I have a form with inputs (each with ID param
, param1
, param3
respectively) and an external php file with a function called getform()
that takes three parameters from the form ( param
, param1
, param3
respectively).
A text feild with id results
to display the response from php file.
I have placed onclick
function on param3
.
I need whenever a user types something in param3
it should be sent to php file and the results displayed in the text filed of id results
.
here is my code
<script>
function post(){
var param = $('#param').val();
var param2 = $('#param2').val();
var param3 = $('#param3').val();
$.post('curencyconvert.php', {
postparam: param,
postparam2: param2,
postparam3:param3
}, function(data){
$('#results').html(data);
});
}
</script>
my php function in the php file
function Conv($param,$param2,$param3){
//my code here
return $output;
}
if(isset($_POST)){
//Calling the defined function here
Conv($_POST['postparam'], $_POST['postparam2'], $_POST['postparam3']);
}
Add these line below your function code.. and better echo the output in your function instead of returning it.
jQuery(document).ready(function($) { $('#param3').change(function() { // put here the code // } ); })
What errors you get in Console(firbug)?
As stated by other answers $_POST should be used.
What your php code returns if it returns an array or returns an object It can not be put into the Input. Return value must be a string
PHP code must be :
function Conv($param,$param2,$param3){
//my code here
// take particular field of the DB results or resultset that you want to show in the input.
return $output['name'];
}
I know answer is incomplete because your question is incomplete. Please let me know errors from the console so that i can help you further
echo instead of return.
if(isset($_POST)){
echo Conv($_POST['postparam'], $_POST['postparam2'], $_POST['postparam3']);
}
Do something like this, it is more clean:
Conv($param, $param2, $param3){
// your code here
return $output;
}
As for the javascript part, jquery ajax is your friend
function post(){
var param = $('#param').val();
var param2 = $('#param2').val();
var param3 = $('#param3').val();
$.ajax({
url: '/path/to/file',
type: 'POST',
data : { postparam: param, postparam2: param2, postparam3: param3 },
}).done(function(data) {
$('#results').html(data);
});
}
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