php 获取单选按钮值并通过ajax发送到php
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Getting radio button value and sending through ajax to php
提问by Paul 'Macca' McGill
I have a poll on my website which displays radio buttons next to each answer. When the user selects an option and submits, im running a a php script via ajax to insert the value or the selected radio button into a table.
我的网站上有一项民意调查，每个答案旁边都显示单选按钮。当用户选择一个选项并提交时，我会通过 ajax 运行 php 脚本以将值或选定的单选按钮插入到表格中。
My Ajax is running but is currently inserting a 0 row each row, so it's not picking up the value from the radio button. Any help would be appreciated.
我的 Ajax 正在运行，但当前每行插入 0 行，因此它没有从单选按钮中获取值。任何帮助，将不胜感激。
HTML:
HTML：
<form id="poll_form" method="post" accept-charset="utf-8">
<input type="radio" name="poll_option" value="1" id="poll_option" /><label for='1'> Arts</label><br />
<input type="radio" name="poll_option" value="2" id="poll_option" /><label for='2'> Film</label><br />
<input type="radio" name="poll_option" value="3" id="poll_option" /><label for='3'> Games</label><br />
<input type="radio" name="poll_option" value="4" id="poll_option" /><label for='4'> Music</label><br />
<input type="radio" name="poll_option" value="5" id="poll_option" /><label for='5'> Sports</label><br />
<input type="radio" name="poll_option" value="6" id="poll_option" /><label for='6'> Television</label><br />
<input type="submit" value="Vote →" id="submit_vote" class="poll_btn"/>
</form>
AJAX:
阿贾克斯：
$("#submit_vote").click(function(e)
{
var option=$('input[type="radio"]:checked').val();
$optionID = "="+optionID;
$.ajax({
type: "POST",
url: "ajax_submit_vote.php",
data: {"optionID" : $optionID}
});
});
PHP: (shortened version)
PHP：（缩短版）
if($_SERVER['REQUEST_METHOD'] == "POST"){
//Get value from posted form
$option = $_POST['poll_option'];
//Insert into db
$insert_vote = "INSERT into poll (userip,categoryid) VALUES ('$ip','$option')";
Thanks in advance!
提前致谢！
回答by StuR
$("#submit_vote").click(function(e){
$.ajax( {
type: "POST",
url: "ajax_submit_vote.php",
data: $('#poll_form').serialize(),
success: function( response ) {}
});
});
You should then have the POST variable "poll_option" accessible in your PHP script.
然后，您应该可以在 PHP 脚本中访问 POST 变量“poll_option”。
回答by MrCode
var option = $('input[type="radio"]:checked').val();
$.ajax({
type: "POST",
url: "ajax_submit_vote.php",
data: { poll_option : option }
});
In the PHP you are reading $_POST['poll_option']therefore you must use poll_optionas the key in your data object. Also, the value is stored in optionnot $optionIDas you were trying to use.
在您正在阅读的 PHP 中，$_POST['poll_option']因此您必须将其poll_option用作数据对象中的键。此外，值存储在option不$optionID作为你试图使用。
The $is a valid character in variable names in Javascript, it doesn't do anything special itself but some coders prefix anything that is a jQuery object with $so they can glance through code and easily see what variables already have the jQuery wrapper.
该$是在Javascript中的变量名有效的字符，它不会做什么特别的事情本身，而是一些程序员任何前缀是一个jQuery对象与$这样他们就可以翻阅代码，并很容易地看到哪些变量已经拥有jQuery的包装。
For example:
例如：
var $option = $('input[type="radio"]:checked'); // $option is the jQuery wrapped HTML element
var myValue = $option.val(); // we know $option already has the jQuery wrapper so no need to use the $(..) syntax.
回答by Naryl
$optionID = "="+optionID;
I don't quite understand what you are trying to do here o.O, in javascript you don't define your variables using $.
我不太明白你在这里尝试做什么 oO，在 javascript 中你没有使用$.
data: { optionID : option}
using it like this should work. You would retrieve it like this in PHP:
像这样使用它应该可以工作。您可以在 PHP 中像这样检索它：
$option_value=$_POST['optionID'];
