[jQuery] Re: How to handle responses from PHP with $.ajax()
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[jQuery] Re: How to handle responses from PHP with $.ajax()
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[jQuery] Re: How to handle responses from PHP with $.ajax()
Hi,
Many thanks for the help. I had a go at implementing what you
suggested but I'm clearly still missing something.
Here is my $.ajax() call:
--
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(del){
if(del.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
--
And here is my json array in PHP:
--
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1).
} else {
$json_data = array('result' = 0, 'error' = 'This is an error').
}
json_encode($json_data);
--
I'm obviously not accessing del.result correctly because $
('div.error').show(); is always executed regardless of the value of
result. Where am I going wrong?
Thanks in advance,
Rob.
On Jan 4, 2:41 pm, donb falconwatc...@comcast.net wrote:
Return a JSON object.
Construct a PHP array such as $json_data = array('result' = 0,
'error' = 'This is an error'). End your PHP script with json_encode
($json_data). Then you can reference del.result and del.error (I'm
referring to your definition of the' success' function from your
sample code)
On Jan 4, 7:16 am, rob303 r...@cube33.com wrote:
Hi,
I'm new to ajax and jquery but I'm not new to PHP. The following
example seems to work okay to a point but I can't figure out how to
handle data validation errors generated in my PHP.
In my example I want to post some data to a script called ajax.php.
That script will check the data for validity and then return true or
false depending on the outcome of the checks. It will also set an
appropriate error message.
How can I handle the returned data and display an error message if
needed in ajax?
--
HTML / Ajax - test.php
--
?php
require_once(includes.php);
?
html
head
script src=js/jquery.js type=text/javascript/script
script type=text/javascript
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(del){
$('form#submit').hide();
$('div.success').fadeIn('medium');
// what if the data failed validation in PHP?
// we need to show 'div.error.
}
});
return false;
});
});
/script
/head
body
form id=submit method=post
Email:
input id=email class=text name=email size=20
type=text /
input type=submit value=send mail /
/form
div class=success style=display:none;
Email sent.
/div
div class=error
?php echo $userError; ?
/div
/body
/html
--
PHP - ajax.php
--
?php
require_once(includes.php);
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
return 1;} else {
$userError = Please enter your email address;
return 0;
}
?
I've been searching the web for an answer but can't find one
anywhere! Any help would be greatly appreciated.
[jQuery] Re: How to handle responses from PHP with $.ajax()
You must 'echo' the JSON output. Also you indicate periods ending
code lines, instead of semicolons but perhaps that's just an error in
your post, not the actual code?
On Jan 4, 10:38 am, rob303 r...@cube33.com wrote:
Hi,
Many thanks for the help. I had a go at implementing what you
suggested but I'm clearly still missing something.
Here is my $.ajax() call:
--
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(del){
if(del.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
--
And here is my json array in PHP:
--
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1).} else {
$json_data = array('result' = 0, 'error' = 'This is an error').
}
json_encode($json_data);
--
I'm obviously not accessing del.result correctly because $
('div.error').show(); is always executed regardless of the value of
result. Where am I going wrong?
Thanks in advance,
Rob.
On Jan 4, 2:41 pm, donb falconwatc...@comcast.net wrote:
Return a JSON object.
Construct a PHP array such as $json_data = array('result' = 0,
'error' = 'This is an error'). End your PHP script with json_encode
($json_data). Then you can reference del.result and del.error (I'm
referring to your definition of the' success' function from your
sample code)
On Jan 4, 7:16 am, rob303 r...@cube33.com wrote:
Hi,
I'm new to ajax and jquery but I'm not new to PHP. The following
example seems to work okay to a point but I can't figure out how to
handle data validation errors generated in my PHP.
In my example I want to post some data to a script called ajax.php.
That script will check the data for validity and then return true or
false depending on the outcome of the checks. It will also set an
appropriate error message.
How can I handle the returned data and display an error message if
needed in ajax?
--
HTML / Ajax - test.php
--
?php
require_once(includes.php);
?
html
head
script src=js/jquery.js type=text/javascript/script
script type=text/javascript
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(del){
$('form#submit').hide();
$('div.success').fadeIn('medium');
// what if the data failed validation in PHP?
