jQuery带参跳转新页面，新页面获取url多个参数的办法

淡淡的烟草味﹌ 2022-10-12 01:25 61阅读 0赞

需求：  
a页面有多个参数  
跳转到b页面  
在b页面获取a页面带过来的参数  
a.html

<!DOCTYPE HTML>
    <html>
        <head>
            <title></title>
          
            <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
            <link rel="stylesheet" href="https://cdn.bootcss.com/twitter-bootstrap/4.3.1/css/bootstrap.min.css">
            <script src="https://cdn.bootcss.com/jquery/3.4.1/jquery.min.js"></script>
            <script src="https://cdn.bootcss.com/twitter-bootstrap/4.3.1/js/bootstrap.min.js"></script>
            <style>
              
        </style>
        </head>
        <body>
           <button  type="button" class="a">跳转到b界面</button>
        </body>
     <script>
    	 //底部分类复选框取消和选中触发事件
    	         $(".a").click(function() {
    	             window.location.href = "b.html?username=super&workNumber=M202105241431&type=2&beginTime=2021-05-24 14:31:00&endTime=2021-05-24 14:31:00";
    	         });
    
     </script>
    </html>

b.html

<!DOCTYPE HTML>
    <html>
    	<head>
    		<title></title>
    
    		<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    		<link rel="stylesheet" href="https://cdn.bootcss.com/twitter-bootstrap/4.3.1/css/bootstrap.min.css">
    		<script src="https://cdn.bootcss.com/jquery/3.4.1/jquery.min.js"></script>
    		<script src="https://cdn.bootcss.com/twitter-bootstrap/4.3.1/js/bootstrap.min.js"></script>
    		<style>
    
    		</style>
    	</head>
    	<body>
    
    	</body>
    
    	<script>
    		//获取对方传过来的参数
    		function jqueryUrl(url) {
    			debugger
    			//将地址从"？"位置分割成两部分
    			var arr = url.split('?');
    			//取地址右边参数部分从"&"位置继续分割，成为单独参数列表
    			var params = arr[1].split('&'); //得到[a=1,b=2,c=3]
    			//定义一个空对象
    			var obj = {};
    			for (var i = 0; i < params.length; i++) {
    				var param = params[i].split('='); //得到[a,1]、[b,2]、[c,3]
    				obj[param[0]] = param[1]; //为对象赋值
    			}
    			return obj;
    		}
    		// url参数列表		
    
    		var urlParams = jqueryUrl(location.href)
    		console.log(urlParams)
    	</script>
    
    </html>

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[18821e57d3939b70a09ebbfd7c2fac60.png]: /images/20221005/f09435d662e84f9b91b6450292f80597.png
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