Mixing
During an $.ajax call some data is lost
I have a form with a multiple select box. When I select more than one option and send the form via an ajax cal to a php file, only the last option will be saved.
Why is that so?
This is the AJAX call:
var daten = $('#formulargesamt').serialize();
console.log(daten);
$.ajax({
url: "/ajax/neukundenanlage/get_formular.php",
data: daten,
method: "POST",
});
In the console.log prompt all data is there (as a string). However in the PHP file it is an array with only the last selected option.
Does anyone have a ideas how I can fix this? Thanks
edit:
the html code:
<div class="container">
<form action="/ajax/neukundenanlage/get_formular.php" method="post"
name="formular" id="formulargesamt">
(...)
<div class="alert fade in" role="alert" id="resultmsg"
style="display:none;">
<button type="button" class="close" data-dismiss="alert" aria-
label="Close"><span aria-hidden="true">×</span></button> <span
id="resulttxt"></span>
</div>
</div>
the PHP code:
<?php
print_r($_POST);
the prompt from console.log:
ed_kid=107&medienverzeichnis=konradin&ed_kbid=&ma_id=0&job_id=0&absendername=&fromaddress=&replyto=&feedbackadresse=&login=agotzens%40mbmedien.de&login=bcioba%40mbmedien.de
jquery html ajax
edited Nov 20 '18 at 12:39
jnuK
1,5341325
asked Nov 20 '18 at 10:00
PeterPeter
135
|
show 3 more comments
I have a form with a multiple select box. When I select more than one option and send the form via an ajax cal to a php file, only the last option will be saved.
Why is that so?
This is the AJAX call:
var daten = $('#formulargesamt').serialize();
console.log(daten);
$.ajax({
url: "/ajax/neukundenanlage/get_formular.php",
data: daten,
method: "POST",
});
In the console.log prompt all data is there (as a string). However in the PHP file it is an array with only the last selected option.
Does anyone have a ideas how I can fix this? Thanks
edit:
the html code:
<div class="container">
<form action="/ajax/neukundenanlage/get_formular.php" method="post"
name="formular" id="formulargesamt">
(...)
<div class="alert fade in" role="alert" id="resultmsg"
style="display:none;">
<button type="button" class="close" data-dismiss="alert" aria-
label="Close"><span aria-hidden="true">×</span></button> <span
id="resulttxt"></span>
</div>
</div>
the PHP code:
<?php
print_r($_POST);
the prompt from console.log:
ed_kid=107&medienverzeichnis=konradin&ed_kbid=&ma_id=0&job_id=0&absendername=&fromaddress=&replyto=&feedbackadresse=&login=agotzens%40mbmedien.de&login=bcioba%40mbmedien.de
jquery html ajax
edited Nov 20 '18 at 12:39
jnuK
1,5341325
asked Nov 20 '18 at 10:00
PeterPeter
135
The important parts we need to see here is the string generated fromserialize()and your PHP.
– Rory McCrossan
Nov 20 '18 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 '18 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 '18 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 '18 at 12:16
If ever you have input fields with the same name, adding an "array-bracket" will store it in an array. Without it, serialize will use the last input value.
– claw68
Nov 20 '18 at 12:48
|
show 3 more comments
I have a form with a multiple select box. When I select more than one option and send the form via an ajax cal to a php file, only the last option will be saved.
Why is that so?
This is the AJAX call:
var daten = $('#formulargesamt').serialize();
console.log(daten);
$.ajax({
url: "/ajax/neukundenanlage/get_formular.php",
data: daten,
method: "POST",
});
In the console.log prompt all data is there (as a string). However in the PHP file it is an array with only the last selected option.
Does anyone have a ideas how I can fix this? Thanks
edit:
the html code:
<div class="container">
<form action="/ajax/neukundenanlage/get_formular.php" method="post"
name="formular" id="formulargesamt">
(...)
