使用easyUI添加查询功能在datagrid

@author  YHC

在这个示例中我们将向你展示如何从数据库得到数据显示他们到datagrid,然后演示如何根据用户输入的搜索关键词搜寻显示结果.

查看 Demo

创建 DataGrid

创建 datagrid和分页功能然后添加工具栏到其中.

<table id="tt" class="easyui-datagrid" style="width:600px;height:250px"  
        url="datagrid24_getdata.php" toolbar="#tb"  
        title="Load Data" iconCls="icon-save"  
        rownumbers="true" pagination="true">  
    <thead>  
        <tr>  
            <th field="itemid" width="80">Item ID</th>  
            <th field="productid" width="80">Product ID</th>  
            <th field="listprice" width="80" align="right">List Price</th>  
            <th field="unitcost" width="80" align="right">Unit Cost</th>  
            <th field="attr1" width="150">Attribute</th>  
            <th field="status" width="60" align="center">Stauts</th>  
        </tr>  
    </thead>  
</table>  

工具栏定义如下:

<div id="tb" style="padding:3px">  
    <span>Item ID:</span>  
    <input id="itemid" style="line-height:26px;border:1px solid #ccc">  
    <span>Product ID:</span>  
    <input id="productid" style="line-height:26px;border:1px solid #ccc">  
    <a href="#" class="easyui-linkbutton" plain="true" onclick="doSearch()">Search</a>  
</div>  

当用户输入查询值和按下查询按钮' doSearch'函数将被调用.

function doSearch(){  
    $('#tt').datagrid('load',{  
        itemid: $('#itemid').val(),  
        productid: $('#productid').val()  
    });  
}  

以上代码我们调用了 load方法去加载新的datagrid数据,我们需要传入itemid和productid参数到服务器:

服务器端代码

include 'conn.php';  
  
$page = isset($_POST['page']) ? intval($_POST['page']) : 1;  
$rows = isset($_POST['rows']) ? intval($_POST['rows']) : 10;  
$itemid = isset($_POST['itemid']) ? mysql_real_escape_string($_POST['itemid']) : '';  
$productid = isset($_POST['productid']) ? mysql_real_escape_string($_POST['productid']) : '';  
  
$offset = ($page-1)*$rows;  
  
$result = array();  
  
$where = "itemid like '$itemid%' and productid like '$productid%'";  
$rs = mysql_query("select count(*) from item where " . $where);  
$row = mysql_fetch_row($rs);  
$result["total"] = $row[0];  
  
$rs = mysql_query("select * from item where " . $where . " limit $offset,$rows");  
  
$items = array();  
while($row = mysql_fetch_object($rs)){  
    array_push($items, $row);  
}  
$result["rows"] = $items;  
  
echo json_encode($result);  

下载 EasyUI 示例代码:

easyui-datagrid-demo.zip