php ajax return false issue
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
I am using the following ajax function
function validatec()
{
var coupon = jQuery('#validateC').val();
var data = "coupon="+coupon;
$.ajax({
type : 'POST', //Method type
url : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
data : data,
async : false,
success: function(msg) {
if(msg > 0)
{
}
else
{
alert("Please enter a valid Coupon Code");
}
return false;
}
});
}
What I want to do is that when the result is received from ajax call then I dont want to load the page. but this code is not working how can I do this ?
php ajax
asked Nov 22 '18 at 15:42
Rohitashv SinghalRohitashv Singhal
2,875104484
add a comment |
I am using the following ajax function
function validatec()
{
var coupon = jQuery('#validateC').val();
var data = "coupon="+coupon;
$.ajax({
type : 'POST', //Method type
url : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
data : data,
async : false,
success: function(msg) {
if(msg > 0)
{
}
else
{
alert("Please enter a valid Coupon Code");
}
return false;
}
});
}
What I want to do is that when the result is received from ajax call then I dont want to load the page. but this code is not working how can I do this ?
php ajax
asked Nov 22 '18 at 15:42
Rohitashv SinghalRohitashv Singhal
2,875104484
There is nothing in the code you've shown that would cause a page refresh.
– Utkanos
Nov 22 '18 at 15:43
I dont want to load page on success
– Rohitashv Singhal
Nov 22 '18 at 15:47
that's what @Utkanos said that in your code there is nothing that would cause a page refresh. Provide more code please (your html would help)
– pr1nc3
Nov 22 '18 at 16:12
add a comment |
I am using the following ajax function
function validatec()
{
var coupon = jQuery('#validateC').val();
var data = "coupon="+coupon;
$.ajax({
type : 'POST', //Method type
url : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
data : data,
async : false,
success: function(msg) {
if(msg > 0)
{
}
else
{
alert("Please enter a valid Coupon Code");
}
return false;
}
});
}
What I want to do is that when the result is received from ajax call then I dont want to load the page. but this code is not working how can I do this ?
php ajax
asked Nov 22 '18 at 15:42
Rohitashv SinghalRohitashv Singhal
2,875104484
I am using the following ajax function
function validatec()
{
var coupon = jQuery('#validateC').val();
var data = "coupon="+coupon;
$.ajax({
type : 'POST', //Method type
url : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
data : data,
async : false,
success: function(msg) {
if(msg > 0)
{
}
else
{
alert("Please enter a valid Coupon Code");
}
return false;
}
});
}
What I want to do is that when the result is received from ajax call then I dont want to load the page. but this code is not working how can I do this ?
php ajax
php ajax
asked Nov 22 '18 at 15:42
Rohitashv SinghalRohitashv Singhal
2,875104484
asked Nov 22 '18 at 15:42
Rohitashv SinghalRohitashv Singhal
2,875104484
edited Nov 22 '18 at 15:44
Rohitashv Singhal
asked Nov 22 '18 at 15:42
Rohitashv SinghalRohitashv Singhal
2,875104484
asked Nov 22 '18 at 15:42
Rohitashv SinghalRohitashv Singhal
2,875104484
asked Nov 22 '18 at 15:42
Rohitashv SinghalRohitashv Singhal
2,875104484
2,875104484
There is nothing in the code you've shown that would cause a page refresh.
– Utkanos
Nov 22 '18 at 15:43
I dont want to load page on success
– Rohitashv Singhal
Nov 22 '18 at 15:47
that's what @Utkanos said that in your code there is nothing that would cause a page refresh. Provide more code please (your html would help)
– pr1nc3
Nov 22 '18 at 16:12
add a comment |
There is nothing in the code you've shown that would cause a page refresh.
– Utkanos
Nov 22 '18 at 15:43
I dont want to load page on success
– Rohitashv Singhal
Nov 22 '18 at 15:47
that's what @Utkanos said that in your code there is nothing that would cause a page refresh. Provide more code please (your html would help)
– pr1nc3
Nov 22 '18 at 16:12
There is nothing in the code you've shown that would cause a page refresh.
– Utkanos
Nov 22 '18 at 15:43
There is nothing in the code you've shown that would cause a page refresh.
– Utkanos
Nov 22 '18 at 15:43
I dont want to load page on success
– Rohitashv Singhal
Nov 22 '18 at 15:47
I dont want to load page on success
– Rohitashv Singhal
Nov 22 '18 at 15:47
that's what @Utkanos said that in your code there is nothing that would cause a page refresh. Provide more code please (your html would help)
– pr1nc3
Nov 22 '18 at 16:12
that's what @Utkanos said that in your code there is nothing that would cause a page refresh. Provide more code please (your html would help)
– pr1nc3
Nov 22 '18 at 16:12
add a comment |
2 Answers
2
active
oldest
votes
Your code not showing that there is a cause to page refresh, i think that you are calling the ajax when you click o button of type 'submit',
if yes, you have to change the type from submit to button and in your success reponse you have to add somthing like:
jQuery('#form').submit();
answered Nov 22 '18 at 15:47
hamzanatekhamzanatek
19611
add a comment |
Try the below code, I have made few tweaks.