// we need to show 'div.error.
}
});
return false;
});
});
/script
/head
body
form id=submit method=post
Email:
input id=email class=text name=email size=20
type=text /
input type=submit value=send mail /
/form
div class=success style=display:none;
Email sent.
/div
div class=error
?php echo $userError; ?
/div
/body
/html
--
PHP - ajax.php
--
?php
require_once(includes.php);
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
return 1;} else {
$userError = Please enter your email address;
return 0;
}
?
I've been searching the web for an answer but can't find one
anywhere! Any help would be greatly appreciated.
[jQuery] Re: How to handle responses from PHP with $.ajax()
Yes, the missing semicolons are an error in the post. I tried to echo
the json_encode() call but I still can't get it to work. If I leave
the email input blank no email is sent. If I enter a string the email
is sent. The 'div.error' is always displayed regardless what's posted.
Here's my full code again:
--
HTML / Ajax
--
?php
require_once(../inc/site_config.php);
require_once(SITE . /includes.php);
?
html
head
script src=js/jquery.js type=text/javascript/script
script type=text/javascript
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php?v=1,
data: email=+ email,
success: function(res){
if(res.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
/script
/head
body
form id=submit method=post
Email:
input id=email class=text name=email size=20
type=text /
input type=submit value=send mail /
/form
div class=success style=display:none;
Email sent.
/div
div class=error style=display:none;err/div
/body
/html
--
PHP
--
?php
require_once(includes.php);
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1);
} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
$message = $_POST['email'];
if($message != '') {
mail('r...@cube33.com', 'Ajax test', $message);
$json_data = array('result' = 1);
} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
--
On Jan 4, 5:44 pm, donb falconwatc...@comcast.net wrote:
You must 'echo' the JSON output. Also you indicate periods ending
code lines, instead of semicolons but perhaps that's just an error in
your post, not the actual code?
On Jan 4, 10:38 am, rob303 r...@cube33.com wrote:
Hi,
Many thanks for the help. I had a go at implementing what you
suggested but I'm clearly still missing something.
Here is my $.ajax() call:
--
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(del){
if(del.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
--
And here is my json array in PHP:
--
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1).} else {
$json_data = array('result' = 0, 'error' = 'This is an error').
}
json_encode($json_data);
--
I'm obviously not accessing del.result correctly because $
('div.error').show(); is always executed regardless of the value of
result. Where am I going wrong?
Thanks in advance,
Rob.
On Jan 4, 2:41 pm, donb falconwatc...@comcast.net wrote:
Return a JSON object.
Construct a PHP array such as $json_data = array('result' = 0,
'error' = 'This is an error'). End your PHP script with json_encode
($json_data). Then you can reference del.result and del.error (I'm
referring to your definition of the' success' function from your
sample code)
On Jan 4, 7:16 am, rob303 r...@cube33.com wrote:
Hi,
I'm new to ajax and jquery but I'm not new to PHP. The following
example seems to work okay to a point but I can't figure out how to
handle data validation errors generated in my PHP.
In my example I want to post some data to a script called ajax.php.
That script will check the data for validity and then return true or
false depending on the outcome of the checks. It will also set an
appropriate error message.
How can I handle the returned data and display an error message if
needed in ajax?
--
HTML / Ajax - test.php
--
?php
require_once(includes.php);
?
html
head
script src=js/jquery.js type=text/javascript/script
script type=text/javascript
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url:
[jQuery] Re: How to handle responses from PHP with $.ajax()
Interestingly, in the Firebug console I'm seeing the correct
responses:
{result:1}
or
{result:0}
How can I access these inside my $.ajax() call?
Many thanks again for all the help!
Rob.
On Jan 4, 5:57 pm, rob303 r...@cube33.com wrote:
Yes, the missing semicolons are an error in the post. I tried to echo
the json_encode() call but I still can't get it to work. If I leave
the email input blank no email is sent. If I enter a string the email
is sent. The 'div.error' is always displayed regardless what's posted.
