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 at 12:39
jnuK
1,5161325
asked Nov 20 at 10:00
Peter
104
|
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 at 12:39
jnuK
1,5161325
asked Nov 20 at 10:00
Peter
104
The important parts we need to see here is the string generated fromserialize()and your PHP.
– Rory McCrossan
Nov 20 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 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 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 at 12:39
jnuK
1,5161325
asked Nov 20 at 10:00
Peter
104
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 at 12:39
jnuK
1,5161325
asked Nov 20 at 10:00
Peter
104
edited Nov 20 at 12:39
jnuK
1,5161325
asked Nov 20 at 10:00
Peter
104
edited Nov 20 at 12:39
jnuK
1,5161325
edited Nov 20 at 12:39
jnuK
1,5161325
edited Nov 20 at 12:39
jnuK
1,5161325
1,5161325
asked Nov 20 at 10:00
Peter
104
asked Nov 20 at 10:00
Peter
104
asked Nov 20 at 10:00
Peter
104
104
The important parts we need to see here is the string generated fromserialize()and your PHP.
– Rory McCrossan
Nov 20 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 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 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 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 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 at 12:48
The important parts we need to see here is the string generated from
serialize() and your PHP.– Rory McCrossan
Nov 20 at 10:02
The important parts we need to see here is the string generated from
serialize() and your PHP.– Rory McCrossan
Nov 20 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 at 10:05
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 at 10:07
please provide your html code
– Bhoomi patel
Nov 20 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 at 12:16
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 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 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 at 12:48
|
show 3 more comments
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The important parts we need to see here is the string generated from
serialize()and your PHP.– Rory McCrossan
Nov 20 at 10:02
How is your html? The name attribute should be something like: name="date" instead of only name="date"
– Eva
Nov 20 at 10:05
please provide your html code
– Bhoomi patel
Nov 20 at 10:07
HI folks, thanks for the replys. @Eva Why should I add an array-bracket to the name?
– Peter
Nov 20 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 at 12:48