In Ajax How to pass country name in html page to php page?
Question
How to pass country name in html page to php page using ajax.
demo.php
<html>
<head>
<title>Dynamic Form</title>
<script src="http://code.jquery.com/jquery-1.10.1.min.js" ></script>
<script>
$(document).ready(function(){
$("form").on('submit',function(event){
event.preventDefault();
data = $(this).serialize();
$.ajax({
type: "GET",
url: "post.php",
data: data
}).done(function( msg ) {
alert( "Data Saved: " + msg );
});
});
});
</script>
</head>
<body>
<form>
<table>
<tr>
<td>
<select name="one" onchange="if (this.value=='other'){this.form['other'].style.visibility='visible';this.form['submit'].style.visibility='visible'}else {this.form['other'].style.visibility='hidden';this.form['submit'].style.visibility='hidden'};"
<option value="" selected="selected">Select...</option>
<option value="India">India</option>
<option value="Pakistan">Pakistan</option>
<option value="Us">Us</option>
<option value="other">Other</option>
</select>
<input type="textbox" name="other" id="other" style="visibility:hidden;"/>
<input type="submit" name="submit" value="Add Country" style="visibility:hidden;"/>
</td>
</tr>
</table>
</form>
</body>
post.php
<?php
if(isset($_POST['submit']))
{
$Country = $_POST['other'];
echo $Country;
}
?>
When user select other from drop down list at that time display one textbox and one submit button.when user click on submit button at that time pass country name of demo.php to post.php using ajax.
2 answers
solution1
0 ACCPTED 2013-11-29 13:35:33
You have a few errors. $_POST['submit'] is never posted with the data = $(this).serialize(); . So you need to check something else. Also, you define GET as your method in AJAX but check $_POST in PHP. So change one of the two. Something like:
HTML / JavaScript:
<html>
<head>
<title>Dynamic Form</title>
<script src="http://code.jquery.com/jquery-1.10.1.min.js" ></script>
<script>
$(document).ready(function(){
$("form").on('submit',function(event){
event.preventDefault();
data = $(this).serialize();
$.ajax({
type: "POST",
url: "post.php",
data: data
}).done(function( msg ) {
alert( "Data Saved: " + msg );
});
});
});
</script>
</head>
<body>
<form>
<table>
<tr>
<td>
<select name="one" onchange="if (this.value=='other'){this.form['other'].style.visibility='visible';this.form['submit'].style.visibility='visible'}else {this.form['other'].style.visibility='hidden';this.form['submit'].style.visibility='hidden'};">
<option value="" selected="selected">Select...</option>
<option value="India">India</option>
<option value="Pakistan">Pakistan</option>
<option value="Us">Us</option>
<option value="other">Other</option>
</select>
<input type="textbox" name="other" id="other" style="visibility:hidden;"/>
<input type="submit" name="submit" value="Add Country" style="visibility:hidden;"/>
</td>
</tr>
</table>
</form>
</body>
PHP:
<?php
if(isset($_POST['other'])) {
$Country = $_POST['other'];
echo $Country;
}
?>
solution2
0 2013-11-29 13:37:30
you can try this
$(document).ready(function(){
$("form").on('submit',function(event){
var country = $("select[name=one]").val();
$.ajax({
type: 'POST',
dataType: 'json',
url: "post.php",
data: { country:country},
success: function(json) {
}
});
});
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