Here's my full code again:
--
HTML / Ajax
--
?php
require_once(../inc/site_config.php);
require_once(SITE . /includes.php);
?
html
head
script src=js/jquery.js type=text/javascript/script
script type=text/javascript
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php?v=1,
data: email=+ email,
success: function(res){
if(res.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
/script
/head
body
form id=submit method=post
Email:
input id=email class=text name=email size=20
type=text /
input type=submit value=send mail /
/form
div class=success style=display:none;
Email sent.
/div
div class=error style=display:none;err/div
/body
/html
--
PHP
--
?php
require_once(includes.php);
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1);} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
$message = $_POST['email'];
if($message != '') {
mail('@cube33.com', 'Ajax test', $message);
$json_data = array('result' = 1);} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
--
On Jan 4, 5:44 pm, donb falconwatc...@comcast.net wrote:
You must 'echo' the JSON output. Also you indicate periods ending
code lines, instead of semicolons but perhaps that's just an error in
your post, not the actual code?
On Jan 4, 10:38 am, rob303 r...@cube33.com wrote:
Hi,
Many thanks for the help. I had a go at implementing what you
suggested but I'm clearly still missing something.
Here is my $.ajax() call:
--
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(del){
if(del.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
--
And here is my json array in PHP:
--
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1).} else {
$json_data = array('result' = 0, 'error' = 'This is an error').
}
json_encode($json_data);
--
I'm obviously not accessing del.result correctly because $
('div.error').show(); is always executed regardless of the value of
result. Where am I going wrong?
Thanks in advance,
Rob.
On Jan 4, 2:41 pm, donb falconwatc...@comcast.net wrote:
Return a JSON object.
Construct a PHP array such as $json_data = array('result' = 0,
'error' = 'This is an error'). End your PHP script with json_encode
($json_data). Then you can reference del.result and del.error (I'm
referring to your definition of the' success' function from your
sample code)
On Jan 4, 7:16 am, rob303 r...@cube33.com wrote:
Hi,
I'm new to ajax and jquery but I'm not new to PHP. The following
example seems to work okay to a point but I can't figure out how to
handle data validation errors generated in my PHP.
In my example I want to post some data to a script called ajax.php.
That script will check the data for validity and then return true or
false depending on the outcome of the checks. It will also set an
appropriate error message.
How can I handle the returned data and display an error message if
needed in ajax?
--
HTML / Ajax - test.php
--
[jQuery] Re: How to handle responses from PHP with $.ajax()
On another thread, someone pointed out to me that json_encode was
supposed to 'know' if the data values were numeric or string and only
quote the latter. That was not my experience, and perhaps not yours,
either. Try '1' as the value you compare to instead of 1.
On Jan 4, 1:05 pm, rob303 r...@cube33.com wrote:
Interestingly, in the Firebug console I'm seeing the correct
responses:
{result:1}
or
{result:0}
How can I access these inside my $.ajax() call?
Many thanks again for all the help!
Rob.
On Jan 4, 5:57 pm, rob303 r...@cube33.com wrote:
Yes, the missing semicolons are an error in the post. I tried to echo
the json_encode() call but I still can't get it to work. If I leave
the email input blank no email is sent. If I enter a string the email
is sent. The 'div.error' is always displayed regardless what's posted.
Here's my full code again:
--
HTML / Ajax
--
?php
require_once(../inc/site_config.php);
require_once(SITE . /includes.php);
?
html
head
script src=js/jquery.js type=text/javascript/script
script type=text/javascript
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php?v=1,
data: email=+ email,
success: function(res){
if(res.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
/script
/head
body
form id=submit method=post
Email:
input id=email class=text name=email size=20
type=text /
input type=submit value=send mail /
/form
div class=success style=display:none;
Email sent.
/div
div class=error style=display:none;err/div
/body
/html
--
PHP
--
?php
require_once(includes.php);
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1);} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
$message = $_POST['email'];
if($message != '') {
mail('@cube33.com', 'Ajax test', $message);
$json_data = array('result' = 1);} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
--
On Jan 4, 5:44 pm, donb falconwatc...@comcast.net wrote:
You must 'echo' the JSON output. Also you indicate periods ending
code lines, instead of semicolons but perhaps that's just an error in
your post, not the actual code?
On Jan 4, 10:38 am, rob303 r...@cube33.com wrote:
Hi,
Many thanks for the help. I had a go at implementing what you
suggested but I'm clearly still missing something.