On a side note: I think that return false; before the success function closes might also be a potential issue.
function validatec()
{
var couponCode = jQuery('#validateC').val();
var dataString = {coupon:couponCode} //change over here
$.ajax({
method : 'POST', //change over here
url : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
data : dataString,
async : false,
success : function(msg) {
alert("Enter success") //change over here
if(msg > 0)
{
alert("Coupon Valid"); //change over here
}
else
{
alert("Please enter a valid Coupon Code");
return false; //change over here
}
alert("Exit success") //change over here
}
});
}
answered Nov 22 '18 at 19:58
dexterdexter
1,32811020
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your code not showing that there is a cause to page refresh, i think that you are calling the ajax when you click o button of type 'submit',
if yes, you have to change the type from submit to button and in your success reponse you have to add somthing like:
jQuery('#form').submit();
answered Nov 22 '18 at 15:47
hamzanatekhamzanatek
19611
add a comment |
Your code not showing that there is a cause to page refresh, i think that you are calling the ajax when you click o button of type 'submit',
if yes, you have to change the type from submit to button and in your success reponse you have to add somthing like:
jQuery('#form').submit();
answered Nov 22 '18 at 15:47
hamzanatekhamzanatek
19611
add a comment |
Your code not showing that there is a cause to page refresh, i think that you are calling the ajax when you click o button of type 'submit',
if yes, you have to change the type from submit to button and in your success reponse you have to add somthing like:
jQuery('#form').submit();
answered Nov 22 '18 at 15:47
hamzanatekhamzanatek
19611
Your code not showing that there is a cause to page refresh, i think that you are calling the ajax when you click o button of type 'submit',
if yes, you have to change the type from submit to button and in your success reponse you have to add somthing like:
jQuery('#form').submit();
answered Nov 22 '18 at 15:47
hamzanatekhamzanatek
19611
answered Nov 22 '18 at 15:47
hamzanatekhamzanatek
19611
answered Nov 22 '18 at 15:47
hamzanatekhamzanatek
19611
answered Nov 22 '18 at 15:47
hamzanatekhamzanatek
19611
19611
add a comment |
add a comment |
Try the below code, I have made few tweaks.
On a side note: I think that return false; before the success function closes might also be a potential issue.
function validatec()
{
var couponCode = jQuery('#validateC').val();
var dataString = {coupon:couponCode} //change over here
$.ajax({
method : 'POST', //change over here
url : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
data : dataString,
async : false,
success : function(msg) {
alert("Enter success") //change over here
if(msg > 0)
{
alert("Coupon Valid"); //change over here
}
else
{
alert("Please enter a valid Coupon Code");
return false; //change over here
}
alert("Exit success") //change over here
}
});
}
answered Nov 22 '18 at 19:58
dexterdexter
1,32811020
add a comment |
Try the below code, I have made few tweaks.
On a side note: I think that return false; before the success function closes might also be a potential issue.
function validatec()
{
var couponCode = jQuery('#validateC').val();
var dataString = {coupon:couponCode} //change over here
$.ajax({
method : 'POST', //change over here
url : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
data : dataString,
async : false,
success : function(msg) {
alert("Enter success") //change over here
if(msg > 0)
{
alert("Coupon Valid"); //change over here
}
else
{
alert("Please enter a valid Coupon Code");
return false; //change over here
}
alert("Exit success") //change over here
}
});
}
answered Nov 22 '18 at 19:58
dexterdexter
1,32811020
add a comment |
Try the below code, I have made few tweaks.
On a side note: I think that return false; before the success function closes might also be a potential issue.
function validatec()
{
var couponCode = jQuery('#validateC').val();
var dataString = {coupon:couponCode} //change over here
$.ajax({
method : 'POST', //change over here
url : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
data : dataString,
async : false,
success : function(msg) {
alert("Enter success") //change over here
if(msg > 0)
{
alert("Coupon Valid"); //change over here
}
else
{
alert("Please enter a valid Coupon Code");
return false; //change over here
}
alert("Exit success") //change over here
}
});
}
answered Nov 22 '18 at 19:58
dexterdexter
1,32811020
Try the below code, I have made few tweaks.
On a side note: I think that return false; before the success function closes might also be a potential issue.
function validatec()
{
var couponCode = jQuery('#validateC').val();
var dataString = {coupon:couponCode} //change over here
$.ajax({
method : 'POST', //change over here
url : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
data : dataString,
async : false,
success : function(msg) {
alert("Enter success") //change over here
if(msg > 0)
{
alert("Coupon Valid"); //change over here
}
else
{
alert("Please enter a valid Coupon Code");
return false; //change over here
}
alert("Exit success") //change over here
}
});
}
answered Nov 22 '18 at 19:58
dexterdexter
1,32811020
answered Nov 22 '18 at 19:58
dexterdexter
1,32811020
answered Nov 22 '18 at 19:58
dexterdexter
1,32811020
answered Nov 22 '18 at 19:58
dexterdexter
1,32811020
1,32811020
add a comment |
add a comment |
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There is nothing in the code you've shown that would cause a page refresh.
– Utkanos
Nov 22 '18 at 15:43
I dont want to load page on success
– Rohitashv Singhal
Nov 22 '18 at 15:47
that's what @Utkanos said that in your code there is nothing that would cause a page refresh. Provide more code please (your html would help)
– pr1nc3
Nov 22 '18 at 16:12