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jqGrid id undefined, why?

开发者 https://www.devze.com 2023-03-27 06:46 出处:网络
I have this code: $(grid).jqGrid(\'addRowData\', \'foo\', d); $(grid).jqGrid(\'addRowData\', \'bar\', \'baz\');

I have this code:

$(grid).jqGrid('addRowData', 'foo', d);
$(grid).jqGrid('addRowData', 'bar', 'baz');

As viewed in FireBug, it creates this DOM:

<tr id="undefined" class="ui-widget-content jqgrow ui-row-ltr" tabindex="-1" role="row">
<tr id="bar" class="ui-widget-content jqgrow ui-row-ltr" tabindex="-1" role="row">

The data expressed in 'd' matches the column model and it displays correctly (so I know the grid basically works fine). BUT, the 'id' on the first row is not set. On the other hand, 'baz' is garbage; this creates a blank row, BUT, the 'id' is correctly set to 'bar'.

In both cases I am sure the ID's are unique within my DOM, plus, I have observed that using non-unique ID's does not cause any immediate problems.

SO; I can't see开发者_高级运维 why the id on the first row is not set. Any ideas?


Try this:-

   $(grid).jqGrid('addRowData', 'foo', 'd');
   $(grid).jqGrid('addRowData', 'bar', 'baz');

data should be in quote otherwise you can declare the variable

like

  var d='sample';
  $(grid).jqGrid('addRowData', 'foo', d);
  $(grid).jqGrid('addRowData', 'bar', 'baz');

Suppose d is an array check d and d[0] if d[0] have a value then go with d[0] :-

  $(grid).jqGrid('addRowData', 'foo', d[0]);
             (or)
  $(grid).jqGrid('addRowData', 'foo', d[id]);

it will be work fine...

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