Here is my $.ajax() call:
--
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(del){
if(del.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
--
And here is my json array in PHP:
--
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1).} else {
$json_data = array('result' = 0, 'error' = 'This is an error').
}
json_encode($json_data);
--
I'm obviously not accessing del.result correctly because $
('div.error').show(); is always executed regardless of the value of
result. Where am I going wrong?
Thanks in advance,
Rob.
On Jan 4, 2:41 pm, donb falconwatc...@comcast.net wrote:
Return a JSON object.
Construct a PHP array such as $json_data = array('result' = 0,
'error' = 'This is an error'). End your PHP script with json_encode
($json_data). Then you can reference del.result and del.error (I'm
referring to your definition of the' success' function from your
sample code)
On Jan 4, 7:16 am, rob303 r...@cube33.com wrote:
Hi,
I'm new to ajax and jquery but I'm not new to PHP. The following
example seems to work okay to a point but I can't figure out how to
handle data validation
[jQuery] Re: How to handle responses from PHP with $.ajax()
Hi again,
I tried that but got the same results.
if(res.result == '1') {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
if(res.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
res.result is always false even though firebug says the response is
{result:1}.
Rob.
On Jan 4, 7:01 pm, donb falconwatc...@comcast.net wrote:
On another thread, someone pointed out to me that json_encode was
supposed to 'know' if the data values were numeric or string and only
quote the latter. That was not my experience, and perhaps not yours,
either. Try '1' as the value you compare to instead of 1.
On Jan 4, 1:05 pm, rob303 r...@cube33.com wrote:
Interestingly, in the Firebug console I'm seeing the correct
responses:
{result:1}
or
{result:0}
How can I access these inside my $.ajax() call?
Many thanks again for all the help!
Rob.
On Jan 4, 5:57 pm, rob303 r...@cube33.com wrote:
Yes, the missing semicolons are an error in the post. I tried to echo
the json_encode() call but I still can't get it to work. If I leave
the email input blank no email is sent. If I enter a string the email
is sent. The 'div.error' is always displayed regardless what's posted.
Here's my full code again:
--
HTML / Ajax
--
?php
require_once(../inc/site_config.php);
require_once(SITE . /includes.php);
?
html
head
script src=js/jquery.js type=text/javascript/script
script type=text/javascript
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php?v=1,
data: email=+ email,
success: function(res){
if(res.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
/script
/head
body
form id=submit method=post
Email:
input id=email class=text name=email size=20
type=text /
input type=submit value=send mail /
/form
div class=success style=display:none;
Email sent.
/div
div class=error style=display:none;err/div
/body
/html
--
PHP
--
?php
require_once(includes.php);
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1);} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
$message = $_POST['email'];
if($message != '') {
mail('@cube33.com', 'Ajax test', $message);
$json_data = array('result' = 1);} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
--
On Jan 4, 5:44 pm, donb falconwatc...@comcast.net wrote:
You must 'echo' the JSON output. Also you indicate periods ending
code lines, instead of semicolons but perhaps that's just an error in
your post, not the actual code?
On Jan 4, 10:38 am, rob303 r...@cube33.com wrote:
Hi,
Many thanks for the help. I had a go at implementing what you
suggested but I'm clearly still missing something.
Here is my $.ajax() call:
--
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(del){
if(del.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
--
And here is my json array in PHP:
--
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1).} else {
$json_data = array('result' = 0, 'error' = 'This is an error').
}
json_encode($json_data);
--
I'm obviously not accessing del.result correctly because $
('div.error').show(); is always executed regardless of the value of
result. Where am I going wrong?
Thanks in advance,
Rob.
On Jan 4, 2:41 pm, donb falconwatc...@comcast.net wrote:
[jQuery] Re: How to handle responses from PHP with $.ajax()
Return a JSON object.
Construct a PHP array such as $json_data = array('result' = 0,
'error' = 'This is an error'). End your PHP script with json_encode
($json_data). Then you can reference del.result and del.error (I'm
referring to your definition of the' success' function from your
sample code)
On Jan 4, 7:16 am, rob303 r...@cube33.com wrote:
Hi,
I'm new to ajax and jquery but I'm not new to PHP. The following
example seems to work okay to a point but I can't figure out how to
handle data validation errors generated in my PHP.
