在deleteItem之后,Slickgrid重新显示网格

时间:2013-03-04 18:39:38

标签: jquery slickgrid

我有几个允许User删除Item(实际上是行)的SlickGrids。这工作正常但当用户想要取消删除行时,我无法重新显示整个网格。这是删除行的相关代码:

//  Modal Grid2 Remove rows
$('#removeBtnGrid2').click(function() {
    for (var i = 1; i <= 2; i++) {
        dataViewMG2.deleteItem(i);
    }
    invmodalgrid2.invalidate();
    invmodalgrid2.render();
});

这是我用来重新显示网格的代码(这不起作用):

// XXXXXXXXXXXXXXXXXXXXXXX Load selected data into MODAL GRID 2 
$('#addBtnGrid1').click(function() {  /// this acts as a cancel btn
    dataViewMG2.beginUpdate();
    dataViewMG2.setItems(modaldata2);
    dataViewMG2.refresh();
    invmodalgrid2.render();
    dataViewMG2.endUpdate();
}

我做错了什么?感谢

1 个答案:

答案 0 :(得分:1)

首先,当我认为你的意思是({2} addBtnGrid1时,你已经放置了addBtnGrid2,其次,你是在处理服务器端来删除你的行吗?因为这似乎不是处理它的正确方法(虽然我不得不承认我没有尝试过你的代码)。我正在处理删除的方式直接将操作嵌入到接受/取消定义的columns定义中的ID本身,例如:

columns1 = [
            {id:'accnt_name', field:'accnt_name', name: "Account Name", width:90, sortable:true, sorter:sorterStringCompare} },
            {id:"id", name:"", field:"id", width:20, formatter: function (r,c,id,def,datactx){ return "<a href='#' onclick='if (confirm(\"Really Delete?\")) removeClick("+id+","+r+")'><img src='/images/deletex_16.png' border='0' /></a>";} }
        ];

然后删除的函数(使用AJAX在PHP中调用服务器端)如下所示:

function removeClick(databaseId, gridRow) {
        $.post('myajaxfile.php?action=delete', {id:databaseId}, function(ServerResponse) {
            // if database return an error then display it to the user and undo the cell change
            if(ServerResponse  === undefined) {
                alert("Deleting from database failed...");
            }else {
                if( parseInt(ServerResponse.affected_rows) < 1 || ServerResponse.error_msg != "" ) {                             
                    alert("Deleting from database failed...\n\nERROR Returned: "+ServerResponse.error_msg);
                }else {
                    // success, update the datagrid view on screen
                    var item = dataView1.getItem(gridRow);//RowNum is the number of the row
                    dataView1.deleteItem( item.id );//RowID is the actual ID of the row and not the row number
                    grid1.invalidate();
                    grid1.render();          
                }    
            }   
            $('#loading').hide(); // hide the loading logo                                                                        
        }, "json")
        .fail(function(ServerResponse) { 
            alert("Deleting from database failed...\n\nERROR Returned: "+ServerResponse.responseText);  
            $('#loading').hide(); // hide the loading logo             
        });      
    }

因此,如果DELETE在服务器端运行良好,我只刷新(渲染)网格...希望它有所帮助!我不认为向你展示要删除的PHP文件(服务器端),但如果你这样做,那么再问一次......