php - 我需要向我的动态数据表添加超链接

问题描述

我有一个表，它从我的数据库中提取数据并从数据库表中显示它。我想为狗名添加超链接。我找到的所有文档都显示了如何使用静态数据而不是下面的动态示例来执行此操作：

var responseObj = [
    { "information": "A1", "weblink": "http://www.microsoft.com" },
    { "information": "A2", "weblink": "http://www.yahoo.com" },
    { "information": "A3", "weblink": "http://www.google.com" },
    { "information": "A4", "weblink": "http://www.duckduckgo.com" }
];

$('#example').dataTable({
   "data": responseObj,
   "columns": [
      { "data": "information" }, 
      { 
         "data": "weblink",
         "render": function(data, type, row, meta){
            if(type === 'display'){
                data = '<a href="' + data + '">' + data + '</a>';
            }

            return data;
         }
      } 
   ]
});

我需要能够选择数据并将其转到我的链接。这是我当前的代码。

$(document).ready( function () {
    var dataTable=$('#example').DataTable({

        "processing": true,
        "serverSide": true,
        "ajax": {
            url:"fetch.php",
            type:"post"
        }

    });

} );
</script

在我的 fetch.php 我有这个。

$request = $_REQUEST;
$col = array (
    1 => 'Dog_Name',
    0 => 'Dog_Number',
    2 => 'Breed',
    3 => 'Sex'
    );

    $sql ="SELECT * FROM dog_profiles";

    $query=mysqli_query($conn,$sql);

    $totalData=mysqli_num_rows($query);

    $totalFilter=$totalData;

//Search

$sql =("SELECT * FROM dog_profiles");
if(!empty($request['search']['value'])){
    $sql.=" WHERE (Dog_Number Like '".$request['search']['value']."%' ";
    $sql.=" OR Dog_Name Like '".$request['search']['value']."%' ";
    $sql.=" OR Breed Like '".$request['search']['value']."%' ";
    $sql.=" OR Sex Like '".$request['search']['value']."%' )";
}
$query=mysqli_query($conn,$sql);
$totalData=mysqli_num_rows($query);

//Order
$sql.=" ORDER BY ".$col[$request['order'][0]['column']]."   ".$request['order'][0]['dir']."  LIMIT ".
    $request['start']."  ,".$request['length']."  ";

$query=mysqli_query($conn,$sql);


$data = array();



while($row=mysqli_fetch_array($query)){
    $subdata=array();
    $subdata[]=$row[0]; //dog number
    $subdata[]=$row[1]; //name
    $subdata[]=$row[2]; //breed
    $subdata[]=$row[3]; //sex 
    $data[]=$subdata;
}


$json_data=array(
    "draw"              =>  intval($request['draw']),
    "recordsTotal"      =>  intval($totalData),
    "recordsFiltered"   =>  intval($totalFilter),
    "data"              =>  $data

);
echo json_encode($json_data);

任何见解都会有所帮助我对使用数据表很陌生。

标签： phpajaxdatatablesjquery-datatables-editor

while($row=mysqli_fetch_array($query)){ $subdata=array(); $subdata[]=$row[0]; //dog number $subdata[]="<a href='".$row[1]."'>".$row[1]."</a>"; $subdata[]=$row[2]; //breed $subdata[]=$row[3]; //sex $data[]=$subdata; }

php - 我需要向我的动态数据表添加超链接

问题描述

解决方案

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