In my example I want to post some data to a script called ajax.php.
That script will check the data for validity and then return true or
false depending on the outcome of the checks. It will also set an
appropriate error message.
How can I handle the returned data and display an error message if
needed in ajax?
--
HTML / Ajax - test.php
--
?php
require_once(includes.php);
?
html
head
script src=js/jquery.js type=text/javascript/script
script type=text/javascript
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(del){
$('form#submit').hide();
$('div.success').fadeIn('medium');
// what if the data failed validation in PHP?
// we need to show 'div.error.
}
});
return false;
});
});
/script
/head
body
form id=submit method=post
Email:
input id=email class=text name=email size=20
type=text /
input type=submit value=send mail /
/form
div class=success style=display:none;
Email sent.
/div
div class=error
?php echo $userError; ?
/div
/body
/html
--
PHP - ajax.php
--
?php
require_once(includes.php);
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
return 1;} else {
$userError = Please enter your email address;
return 0;
}
?
I've been searching the web for an answer but can't find one
anywhere! Any help would be greatly appreciated.
[jQuery] Re: How to handle responses from PHP with $.ajax()
After a little more twiddling ... This works:
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(res){
if(res == {\result\:1}) {
$('form#submit').hide();
$('div.error').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
The json object contains what I need but I can only test it as if it
were a string. res.result is undefined. Why?
Thanks in advance for any help!
Rob.
rob303 wrote:
Hi again,
I tried that but got the same results.
if(res.result == '1') {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
if(res.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
res.result is always false even though firebug says the response is
{result:1}.
Rob.
On Jan 4, 7:01�pm, donb falconwatc...@comcast.net wrote:
On another thread, someone pointed out to me that json_encode was
supposed to 'know' if the data values were numeric or string and only
quote the latter. �That was not my experience, and perhaps not yours,
either. �Try '1' as the value you compare to instead of 1.
On Jan 4, 1:05�pm, rob303 r...@cube33.com wrote:
Interestingly, in the Firebug console I'm seeing the correct
responses:
{result:1}
or
{result:0}
How can I access these inside my $.ajax() call?
Many thanks again for all the help!
Rob.
On Jan 4, 5:57�pm, rob303 r...@cube33.com wrote:
Yes, the missing semicolons are an error in the post. I tried to echo
the json_encode() call but I still can't get it to work. If I leave
the email input blank no email is sent. If I enter a string the email
is sent. The 'div.error' is always displayed regardless what's posted.
Here's my full code again:
--
HTML / Ajax
--
?php
require_once(../inc/site_config.php);
require_once(SITE . /includes.php);
?
html
� head
� � � � script src=js/jquery.js type=text/javascript/script
� � script type=text/javascript
� � � $(document).ready(function(){
� � � � $(form#submit).submit(function() {
� � � � � var email = $('#email').attr('value');
� � � � � $.ajax({
� � � � � � type: POST,
� � � � � � url: ajax.php?v=1,
� � � � � � data: email=+ email,
� � � � � � success: function(res){
� � � � � � � if(res.result == 1) {
� � � � � � � � $('form#submit').hide();
� � � � � � � � $('div.success').fadeIn('medium');
� � � � � � � } else {
� � � � � � � � $('div.error').show();
� � � � � � � }
� � � � � � }
� � � � � });
� � � � � return false;
� � � � });
� � � });
� � /script
� /head
� body
� � � form id=submit method=post
� � � � Email:
� � � � input id=email class=text name=email size=20
type=text /
� � � � � � input type=submit value=send mail /
� � � � � /form
� � � div class=success style=display:none;
� � � � Email sent.
� � � /div
� � � div class=error style=display:none;err/div
� /body
/html
--
PHP
--
?php
require_once(includes.php);
$message = $_POST['email'];
if($message != '') {
� mail('some-em...@email.com', 'Ajax test', $message);
� $json_data = array('result' = 1);} else {
� $json_data = array('result' = 0);
}
echo json_encode($json_data);
?
$message = $_POST['email'];
if($message != '') {
� mail('@cube33.com', 'Ajax test', $message);
� $json_data = array('result' = 1);} else {
� $json_data = array('result' = 0);
}
echo json_encode($json_data);
?
