How to display success message after ajax form summit
1
I've been working on this code I found somewhere. I use it to submit the Send to Friend form from a popup. It works but I can't get it to display the success message after it submits the form. Right now after it submits the form it just disappears. Maybe someone good at ajax can see what's the matter with it. Thanks
<script type="text/javascript">
//<![CDATA[
var formId = 'product_sendtofriend_form';
var sendfriendForm = new VarienForm(formId, false);
var postUrl = '<?php echo $this->getSendUrl() ?>';
function doAjax() {
if (sendfriendForm.validator.validate()) {
new Ajax.Updater(
{ success:'formSuccess' }, postUrl, {
method:'post',
asynchronous:true,
evalScripts:false,
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess');
},
onLoading:function(request, json){
Element.show('formLoader');
},
parameters: $(formId).serialize(true),
}
);
}
}
new Event.observe(formId, 'submit', function(e){
e.stop();
doAjax();
});
//]]>
</script>
magento-1.9 ajax forms
asked Apr 13 '16 at 15:57
FreejoyFreejoy
444924
bumped to the homepage by Community♦ 8 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
1
I've been working on this code I found somewhere. I use it to submit the Send to Friend form from a popup. It works but I can't get it to display the success message after it submits the form. Right now after it submits the form it just disappears. Maybe someone good at ajax can see what's the matter with it. Thanks
<script type="text/javascript">
//<![CDATA[
var formId = 'product_sendtofriend_form';
var sendfriendForm = new VarienForm(formId, false);
var postUrl = '<?php echo $this->getSendUrl() ?>';
function doAjax() {
if (sendfriendForm.validator.validate()) {
new Ajax.Updater(
{ success:'formSuccess' }, postUrl, {
method:'post',
asynchronous:true,
evalScripts:false,
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess');
},
onLoading:function(request, json){
Element.show('formLoader');
},
parameters: $(formId).serialize(true),
}
);
}
}
new Event.observe(formId, 'submit', function(e){
e.stop();
doAjax();
});
//]]>
</script>
magento-1.9 ajax forms
asked Apr 13 '16 at 15:57
FreejoyFreejoy
444924
bumped to the homepage by Community♦ 8 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
1
1
1
2
I've been working on this code I found somewhere. I use it to submit the Send to Friend form from a popup. It works but I can't get it to display the success message after it submits the form. Right now after it submits the form it just disappears. Maybe someone good at ajax can see what's the matter with it. Thanks
<script type="text/javascript">
//<![CDATA[
var formId = 'product_sendtofriend_form';
var sendfriendForm = new VarienForm(formId, false);
var postUrl = '<?php echo $this->getSendUrl() ?>';
function doAjax() {
if (sendfriendForm.validator.validate()) {
new Ajax.Updater(
{ success:'formSuccess' }, postUrl, {
method:'post',
asynchronous:true,
evalScripts:false,
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess');
},
onLoading:function(request, json){
Element.show('formLoader');
},
parameters: $(formId).serialize(true),
}
);
}
}
new Event.observe(formId, 'submit', function(e){
e.stop();
doAjax();
});
//]]>
</script>
magento-1.9 ajax forms
asked Apr 13 '16 at 15:57
FreejoyFreejoy
444924
I've been working on this code I found somewhere. I use it to submit the Send to Friend form from a popup. It works but I can't get it to display the success message after it submits the form. Right now after it submits the form it just disappears. Maybe someone good at ajax can see what's the matter with it. Thanks
<script type="text/javascript">
//<![CDATA[
var formId = 'product_sendtofriend_form';
var sendfriendForm = new VarienForm(formId, false);
var postUrl = '<?php echo $this->getSendUrl() ?>';
function doAjax() {
if (sendfriendForm.validator.validate()) {
new Ajax.Updater(
{ success:'formSuccess' }, postUrl, {
method:'post',
asynchronous:true,
evalScripts:false,
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess');
},
onLoading:function(request, json){
Element.show('formLoader');
},
parameters: $(formId).serialize(true),
}
);
}
}
new Event.observe(formId, 'submit', function(e){
e.stop();
doAjax();
});
//]]>
</script>
magento-1.9 ajax forms
magento-1.9 ajax forms
asked Apr 13 '16 at 15:57
FreejoyFreejoy
444924
asked Apr 13 '16 at 15:57
FreejoyFreejoy
444924
asked Apr 13 '16 at 15:57
FreejoyFreejoy
444924
asked Apr 13 '16 at 15:57
FreejoyFreejoy
444924
asked Apr 13 '16 at 15:57
FreejoyFreejoy
444924
444924
bumped to the homepage by Community♦ 8 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 8 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
0
