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1  General Category / EasyUI for jQuery / column visibility false or true status on: May 08, 2013, 08:00:07 PM
I need to bid_status column visible false. Please give me suggestion. I also gave visible="fasle". But It didnt work

<table id="dg" title="Conops Requests" class="easyui-datagrid" style="width:850px;height:330px"
                   url="/get_conops_request.php"
                   toolbar="#toolbar" pagination="true"
                   rownumbers="true"  singleSelect="true" fitcolumns="false">
                <thead >
                    <tr>
                        <th field = "description" width = "150">Description</th>

                            
                      
                        <th field = "conops_file" width = "150" formatter="formatDetail_csp_conops">CONOPS</th>
                        <th field ="date_created" width = "100" >Date</th>
                        
                        <th field = "bid_count" width = "100" formatter="format_csp_conops_bid_count">No Of Bids</th>
                        <th field ="bid_status" width = "100" visible="false">Bid Status</th>
                        

                    </tr>
                </thead>
            </table>
2  General Category / EasyUI for jQuery / file download root path info on: March 20, 2013, 05:55:45 AM
Hi I used jquery easy ui datagrid. I can download file from the Download SSP link on datagrid. I used local path of the file which is needed to download. It works fine. But when I hosting the file on the server, it doesnt work as local path. I want to give server path to the href value. I wrote the code below which I used. Please give suggestion. I also give the wrong code which I try to give path on server.

function formatDetail_csp(value, row) {

//$_SERVER [ 'DOCUMENT_ROOT' ] . '/core/init.php';
//var href = '$_SERVER [ DOCUMENT_ROOT ] /dashboard/upload/' + row.ssp_file;
   
var href = 'http://localhost:8080/dashboard/upload/' + row.ssp_file;
    return '<a  href="' + href + '">Download SSP</a>';
}
3  General Category / EasyUI for jQuery / Re: datagrid hyperlink event on: March 13, 2013, 03:24:47 AM
Hi stworthy,

Thank you very much. It works fine.

Thanks & Regards
4  General Category / EasyUI for jQuery / datagrid hyperlink event on: March 13, 2013, 01:04:04 AM
Hi I used easyUI datagrid to view the user details. User has 10 attributes. I need to show only 6(columns)  attributes in the datagrid. I want to add new column as "View Details". I need to display the 10 user details in another page as user_details.php when I click the viewdetails hyper link at any row. I retrieved data from get_details.php page.

I attached code which I used in the project below

//datagrid page
<table id="dg" title=" Requests" class="easyui-datagrid" style="width:850px;height:330px"
         url="/get_details.php"
         toolbar="#toolbar" pagination="true"
         rownumbers="true"  singleSelect="true">
      <thead >
         <tr>

                            <th field = "name" width = "120">Name</th>
                            <th field = "address" width = "120">Address</th>
                            <th field = "phone" width = "120">Phone</th>
                            <th field = "expected_time_period" width = "120">Expect Time</th>
                            <th field ='' " width = "120">View Details</th>
                           
         </tr>
      </thead>
   </table>
//In the datagrid I attached View Details column. But I dont know how give the hyper link event to the column.All the columns get values from get_details.php page except this column. I need to send user Id throw the hyperlink to user_details.php. So please suggestion to the line "<th field ='' " width = "120">View Details</th>"

//get_details.php
<?php
   $page = isset($_POST['page']) ? intval($_POST['page']) : 1;
   $rows = isset($_POST['rows']) ? intval($_POST['rows']) : 10;
   $offset = ($page-1)*$rows;
   $result = array();
       

       
        $rs = mysql_query("select count(*) from user where  status = 1");
   $row = mysql_fetch_row($rs);
   $result["total"] = $row[0];
   $rs = mysql_query("select * from user where  status = 1 limit $offset,$rows");
   
   $items = array();
   while($row = mysql_fetch_object($rs)){
      array_push($items, $row);
   }
   $result["rows"] = $items;
//print_r($Items);
   echo json_encode($result);

?> 
 I am new to php coding. I also attached screen shot of the datagrid. Please check it.

Thanks & Regards
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