我想用一个按钮来创建一个表单,要求用户输入并将其发送到php文件而无需刷新,但是在我将表单附加到html文件并试图提交它之后,java函数没有跑。jQuery追加表格
,我用于追加形式的函数:
function reporterror(){
if (document.getElementById("report")){
console.log("Still have report");
}else{
$("<form id='report' action='report.php' method='post'><div ><h2 style='text-align:center;'>Report Error</h2><h3>Description:</h3><textarea rows='10' cols='60' name='report' id='description' placeholder='No more than 500 words' style='border-radius:10px;border:1px solid #000;'></textarea><input type='submit' style='border-radius:5px;'></input><input type='button' onclick='reportclose()' value='Close' style='border-radius:5px;'></input></div></form>").appendTo("#content");
}
};
是didnt运行的函数:
$(document).ready(function() {
$("#report").on("submit",function(event){
event.preventDefault();
if (document.getElementById("description").value==""){
$(".warning").remove();
$("#description").after("<p class='warning' style='color:red'>Description is required</p>");
console.log("yes");
}else{
$.ajax({
type:"POST",
url:"report.php",
data:$("#report").serialize(),
success:function(){
console.log("Great");
document.getElementById("report").innerHTML="<h2>Thank You for helping us to improve!</h2><br><p>Your message have been successfully sented to this awesome website!</p><input type='button' onclick='reportclose()' value='Close' style='border-radius:5px;'></input>";
}
})
}
});
})
这是它看起来像现在:http://cs20p.site11.com/
我想你需要委托onsubmit事件:http://learn.jquery.com/events/event-delegation/但你怎么称'reporterror()'方法? – 2014-12-03 20:54:18