Unable to make a ajax async post request to the same page
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I'm trying to make an async call to the same page but for some reason, the async call is not being made and hence getting "userinfo" undefined error! could someone please tell me what I'm doing wrong in this thank you!
<script src="https://code.jquery.com/jquery-3.3.1.min.js"></script>
<script>
$( document ).ready(function() {
//getting location of the user and assigning it to the variable
$.get("http://ipinfo.io", function (response) {
var usercity=response.city;
console.log(usercity);
$.ajax({
//no need to provide url since we are making async call on the same page
type: "post",
data: {cityinfo:usercity},
});
}, "jsonp");
});
</script>
<?php
session_start();
$usercity=$_POST['cityinfo'];
echo $usercity;
?>
php ajax
edited Jan 3 at 22:40
Quentin
658k728951057
asked Jan 3 at 22:32
Faizal Mohaideen KadershaFaizal Mohaideen Kadersha
176
add a comment |
I'm trying to make an async call to the same page but for some reason, the async call is not being made and hence getting "userinfo" undefined error! could someone please tell me what I'm doing wrong in this thank you!
<script src="https://code.jquery.com/jquery-3.3.1.min.js"></script>
<script>
$( document ).ready(function() {
//getting location of the user and assigning it to the variable
$.get("http://ipinfo.io", function (response) {
var usercity=response.city;
console.log(usercity);
$.ajax({
//no need to provide url since we are making async call on the same page
type: "post",
data: {cityinfo:usercity},
});
}, "jsonp");
});
</script>
<?php
session_start();
$usercity=$_POST['cityinfo'];
echo $usercity;
?>
php ajax
edited Jan 3 at 22:40
Quentin
658k728951057
asked Jan 3 at 22:32
Faizal Mohaideen KadershaFaizal Mohaideen Kadersha
176
I tried var_dump($_POST) and it shows
– Faizal Mohaideen Kadersha
Jan 3 at 22:34
array (size=0) empty // if im correct the data is not being passed or the post request is not made!
– Faizal Mohaideen Kadersha
Jan 3 at 22:35
add a comment |
I'm trying to make an async call to the same page but for some reason, the async call is not being made and hence getting "userinfo" undefined error! could someone please tell me what I'm doing wrong in this thank you!
<script src="https://code.jquery.com/jquery-3.3.1.min.js"></script>
<script>
$( document ).ready(function() {
//getting location of the user and assigning it to the variable
$.get("http://ipinfo.io", function (response) {
var usercity=response.city;
console.log(usercity);
$.ajax({
//no need to provide url since we are making async call on the same page
type: "post",
data: {cityinfo:usercity},
});
}, "jsonp");
});
</script>
<?php
session_start();
$usercity=$_POST['cityinfo'];
echo $usercity;
?>
php ajax
edited Jan 3 at 22:40
Quentin
658k728951057
asked Jan 3 at 22:32
Faizal Mohaideen KadershaFaizal Mohaideen Kadersha
176
I'm trying to make an async call to the same page but for some reason, the async call is not being made and hence getting "userinfo" undefined error! could someone please tell me what I'm doing wrong in this thank you!
<script src="https://code.jquery.com/jquery-3.3.1.min.js"></script>
<script>
$( document ).ready(function() {
//getting location of the user and assigning it to the variable
$.get("http://ipinfo.io", function (response) {
var usercity=response.city;
console.log(usercity);
$.ajax({
//no need to provide url since we are making async call on the same page
type: "post",
data: {cityinfo:usercity},
});
}, "jsonp");
});
</script>
<?php
session_start();
$usercity=$_POST['cityinfo'];
echo $usercity;
?>
php ajax
php ajax
edited Jan 3 at 22:40
Quentin
658k728951057
asked Jan 3 at 22:32
Faizal Mohaideen KadershaFaizal Mohaideen Kadersha
176
edited Jan 3 at 22:40
Quentin
658k728951057
asked Jan 3 at 22:32
Faizal Mohaideen KadershaFaizal Mohaideen Kadersha
176
edited Jan 3 at 22:40
Quentin
658k728951057
edited Jan 3 at 22:40
Quentin
658k728951057
edited Jan 3 at 22:40
Quentin
658k728951057
658k728951057
asked Jan 3 at 22:32
Faizal Mohaideen KadershaFaizal Mohaideen Kadersha
176
asked Jan 3 at 22:32
Faizal Mohaideen KadershaFaizal Mohaideen Kadersha
176
asked Jan 3 at 22:32
Faizal Mohaideen KadershaFaizal Mohaideen Kadersha
176
176
I tried var_dump($_POST) and it shows
– Faizal Mohaideen Kadersha
Jan 3 at 22:34
array (size=0) empty // if im correct the data is not being passed or the post request is not made!
– Faizal Mohaideen Kadersha
Jan 3 at 22:35
add a comment |
I tried var_dump($_POST) and it shows
– Faizal Mohaideen Kadersha
Jan 3 at 22:34
array (size=0) empty // if im correct the data is not being passed or the post request is not made!
– Faizal Mohaideen Kadersha
Jan 3 at 22:35
I tried var_dump($_POST) and it shows
– Faizal Mohaideen Kadersha
Jan 3 at 22:34
I tried var_dump($_POST) and it shows
– Faizal Mohaideen Kadersha
Jan 3 at 22:34
array (size=0) empty // if im correct the data is not being passed or the post request is not made!
– Faizal Mohaideen Kadersha
Jan 3 at 22:35
array (size=0) empty // if im correct the data is not being passed or the post request is not made!
– Faizal Mohaideen Kadersha
Jan 3 at 22:35
add a comment |
1 Answer
1
active
oldest
votes
First, you make a regular request to the PHP page. $_POST['cityinfo'] is undefined at this stage because you don't set it. The page includes an error message telling you that.
