php ajax表单动态提交 [英] php ajax form submit dynamically
问题描述
我不知道有关websocket进行聊天的信息，我想使用php ajax进行表单提交以进行聊天 我想在不重新加载的情况下动态地求和表格，但是它会重新加载(我不想要). 我创建了一个聊天室，其中php将信息发送到xml并显示所有xml信息，并且当用户在下面提交表单时
i have no knowledge about websocket for chat i want to use php ajax for form submit for chat i want to sumbit form dynamically without reload but it gets reload ( which i dont want ). i have created a chat in which php sends information to xml as and displays all xml information, and when user submits the form below
<form action="action.php" method="post" id="formpost">
<input type="text" id="input" value="php echo">
<input type="submit" value="send">
</form>
它重新加载以显示此php
it reloads to display this php
<div class="msg"><?php print $message->getName() ." : " . $chat->message . ""; ?></div>
其他信息:当我删除聊天室$chat->message .时，没有显示任何消息，因为php循环中的唯一名称显示在上面的<div class="msg">
Additional info : when i remove the chat $chat->message . no msgs display because the php loop only name shows in <div class="msg"> above
我已经尝试过使用javascript动态提交表单，但是当我单击该按钮时，我自己的html <html><body>..</html>会发出警报，并且当我手动重新加载msg显示的页面时会显示
i have tried this to submit form dynamically by javascript but when i click the button a alert comes with my own html <html><body>..</html>, and when i reload the page manually msg shows
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script>
$("#formpost").on('submit', function(event){
event.preventDefault();
var form = $(this);
var url = form.attr('action');
$.ajax({
type: "POST",
url: "club.php",
data: form.serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
});
</script>
推荐答案
$(function() {
$("#formpost").submit(function(event) {
event.preventDefault();
$.ajax({
type: "POST",
url: "club.php",
data: $(this).serialize(),
}).done(function(data) {
var msg = $(data).find('#msg').text();
alert(msg);
});
});
});
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