JDev JDev - 4 years ago 193
jQuery Question

How to set js-GRID default inserting, instead of filtering

I have a js-GRID which when loading is default set to filtering in the upper right section. look at here please

How will I set it to be default in inserting? I mean how will the filtering being displayed to the top instead of the + inserting button?.

This is my code:

<!DOCTYPE html>
<html>
<head>
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<title>jsGrid - Custom View Scenario</title>
<link rel="stylesheet" type="text/css" href="demos.css" />
<link href='http://fonts.googleapis.com/css?family=Open+Sans:300,600,400' rel='stylesheet' type='text/css'>

<link rel="stylesheet" type="text/css" href="../css/jsgrid.css" />
<link rel="stylesheet" type="text/css" href="../css/theme.css" />

<script src="../external/jquery/jquery-1.8.3.js"></script>
<script src="db.js"></script>

<script src="../src/jsgrid.core.js"></script>
<script src="../src/jsgrid.load-indicator.js"></script>
<script src="../src/jsgrid.load-strategies.js"></script>
<script src="../src/jsgrid.sort-strategies.js"></script>
<script src="../src/jsgrid.field.js"></script>
<script src="../src/fields/jsgrid.field.text.js"></script>
<script src="../src/fields/jsgrid.field.number.js"></script>
<script src="../src/fields/jsgrid.field.select.js"></script>
<script src="../src/fields/jsgrid.field.checkbox.js"></script>
<script src="../src/fields/jsgrid.field.control.js"></script>

<style>
.config-panel {
padding: 10px;
margin: 10px 0;
background: #fcfcfc;
border: 1px solid #e9e9e9;
display: inline-block;
}

.config-panel label {
margin-right: 10px;
}
</style>
</head>
<body>
<h1>Custom View</h1>
<div class="config-panel">
<label><input id="heading" type="checkbox" checked /> Heading</label>
<label><input id="filtering" type="checkbox" checked /> Filtering</label>
<label><input id="inserting" type="checkbox" /> Inserting</label>
<label><input id="editing" type="checkbox" checked /> Editing</label>
<label><input id="paging" type="checkbox" checked /> Paging</label>
<label><input id="sorting" type="checkbox" checked /> Sorting</label>
<label><input id="selecting" type="checkbox" checked /> Selecting</label>
</div>

<div id="jsGrid"></div>

<script>
$(function() {

$("#jsGrid").jsGrid({
height: "70%",
width: "100%",
filtering: true,
editing: true,
sorting: true,
paging: true,
autoload: true,
pageSize: 15,
pageButtonCount: 5,
controller: db,
fields: [
{ name: "Name", type: "text", width: 150 },
{ name: "Age", type: "number", width: 50 },
{ name: "Address", type: "text", width: 200 },
{ name: "Country", type: "select", items: db.countries, valueField: "Id", textField: "Name" },
{ name: "Married", type: "checkbox", title: "Is Married", sorting: false },
{ type: "control", modeSwitchButton: false, editButton: false }
]
});

$(".config-panel input[type=checkbox]").on("click", function() {
var $cb = $(this);
$("#jsGrid").jsGrid("option", $cb.attr("id"), $cb.is(":checked"));
});
});
</script>
</body>
</html>

Answer Source

The easiest solution (a bit hacky) would be just click on mode button after grid initialization:

// initialization is here

$("#jsGrid").find(".jsgrid-mode-button").click();

Showing filter by default is the behavior of control field. A better way would be creating a custom field with the expected behavior inheriting from control field.

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