Thanks to you pointing me in the right place to look I figured out something that worked. I changed Element.show('formSuccess'); to Element.show('formSuccess2'); and added the div like this <div id="formSuccess2" style="display:none">hello</div>
<script type="text/javascript">
//<![CDATA[
var formId = 'product_sendtofriend_form';
var sendfriendForm = new VarienForm(formId, false);
var postUrl = '<?php echo $this->getSendUrl() ?>';
function doAjax() {
if (sendfriendForm.validator.validate()) {
new Ajax.Updater(
{ success:'formSuccess' }, postUrl, {
method:'post',
asynchronous:true,
evalScripts:false,
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess2');
},
onLoading:function(request, json){
Element.show('formLoader');
},
parameters: $(formId).serialize(true),
}
);
}
}
new Event.observe(formId, 'submit', function(e){
e.stop();
doAjax();
});
//]]>
</script>
answered Apr 13 '16 at 19:21
FreejoyFreejoy
444924
add a comment |
0
Change code
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess');
}
to:
onComplete:function(request, json) {
$('formSuccess').show();
}
answered Apr 13 '16 at 17:58
Prashant ValandaPrashant Valanda
9,55412353
It works that way but not how it needs to work. If I put `<div id="formSuccess" style="display:none">hello</div>' it replaces the div with the whole products details page and tries to put that into the popup. I would like it to show "hello" or go get another html document. I've tried everything I can think of. Thanks
– Freejoy
Apr 13 '16 at 18:48
updated answer check it
– Prashant Valanda
Apr 13 '16 at 20:07
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
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active
oldest
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active
oldest
votes
0
Thanks to you pointing me in the right place to look I figured out something that worked. I changed Element.show('formSuccess'); to Element.show('formSuccess2'); and added the div like this <div id="formSuccess2" style="display:none">hello</div>
<script type="text/javascript">
//<![CDATA[
var formId = 'product_sendtofriend_form';
var sendfriendForm = new VarienForm(formId, false);
var postUrl = '<?php echo $this->getSendUrl() ?>';
function doAjax() {
if (sendfriendForm.validator.validate()) {
new Ajax.Updater(
{ success:'formSuccess' }, postUrl, {
method:'post',
asynchronous:true,
evalScripts:false,
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess2');
},
onLoading:function(request, json){
Element.show('formLoader');
},
parameters: $(formId).serialize(true),
}
);
}
}
new Event.observe(formId, 'submit', function(e){
e.stop();
doAjax();
});
//]]>
</script>
answered Apr 13 '16 at 19:21
FreejoyFreejoy
444924
add a comment |
0
Thanks to you pointing me in the right place to look I figured out something that worked. I changed Element.show('formSuccess'); to Element.show('formSuccess2'); and added the div like this <div id="formSuccess2" style="display:none">hello</div>
<script type="text/javascript">
//<![CDATA[
var formId = 'product_sendtofriend_form';
var sendfriendForm = new VarienForm(formId, false);
var postUrl = '<?php echo $this->getSendUrl() ?>';
function doAjax() {
if (sendfriendForm.validator.validate()) {
new Ajax.Updater(
{ success:'formSuccess' }, postUrl, {
method:'post',
asynchronous:true,
evalScripts:false,
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess2');
},
onLoading:function(request, json){
Element.show('formLoader');
},
parameters: $(formId).serialize(true),
}
);
}
}
new Event.observe(formId, 'submit', function(e){
e.stop();
doAjax();
});
//]]>
</script>
answered Apr 13 '16 at 19:21
FreejoyFreejoy
444924
add a comment |
0
0
0
Thanks to you pointing me in the right place to look I figured out something that worked. I changed Element.show('formSuccess'); to Element.show('formSuccess2'); and added the div like this <div id="formSuccess2" style="display:none">hello</div>
<script type="text/javascript">
//<![CDATA[
var formId = 'product_sendtofriend_form';
var sendfriendForm = new VarienForm(formId, false);
var postUrl = '<?php echo $this->getSendUrl() ?>';
function doAjax() {
if (sendfriendForm.validator.validate()) {
new Ajax.Updater(
{ success:'formSuccess' }, postUrl, {
method:'post',
asynchronous:true,
evalScripts:false,
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess2');
},
onLoading:function(request, json){
Element.show('formLoader');
},
parameters: $(formId).serialize(true),
}
);
}
}
new Event.observe(formId, 'submit', function(e){
e.stop();
doAjax();
});
//]]>
</script>
answered Apr 13 '16 at 19:21
FreejoyFreejoy
444924
Thanks to you pointing me in the right place to look I figured out something that worked. I changed Element.show('formSuccess'); to Element.show('formSuccess2'); and added the div like this <div id="formSuccess2" style="display:none">hello</div>
