无法在jqgrid中保持选中行

Question

我有一个函数在jqgrid中向下移动一行，并保持选中该行。向上移动一行时，相同的功能可以正常工作。移动该行，但重新加载时未选中该行。

  .navButtonAdd("#" + gridFieldsPagerID, {
  title: "Move down",
  caption: "",
  buttonicon: "ui-icon-arrow-1-s",
  position: "last",
  onClickButton: function () {
    var currentGridFieldIndex = $('#gridFields-grid').jqGrid('getGridParam', 'selrow');
    if (currentGridFieldIndex === null) {
      return;
    }
    var selectedGridField = $('#gridFields-grid').jqGrid('getLocalRow', currentGridFieldIndex);
    var selectedGrid = $("#grid-grid").jqGrid('getLocalRow', row_id);
    var recordCount = $('#gridFields-grid').jqGrid('getGridParam', 'reccount');

    if (currentGridFieldIndex <= recordCount) {
      MoveGridField(selectedGrid.name, selectedGridField.property, "down");

      $('#gridFields-grid')
        .jqGrid('setGridParam', {
          data: GetGridFieldData(gridName)
        })
        .trigger('reloadGrid');

       // The move up with -1 works as expected. 
      $('#gridFields-grid').jqGrid('setSelection', currentGridFieldIndex + 1);
    }
  }

})

编辑： 这是MoveGridField：

  MoveGridField = function (gridName, propertyName, direction) {
var gridIndex = findIndexByKeyValue(this._grids, "name", gridName);
var fieldList = this._grids[gridIndex].fields;
moveField.call(this, fieldList, propertyName, direction);

}，

和moveField：

moveField = function (fieldList, propertyName, direction) {
var index = findIndexByKeyValue(fieldList, 'property', propertyName);
if (direction == "up") {
  if (index != 0) {
    Swap(fieldList, index, index - 1);
    return;
  }
}
if (direction == "down") {
  if (index != fieldList.length - 1) {
    Swap(fieldList, index, index + 1);
    return;
  }
}

}