CODEWITHSUNDEEP
javascriptjqueryjqgrid
Debaprasad Jana
Debaprasad Jana

Reputation: 185

How to set viewrecords property of jqgrid dynamically?

I want to set viewrecords property of jqgrid dynamically. By default this property is set as false. I want to set this to true or false (sometimes to show and at times not to show the recordText at the table footer) depending upon the data that I am populating in the grid dynamicaly. I tried with the following but with no avail-

jQuery("#gridID").jqGrid({viewrecords : true});    
jQuery("#gridID").setGridParam({viewrecords : true});

Upvotes: 2

Views: 4442

Answers (1)

Oleg
Oleg

Reputation: 221997

I recommend you to use viewrecords: true and just hide the div.ui-paging-info inside of loadComplete depend from the current number of records. For example

loadComplete: function (data) {
    if (parseInt(data.records, 10) > 10) {
        $("#pager div.ui-paging-info").show();
    } else {
        $("#pager div.ui-paging-info").hide();
    }
}

The demo demonstrate the approach. If you open on the demo the searching dialog and filter for the client data equal to test you will see only one record and the viewrecords field will be not visible:

enter image description here

Clicking on the "Reload Grid" navigator button will follow to show the viewrecords field back.

Upvotes: 2

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