Pau*_*ill 5 html php ajax jquery radio-button
我的网站上有一个民意调查,每个答案旁边都会显示单选按钮.当用户选择一个选项并提交时,我通过ajax运行一个php脚本,将值或选中的单选按钮插入表中.
我的Ajax正在运行但是当前每行都插入一行0,所以它没有从单选按钮中获取值.任何帮助,将不胜感激.
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>
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AJAX:
$("#submit_vote").click(function(e)
{
var option=$('input[type="radio"]:checked').val();
$optionID = "="+optionID;
$.ajax({
type: "POST",
url: "ajax_submit_vote.php",
data: {"optionID" : $optionID}
});
});
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PHP :(缩短版)
if($_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')";
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提前致谢!
$("#submit_vote").click(function(e){
$.ajax( {
type: "POST",
url: "ajax_submit_vote.php",
data: $('#poll_form').serialize(),
success: function( response ) {}
});
});
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然后,您应该在PHP脚本中访问POST变量"poll_option".
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