--
On Jan 4, 5:44�pm, donb falconwatc...@comcast.net wrote:
You must 'echo' the JSON output. �Also you indicate periods ending
code lines, instead of semicolons but perhaps that's just an error in
your post, not the actual code?
On Jan 4, 10:38�am, rob303 r...@cube33.com wrote:
Hi,
Many thanks for the help. �I had a go at implementing what you
suggested but I'm clearly still missing something.
Here is my $.ajax() call:
--
$(document).ready(function(){
� � � � $(form#submit).submit(function() {
� � � � � var email = $('#email').attr('value');
� � � � � $.ajax({
� � � � � � type: POST,
� � � � � � url: ajax.php,
� � � � � � data: email=+ email,
� � � � � � success: function(del){
� � � � � � � if(del.result == 1) {
[jQuery] Re: How to handle responses from PHP with $.ajax()
I found the problem. I was missing the dataType option:
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
dataType: json,
success: function(res){
if(res.result == 1) {
$('form#submit').hide();
$('div.error').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
It works just fine now.
Cheers,
Rob.
On Jan 4, 10:12 pm, rob303 r...@cube33.com wrote:
After a little more twiddling ... This works:
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php,
data: email=+ email,
success: function(res){
if(res == {\result\:1}) {
$('form#submit').hide();
$('div.error').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
The json object contains what I need but I can only test it as if it
were a string. res.result is undefined. Why?
Thanks in advance for any help!
Rob.
rob303 wrote:
Hi again,
I tried that but got the same results.
if(res.result == '1') {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
if(res.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
res.result is always false even though firebug says the response is
{result:1}.
Rob.
On Jan 4, 7:01 pm, donb falconwatc...@comcast.net wrote:
On another thread, someone pointed out to me that json_encode was
supposed to 'know' if the data values were numeric or string and only
quote the latter. That was not my experience, and perhaps not yours,
either. Try '1' as the value you compare to instead of 1.
On Jan 4, 1:05 pm, rob303 r...@cube33.com wrote:
Interestingly, in the Firebug console I'm seeing the correct
responses:
{result:1}
or
{result:0}
How can I access these inside my $.ajax() call?
Many thanks again for all the help!
Rob.
On Jan 4, 5:57 pm, rob303 r...@cube33.com wrote:
Yes, the missing semicolons are an error in the post. I tried to echo
the json_encode() call but I still can't get it to work. If I leave
the email input blank no email is sent. If I enter a string the email
is sent. The 'div.error' is always displayed regardless what's posted.
Here's my full code again:
--
HTML / Ajax
--
?php
require_once(../inc/site_config.php);
require_once(SITE . /includes.php);
?
html
head
script src=js/jquery.js type=text/javascript/script
script type=text/javascript
$(document).ready(function(){
$(form#submit).submit(function() {
var email = $('#email').attr('value');
$.ajax({
type: POST,
url: ajax.php?v=1,
data: email=+ email,
success: function(res){
if(res.result == 1) {
$('form#submit').hide();
$('div.success').fadeIn('medium');
} else {
$('div.error').show();
}
}
});
return false;
});
});
/script
/head
body
form id=submit method=post
Email:
input id=email class=text name=email size=20
type=text /
input type=submit value=send mail /
/form
div class=success style=display:none;
Email sent.
/div
div class=error style=display:none;err/div
/body
/html
--
PHP
--
?php
require_once(includes.php);
$message = $_POST['email'];
if($message != '') {
mail('some-em...@email.com', 'Ajax test', $message);
$json_data = array('result' = 1);} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
$message = $_POST['email'];
if($message != '') {
mail('@cube33.com', 'Ajax test', $message);
$json_data = array('result' = 1);} else {
$json_data = array('result' = 0);
}
echo json_encode($json_data);
?
--
On Jan 4, 5:44 pm, donb falconwatc...@comcast.net wrote:
You must 'echo' the JSON output. Also you indicate periods ending
code lines, instead of semicolons but perhaps that's just an error
in
your post, not the actual code?
On Jan 4, 10:38 am, rob303 r...@cube33.com wrote:
Hi,