Second, you make a JSONP request to ipinfo.io, and call a callback function when you get a response.
Third, you make a second HTTP request to the PHP page. This time you do define $_POST['cityinfo']. You have no success handler to do anything at all with the response so, although $_POST['cityinfo']; is set, you don't see any effect of that (unless you were to use the developer tools Network tab to examine the response itself).
It is important to note that this is a second, distinct request to the same URL. The JavaScript does not travel back through time and set the $_POST['cityinfo']; variable in the previous request (and it is the response to the previous request that is still displayed in the browser window).
answered Jan 3 at 22:39
QuentinQuentin
658k728951057
that makes a lot a sens thanks for the insight il try to work on the network tab but i have never worked on it so far, so it will be quite an experience thank you!!!
– Faizal Mohaideen Kadersha
Jan 3 at 22:52
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
First, you make a regular request to the PHP page. $_POST['cityinfo'] is undefined at this stage because you don't set it. The page includes an error message telling you that.
Second, you make a JSONP request to ipinfo.io, and call a callback function when you get a response.
Third, you make a second HTTP request to the PHP page. This time you do define $_POST['cityinfo']. You have no success handler to do anything at all with the response so, although $_POST['cityinfo']; is set, you don't see any effect of that (unless you were to use the developer tools Network tab to examine the response itself).
It is important to note that this is a second, distinct request to the same URL. The JavaScript does not travel back through time and set the $_POST['cityinfo']; variable in the previous request (and it is the response to the previous request that is still displayed in the browser window).
answered Jan 3 at 22:39
QuentinQuentin
658k728951057
that makes a lot a sens thanks for the insight il try to work on the network tab but i have never worked on it so far, so it will be quite an experience thank you!!!
– Faizal Mohaideen Kadersha
Jan 3 at 22:52
add a comment |
First, you make a regular request to the PHP page. $_POST['cityinfo'] is undefined at this stage because you don't set it. The page includes an error message telling you that.
Second, you make a JSONP request to ipinfo.io, and call a callback function when you get a response.
Third, you make a second HTTP request to the PHP page. This time you do define $_POST['cityinfo']. You have no success handler to do anything at all with the response so, although $_POST['cityinfo']; is set, you don't see any effect of that (unless you were to use the developer tools Network tab to examine the response itself).
It is important to note that this is a second, distinct request to the same URL. The JavaScript does not travel back through time and set the $_POST['cityinfo']; variable in the previous request (and it is the response to the previous request that is still displayed in the browser window).
answered Jan 3 at 22:39
QuentinQuentin
658k728951057
that makes a lot a sens thanks for the insight il try to work on the network tab but i have never worked on it so far, so it will be quite an experience thank you!!!
– Faizal Mohaideen Kadersha
Jan 3 at 22:52
add a comment |
First, you make a regular request to the PHP page. $_POST['cityinfo'] is undefined at this stage because you don't set it. The page includes an error message telling you that.
Second, you make a JSONP request to ipinfo.io, and call a callback function when you get a response.
Third, you make a second HTTP request to the PHP page. This time you do define $_POST['cityinfo']. You have no success handler to do anything at all with the response so, although $_POST['cityinfo']; is set, you don't see any effect of that (unless you were to use the developer tools Network tab to examine the response itself).
It is important to note that this is a second, distinct request to the same URL. The JavaScript does not travel back through time and set the $_POST['cityinfo']; variable in the previous request (and it is the response to the previous request that is still displayed in the browser window).
answered Jan 3 at 22:39
QuentinQuentin
658k728951057
First, you make a regular request to the PHP page. $_POST['cityinfo'] is undefined at this stage because you don't set it. The page includes an error message telling you that.
Second, you make a JSONP request to ipinfo.io, and call a callback function when you get a response.
Third, you make a second HTTP request to the PHP page. This time you do define $_POST['cityinfo']. You have no success handler to do anything at all with the response so, although $_POST['cityinfo']; is set, you don't see any effect of that (unless you were to use the developer tools Network tab to examine the response itself).
It is important to note that this is a second, distinct request to the same URL. The JavaScript does not travel back through time and set the $_POST['cityinfo']; variable in the previous request (and it is the response to the previous request that is still displayed in the browser window).
answered Jan 3 at 22:39
QuentinQuentin
658k728951057
answered Jan 3 at 22:39
QuentinQuentin
658k728951057
answered Jan 3 at 22:39
QuentinQuentin
658k728951057
answered Jan 3 at 22:39
QuentinQuentin
658k728951057
658k728951057
that makes a lot a sens thanks for the insight il try to work on the network tab but i have never worked on it so far, so it will be quite an experience thank you!!!
– Faizal Mohaideen Kadersha
Jan 3 at 22:52
add a comment |
that makes a lot a sens thanks for the insight il try to work on the network tab but i have never worked on it so far, so it will be quite an experience thank you!!!
– Faizal Mohaideen Kadersha
Jan 3 at 22:52
that makes a lot a sens thanks for the insight il try to work on the network tab but i have never worked on it so far, so it will be quite an experience thank you!!!
– Faizal Mohaideen Kadersha
Jan 3 at 22:52
that makes a lot a sens thanks for the insight il try to work on the network tab but i have never worked on it so far, so it will be quite an experience thank you!!!
– Faizal Mohaideen Kadersha
Jan 3 at 22:52
add a comment |
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I tried var_dump($_POST) and it shows
– Faizal Mohaideen Kadersha
Jan 3 at 22:34
array (size=0) empty // if im correct the data is not being passed or the post request is not made!
– Faizal Mohaideen Kadersha
Jan 3 at 22:35