JQuery 不会运行提交表单
[英]JQuery will not run to submit form
问题描述
我可能错过了一些简单的东西。 此页面上的 JQuery 未运行。 这很明显，因为警报不会执行。 我只是想在不刷新页面的情况下提交表单。 另外，我没有收到控制台错误。 提前致谢。
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script>
$(document).ready(function(e) {
$('submitpicks').on('submit','#submitpicks',function(e){
e.preventDefault(); //this will prevent reloading page
alert('Form submitted Without Reloading');
});
});
</script>
</head>
<body>
<form name="submitpicks" id="submitpicks" action="" method="post">
<script type="">
var v=0;
function acceptpick(thepick,removepick){
var userPick = confirm("You picked " + thepick + ". Accept this pick?");
//var theid = "finalpick" + v;
var removebtn = "btn" + removepick;
//alert(theid);
if(userPick==1){
document.getElementById("finalpick").value=removepick;
document.getElementById(removebtn).disabled = true;
document.getElementById("submitpicks").submit();
v=v+1;
}
}
</script>
<?php
include "Connections/myconn.php";
//$setid = $_SESSION["gbsid"];
$setid = 11;
$setqry = "Select * from grabBagParticipants where gbsid = $setid order by rand()";
$setresult = mysqli_query($conn, $setqry);
$u=0;
if(mysqli_num_rows($setresult)>0){
while($setrow = mysqli_fetch_array($setresult)){
//shuffle($setrow);
echo '<input type="button" name="' . $setrow["gbpid"] . '" id="btn' . $setrow["gbpid"] . '" value="' . $u . '" onClick=\'acceptpick("' . $setrow["gbpname"] . '", ' . $setrow["gbpid"] . ');\' /><br />';
$u=$u+1;
}
}
?>
<input type="text" name="finalpick" id="finalpick" />
<input type="submit" value="Save" />
</form>
<div id="results"> </div>
</body>
</html>
2 个解决方案
解决方案1
0 2015-11-10 03:28:28
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script>
$(document).ready(function(e) {
$('#submitpicks').on('submit',function(e){
e.preventDefault(); //this will prevent reloading page
alert('Form submitted Without Reloading');
});
});
</script>
</head>
<body>
<form name="submitpicks" id="submitpicks" action="test.php" method="post">
<script type="">
var v=0;
function acceptpick(thepick,removepick){
var userPick = confirm("You picked " + thepick + ". Accept this pick?");
//var theid = "finalpick" + v;
var removebtn = "btn" + removepick;
//alert(theid);
if(userPick==1){
document.getElementById("finalpick").value=removepick;
document.getElementById(removebtn).disabled = true;
document.getElementById("submitpicks").submit();
v=v+1;
}
}
</script>
<input type="text" name="finalpick" id="finalpick" />
<input type="submit" value="Save" />
</form>
<div id="results"> </div>
</body>
</html>
解决方案2
0 2015-11-10 04:47:35
这应该可以解决您的问题。 仅供参考，考虑到没有出现错误，只是在这里做 JS 对你不起作用。
<html> <head> <meta charset="utf-8"> <title>Untitled Document</title> <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script> <script> $(document).ready(function(e) { $('#submitpicks').submit(function(e) { e.preventDefault(); alert('Form Submitted without reloading.'); }) }); </script> </head> <body> <form name="submitpicks" id="submitpicks" action="" method="post"> <input type="text" name="finalpick" id="finalpick" /> <input type="submit" value="Save" /> </form> </body>
从jQuery运行PHP脚本，然后提交表单
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.