<div class="alert fade in" role="alert" id="resultmsg"
style="display:none;">
<button type="button" class="close" data-dismiss="alert" aria-
label="Close"><span aria-hidden="true">×</span></button> <span
id="resulttxt"></span>
</div>
</div>
the PHP code:
<?php
print_r($_POST);
the prompt from console.log:
ed_kid=107&medienverzeichnis=konradin&ed_kbid=&ma_id=0&job_id=0&absendername=&fromaddress=&replyto=&feedbackadresse=&login=agotzens%40mbmedien.de&login=bcioba%40mbmedien.de
jquery html ajax
edited Nov 20 '18 at 12:39
jnuK
1,5341325
asked Nov 20 '18 at 10:00
PeterPeter
135
I have a form with a multiple select box. When I select more than one option and send the form via an ajax cal to a php file, only the last option will be saved.
Why is that so?
This is the AJAX call:
var daten = $('#formulargesamt').serialize();
console.log(daten);
$.ajax({
url: "/ajax/neukundenanlage/get_formular.php",
data: daten,
method: "POST",
});
In the console.log prompt all data is there (as a string). However in the PHP file it is an array with only the last selected option.
Does anyone have a ideas how I can fix this? Thanks
edit:
the html code:
<div class="container">
<form action="/ajax/neukundenanlage/get_formular.php" method="post"
name="formular" id="formulargesamt">
(...)
<div class="alert fade in" role="alert" id="resultmsg"
style="display:none;">
<button type="button" class="close" data-dismiss="alert" aria-
label="Close"><span aria-hidden="true">×</span></button> <span
id="resulttxt"></span>
</div>
</div>
the PHP code:
<?php
print_r($_POST);
the prompt from console.log:
ed_kid=107&medienverzeichnis=konradin&ed_kbid=&ma_id=0&job_id=0&absendername=&fromaddress=&replyto=&feedbackadresse=&login=agotzens%40mbmedien.de&login=bcioba%40mbmedien.de
jquery html ajax
jquery html ajax
edited Nov 20 '18 at 12:39
jnuK
1,5341325
asked Nov 20 '18 at 10:00
PeterPeter
135
edited Nov 20 '18 at 12:39
jnuK
1,5341325
asked Nov 20 '18 at 10:00
PeterPeter
135
edited Nov 20 '18 at 12:39
jnuK
1,5341325
edited Nov 20 '18 at 12:39
jnuK
1,5341325
edited Nov 20 '18 at 12:39
jnuK
1,5341325
1,5341325
asked Nov 20 '18 at 10:00
PeterPeter
135
asked Nov 20 '18 at 10:00
PeterPeter
135
asked Nov 20 '18 at 10:00
PeterPeter
135
135
The important parts we need to see here is the string generated fromserialize()and your PHP.
– Rory McCrossan
Nov 20 '18 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 '18 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 '18 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 '18 at 12:16
If ever you have input fields with the same name, adding an "array-bracket" will store it in an array. Without it, serialize will use the last input value.
– claw68
Nov 20 '18 at 12:48
|
show 3 more comments
The important parts we need to see here is the string generated fromserialize()and your PHP.
– Rory McCrossan
Nov 20 '18 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 '18 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 '18 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 '18 at 12:16
If ever you have input fields with the same name, adding an "array-bracket" will store it in an array. Without it, serialize will use the last input value.
– claw68
Nov 20 '18 at 12:48
The important parts we need to see here is the string generated from
serialize() and your PHP.– Rory McCrossan
Nov 20 '18 at 10:02
The important parts we need to see here is the string generated from
serialize() and your PHP.– Rory McCrossan
Nov 20 '18 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 '18 at 10:05
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 '18 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 '18 at 10:07
please provide your html code
– Bhoomi patel
Nov 20 '18 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 '18 at 12:16
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 '18 at 12:16
If ever you have input fields with the same name, adding an "array-bracket" will store it in an array. Without it, serialize will use the last input value.
– claw68
Nov 20 '18 at 12:48
If ever you have input fields with the same name, adding an "array-bracket" will store it in an array. Without it, serialize will use the last input value.
– claw68
Nov 20 '18 at 12:48
|
show 3 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53390456%2fduring-an-ajax-call-some-data-is-lost%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53390456%2fduring-an-ajax-call-some-data-is-lost%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
The important parts we need to see here is the string generated from
serialize()and your PHP.– Rory McCrossan
Nov 20 '18 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 '18 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 '18 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 '18 at 12:16
If ever you have input fields with the same name, adding an "array-bracket" will store it in an array. Without it, serialize will use the last input value.
– claw68
Nov 20 '18 at 12:48