Javascript JQuery将不会运行以提交表单
我可能错过了一些简单的东西。此页面上的JQuery未运行。这是显而易见的,因为警报不会执行。我只是尝试在没有刷新页面的情况下提交表单。此外,我没有得到控制台错误。提前谢谢Javascript JQuery将不会运行以提交表单,javascript,jquery,Javascript,Jquery,我可能错过了一些简单的东西。此页面上的JQuery未运行。这是显而易见的,因为警报不会执行。我只是尝试在没有刷新页面的情况下提交表单。此外,我没有得到控制台错误。提前谢谢 <!doctype html> <html> <head> <meta charset="utf-8"> <title>Untitled Document</title> <script src="https://ajax.googleapis.co
<!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>
无标题文件
$(文档).ready(函数(e){
$('submitpicks')。关于('submit','#submitpicks',函数(e){
e、 preventDefault();//这将阻止重新加载页面
警报(“未重新加载而提交的表单”);
});
});
var v=0;
函数acceptpick(Pick,removepick){
var userPick=confirm(“您选择了“+thepick+”。是否接受此选择?”);
//var theid=“finalpick”+v;
var removebtn=“btn”+removepik;
//警报(theid);
if(userPick==1){
document.getElementById(“finalpick”).value=removepick;
document.getElementById(removebtn).disabled=true;
document.getElementById(“submitpicks”).submit();
v=v+1;
}
}
无标题文件
$(文档).ready(函数(e){
$('submitpicks')。关于('submit',函数(e){
e、 preventDefault();//这将阻止重新加载页面
警报(“未重新加载而提交的表单”);
});
});
var v=0;
函数acceptpick(Pick,removepick){
var userPick=confirm(“您选择了“+thepick+”。是否接受此选择?”);
//var theid=“finalpick”+v;
var removebtn=“btn”+removepik;
//警报(theid);
if(userPick==1){
document.getElementById(“finalpick”).value=removepick;
document.getElementById(removebtn).disabled=true;
document.getElementById(“submitpicks”).submit();
v=v+1;
}
}
这应该可以解决您的问题。仅供参考,在这里只做JS,考虑到没有出现错误,这对你来说不起作用
无标题文件
$(文档).ready(函数(e){
$(#submitpicks')。提交(函数(e){
e、 预防默认值();
警报(“未重新加载而提交的表单”);
})
});
$('submitpicks')
==>$('submitpicks')
。并使用$('submitpicks')。在('submit',function(){
上绑定表单上的提交事件如果使用jQuery,请使用jQuery。不要执行文档.getElementById(“finalpick”).value=removepick;
,执行$('#finalpick').val(removepick)
@Tushar仍然不工作请尝试注释您的php代码,并将标记置于表单之外。无法检查您的php代码是否包含数据库相关代码,否则其工作正常。更新了对我有效的代码,刚刚删除了php代码,我收到了警告。我明白了,那么我如何使用php?php运行正常。
<!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>