我的需求是跳转到第三方网站，看官方是写了如何跳转站内路由，不符合我的要求，在csdn发现了一篇文章，我贴一下代码

 

<template><Table border :columns="ReportColumns" :data="ReportData"><template #name="{ row }"><strong>{{ row.name }}</strong></template><template #action="{ row, index }"><!--                                        <a href="http://wwww.baidu.com/" target="_blank">view</a>-->
<!--                                        <Button type="primary" size="small" style="margin-right: 5px"  target="_blank" >View</Button>--><Button type="error" size="small" @click="remove(index)">Delete</Button></template></Table></template>

 data(){return{ReportColumns:[{title: '生成时间',key: 'createTime',width: 220},{title: '生成用户',key: 'creator',width: 200},{title:'报告地址',key:'reportUrl',width: 800,render: (h, params) => {return h('span',{domProps: {innerHTML: "<a href='" + params.row.reportUrl + "' target='_blank'>" + params.row.reportUrl + '</a>'}})}},{title: '操作',slot: 'action',}],

后端接口报文格式

{"ret": true,"code": 0,"msg": "查询成功","content": [{"id": 522,"reportId": 2561,"creator": "李朝阳","createTime": "2023-11-03 12:05:00","reportUrl":"http://192.168.0.1:1111/view/project/5f1e8e97b6e123455555" },{"id": 523,"reportId": 2562,"creator": "白克玲","createTime": "2023-10-03 11:25:00","reportUrl":"https://blog.csdn.net/qq_42772400/article/details/128386185"}]
}

原贴地址

iview table实现通过render实现点击跳转到相应的网址_iview页面跳转-CSDN博客