<script type="text/javascript">
//<![CDATA[
var formId = 'product_sendtofriend_form';
var sendfriendForm = new VarienForm(formId, false);
var postUrl = '<?php echo $this->getSendUrl() ?>';
function doAjax() {
if (sendfriendForm.validator.validate()) {
new Ajax.Updater(
{ success:'formSuccess' }, postUrl, {
method:'post',
asynchronous:true,
evalScripts:false,
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess2');
},
onLoading:function(request, json){
Element.show('formLoader');
},
parameters: $(formId).serialize(true),
}
);
}
}
new Event.observe(formId, 'submit', function(e){
e.stop();
doAjax();
});
//]]>
</script>
answered Apr 13 '16 at 19:21
FreejoyFreejoy
444924
answered Apr 13 '16 at 19:21
FreejoyFreejoy
444924
answered Apr 13 '16 at 19:21
FreejoyFreejoy
444924
answered Apr 13 '16 at 19:21
FreejoyFreejoy
444924
444924
add a comment |
add a comment |
0
Change code
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess');
}
to:
onComplete:function(request, json) {
$('formSuccess').show();
}
answered Apr 13 '16 at 17:58
Prashant ValandaPrashant Valanda
9,55412353
It works that way but not how it needs to work. If I put `<div id="formSuccess" style="display:none">hello</div>' it replaces the div with the whole products details page and tries to put that into the popup. I would like it to show "hello" or go get another html document. I've tried everything I can think of. Thanks
– Freejoy
Apr 13 '16 at 18:48
updated answer check it
– Prashant Valanda
Apr 13 '16 at 20:07
add a comment |
0
Change code
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess');
}
to:
onComplete:function(request, json) {
$('formSuccess').show();
}
answered Apr 13 '16 at 17:58
Prashant ValandaPrashant Valanda
9,55412353
It works that way but not how it needs to work. If I put `<div id="formSuccess" style="display:none">hello</div>' it replaces the div with the whole products details page and tries to put that into the popup. I would like it to show "hello" or go get another html document. I've tried everything I can think of. Thanks
– Freejoy
Apr 13 '16 at 18:48
updated answer check it
– Prashant Valanda
Apr 13 '16 at 20:07
add a comment |
0
0
0
Change code
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess');
}
to:
onComplete:function(request, json) {
$('formSuccess').show();
}
answered Apr 13 '16 at 17:58
Prashant ValandaPrashant Valanda
9,55412353
Change code
onComplete:function(request, json) {
Element.hide(formId);
Element.show('formSuccess');
}
to:
onComplete:function(request, json) {
$('formSuccess').show();
}
answered Apr 13 '16 at 17:58
Prashant ValandaPrashant Valanda
9,55412353
edited Apr 13 '16 at 20:07
answered Apr 13 '16 at 17:58
Prashant ValandaPrashant Valanda
9,55412353
answered Apr 13 '16 at 17:58
Prashant ValandaPrashant Valanda
9,55412353
answered Apr 13 '16 at 17:58
Prashant ValandaPrashant Valanda
9,55412353
9,55412353
It works that way but not how it needs to work. If I put `<div id="formSuccess" style="display:none">hello</div>' it replaces the div with the whole products details page and tries to put that into the popup. I would like it to show "hello" or go get another html document. I've tried everything I can think of. Thanks
– Freejoy
Apr 13 '16 at 18:48
updated answer check it
– Prashant Valanda
Apr 13 '16 at 20:07
add a comment |
It works that way but not how it needs to work. If I put `<div id="formSuccess" style="display:none">hello</div>' it replaces the div with the whole products details page and tries to put that into the popup. I would like it to show "hello" or go get another html document. I've tried everything I can think of. Thanks
– Freejoy
Apr 13 '16 at 18:48
updated answer check it
– Prashant Valanda
Apr 13 '16 at 20:07
It works that way but not how it needs to work. If I put `<div id="formSuccess" style="display:none">hello</div>' it replaces the div with the whole products details page and tries to put that into the popup. I would like it to show "hello" or go get another html document. I've tried everything I can think of. Thanks
– Freejoy
Apr 13 '16 at 18:48
It works that way but not how it needs to work. If I put `<div id="formSuccess" style="display:none">hello</div>' it replaces the div with the whole products details page and tries to put that into the popup. I would like it to show "hello" or go get another html document. I've tried everything I can think of. Thanks
– Freejoy
Apr 13 '16 at 18:48
updated answer check it
– Prashant Valanda
Apr 13 '16 at 20:07
updated answer check it
– Prashant Valanda
Apr 13 '16 at 20:07
add a comment |
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