Submit" name=submit>
</form>

//page2.php
<form method=post action="pagename.php">
<input type=hidden name=txtAmount value="<?php echo $_POST['txtAmt'];?>">
<input type=submit name=submit value="Confirm Order">
</form>

 

Hope this helps you also check http://www.mysqlandphp.net

[chronister]

The way I understand it is that you want a hidden input to be the value of whatever item is selected in a select menu. These are both in the same form and you cannot use $_POST['fieldName'] to fill it in. So therefore the hidden input needs to be valued with javascript.

 

Here is the best i can come up with. I am very very limited with Javascript, but I think this will do what you want.

 

<?php
echo '<pre>'; print_r($_POST); echo '</pre>';

?>
<script type="text/javascript" language="javascript">

function changeIt(selectedItem)
{
var theField = document.getElementById('testing2');

theField.value = selectedItem.value;
}

</script>
<form name="test" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">

<select name="testing" onChange="javascript:changeIt(this)" >
	<option value="1">1</option>
	<option value="2">2</option>
	<option value="3">3</option>
</select>

<input type="hidden" name="testing2" id="testing2" value="" />
<input type="submit" name="submit" value="Submit" />
</form>

 

Hopefully this is what your looking for.

[redarrow]

if it from a database the value then echo the database value

 

<?php

$sql="select * from what_ever";
$res=mysql_query($sql)or die(mysql_error());

while($date=mysql_fetch_assoc($res)){

echo"<input type='hidden' name='txtAmount' value=' ".$data['database_field']." '>"; 
}

?>

[phpnewbie112]

thank you all, I will try all the proposed options but I think a double form as ashishag67 mentioned is better. thanks again