layui之表单提交到服务器数据库

前几天完成了layui中的图片上传，表单部分就剩下如何提交到数据库了。

说实话，sham又再这个问题上困住了好久，最后好不容易搞定了，赶紧来记录备忘下。

首先是html部分代码

<form class="layui-form" action="">
	<div class="layui-form-item">
		<label class="layui-form-label">文章标题</label>
		<div class="layui-input-block">
			<input type="text" name="title" required  lay-verify="required" placeholder="请输入标题" autocomplete="off" class="layui-input">
		</div>
	</div>
	<div class="layui-form-item">
		<div class="layui-input-block">
			<button class="layui-btn" lay-submit lay-filter="articlesub">立即提交</button>
			<button type="reset" class="layui-btn layui-btn-primary">重置</button>
		</div>
	</div>
</form>

然后是js部分

<script>
	layui.use('form', function(){
		var form = layui.form;
		//监听提交
		form.on('submit(articlesub)', function(data){
			$.ajax({ 
				url:'articlesub.php',   
				method:'post',   
				data:data.field,    
				dataType:'JSON',    
				success:function(res){   
					if(res.code=='0'){   
					layer.msg("提交成功",{icon: 1, time: 3000},function(){location.reload();});
					}        
					else      
					alert(res.msg); 
					},       
					error:function (data) {

					}     
			}); 
			return false;
		});
	});
 </script>

然后是后端PHP文件代码

<?php
	include_once("dbcon.php");
//判断是否有权限
	session_start();
	if($_SESSION['admin']!=="admin"){
	echo "<script>location.href='../adminlogin.php';</script>";
	}
		$title=$_POST['title'];
		
		$sql="INSERT INTO `articles`(`ID`, `title`) VALUES (null, '$title')";
    if($result = $con->query($sql)){
      echo "{".'"code": 0,'.  '"msg": "发不成功"'."}";  //这个用来返回成功数据给layui
     //echo "发不成功";
    }else{
      echo "失败"; $con ->error;
    } 
?>