jqgrid修改在表中显示之前从ajax调用返回的数据

Question

我必须显示一些我从服务器收到的数据作为json对象

{"rowndx":"0","rows":"25","rowstotal":"100","rowsdata":[ 
["00","DEVICE001","T0_IHOME","1","***","1","10"], 
["01","DEVICE002","NO_DEVICE","1","***","1","10"],
["02","DEVICE003","NO_DEVICE","0","***","1","10"],
.....

在表格中显示收到的数据之前,我想在必要时进行更改,在数字中添加单位或用单词替换数字(例如0 - > OFF 1-> ON)为此,我在ajax选项中关联"成功"我的编码功能.但是,在这种情况下,"Loading ..."消息始终可见,并且不允许其他操作.我将我的重新编码过程移动到"完整"的ajax选项,这次它似乎工作.但我不明白我的错误是什么,我不知道我的程序是否可行.这是我的表ajax配置

    url      : "devtbl.json",
    mtype    : "POST",
    datatype : "json",
    postData : ......

    ajaxGridOptions: {
      type       : 'post',
      contentType: 'application/json',
      async      : false,
      complete   : DEVparse_serverdata,
      error      : function() { alert('Something bad happened. Stopping');},
    },

    jsonReader : {
      root        : "tablerows",
      page        : "currentpage",
      total       : "totalpages",
      records     : "totalrecords",
      cell        : "",
      id          : "0",
      userdata    : "userdata",
      repeatitems : true
    },

和我的编码功能

    function DEVparse_serverdata(js , textStatus) {

  var jsontablereply = {} ;
  var rowsxpage_int  = parseInt(UB.rowsxpage.DEVtable) ;
  var jsonreply =  jQuery.parseJSON(js.responseText) ;

  jsontablereply.currentpage  = "" + (1 + (parseInt(jsonreply.rowndx) / rowsxpage_int));
  jsontablereply.totalpages   = "" + parseInt((parseInt(jsonreply.rowstotal) + (rowsxpage_int-1)) / rowsxpage_int) ;
  jsontablereply.totalrecords = jsonreply.rowstotal;

  jsontablereply.tablerows = [] ;
  $.each(jsonreply.rowsdata, function(ndx, row) {
     var rowarray = [] ;

     rowarray[0] = row[0] ;
     rowarray[1] = row[1] ;
     rowarray[2] = row[2] ;
     rowarray[3] = row[3] ;
     rowarray[4] = row[4] ;

     switch (row[2]) {
       case "NO_DEVICE":
            rowarray[5] = "***" ;
            break ;

       case "T0_IHOME":
            rowarray[5] = "T=" + row[5] + "°C" ;
            break ;
     }
     jsontablereply.tablerows[ndx] = rowarray ;
  }) ; // each

  jQuery("#DEVtbl")[0].addJSONData(jsontablereply);
}

(我是Jquery的初学者,我的编码风格很差)

Answer 1

您必须有许多可能性来实现您的要求.

在许多情况下,可以使用预定义的格式化程序:"select"或formatter: case的"checkbox"0 ->OFF 1-> ON.

另一种可能性是使用自定义格式化程序和unformatter.自定义格式化程序只是回调,jqGrid将在构建相应列的单元格的HTML片段时使用它.如果您需要共同显示某些文本,格式化程序将如下所示

formatter: function (cellValue) { return $.jgrid.htmlEncode(cellValue); }

自定义格式化程序的优点是,您不仅可以对源文本进行多次修改,还可以根据其他列的信息构建单元格(请参阅rawData下面的参数)

formatter: function (cellValue, options, rawData, action) {
    // options is the the object defined as
    //         {rowId:rowId, colModel:cm, gid:gridId, pos:colpos }
    // rawData contains the representation of the WHOLE row
    //         the most problem of the value is that it is in the same
    //         format as the input data. So if will be array of items
    //         in your case or if could be XML fragment for the row.
    //         Additional problem one will have in case of usage of
    //         loadonce:true. Ath the first load the rawData will be
    //         array of strings and on later could be named object
    //         with the properties corresponds to the column names.
    // action is typically non-important and is 'add' or 'edit'
}

例如,第5列的自定义格式化程序可以是

formatter: function (cellValue, options, rawData, action) {
    switch (rawData[2]) {
    case "NO_DEVICE":
        return "***";
    case "T0_IHOME":
        return "T=" + $.jgrid.htmlEncode(cellValue) + "°C" ;
    default:
        return $.jgrid.htmlEncode(cellValue);
    }
}

另一个选择是使用beforeProcessing回调.它大部分都接近您当前代码的逻辑.在beforeProcessing回调内部,您可以在jqGrid 处理数据之前对服务器上的数据进行任何修改.