我想使用超链接创建我的Site_ID列,当单击每个Site_ID时,它将链接到不同的页面。下面是我的代码,有人可以给我一些建议吗?
var infoTable = $('#resultTable').dataTable({
"processing" : true,
"serverSide" : true,
"scrollX" : true,
"jQueryUI" : true,
"deferRender" : true,
"order" : [[ 0, "asc" ]],
"ajax": "./DataTables-1.10.15/examples/server_side/scripts/server_processing.php",
"pagingType" : "full_numbers",
"oLanguage" : {
"sSearch": "Search all columns:"
},
"aoColumns":[
{
"data":"Site_ID",
"render": function ( data,type,full, meta ) {
return '<a href="'+data+'">'+data+'</a>';
}
},
]
});
答案 0 :(得分:0)
我得到了答案,它运行良好。
var infoTable = $(&#39;#resultTable&#39;)。dataTable({
&#34;处理&#34; :真的,
&#34;服务器侧&#34; :真的,
&#34; scrollX&#34; :真的,
&#34; jQueryUI的&#34; :真的,
&#34; deferRender&#34; :真的,
&#34;为了&#34; :[[0,&#34; asc&#34; ]]
&#34; ajax&#34;:&#34; ./ DataTables-1.10.15 / examples / server_side / scripts / server_processing.php&#34;,
&#34; pagingType&#34; :&#34; full_numbers&#34;,
&#34; oLanguage&#34; :{
&#34; sSearch&#34;:&#34;搜索所有列:&#34;
},
columnDefs:[
{目标:[0],
render:function(data){
返回&#39; +数据+&#39;&#39;
}
}
]
});