Brandon
Brandon

Reputation: 456

jquery widget shares data when multiple instance are created

I created a a jquery widget and all was working, until I needed another instance. That's when I noticed the two instances were sharing the same data. The plugin is supposed to keep track of which rows are checked in a table so I dont' have to calc them everytime I need to use them. If i insantiate widget on two tables, click one row on one table - both tables will have the same data. This happens with both tables. I've only created a couple of jquery widgets so I'm not sure how this happens, but I have stepped through the code and can see it happening.

It seems like I'm using the widget factory incorrectly. Thanks for any help in advance!

Here is the widget code.

$.widget('ui.selectAndFilter', {
_init: function () {

},
_create: function () {
    var self = this;
    self.options.$mainTable = $(self.element).addClass(this.options.cssWidgetClass);

    self.options.$mainTable.find('tbody tr').bind('click', function (e) {
        self._onClick(e, $(this), false);
        //Need to determine what todo with last param - redrawTables
    });
},
options: {
    cssWidgetClass: 'select-and-filter',
    cssSelectedClass: 'selected',
    $mainTable: {},
    childTables: [],
    canMultiSelect: false,
    clickFinishedCallbacks: [],
    minCount: 1
},
destroy: function () {
    $(this.element).removeClass(this.options.cssWidgetClass);
},
_checkedIds: [],    
checkRow: function ($tr) {
    if ($.isDigit($tr))
        $tr = $(this.options.$mainTable.find('tbody tr:eq(' + $tr + ')'));
    if ($tr.length) {
        var id = $tr.addClass(this.options.cssSelectedClass).find(':checkbox').attr('checked', true).val();
        this._addId(id);
    }
    return this;
},
_uncheckRow: function ($tr) {
    if ($tr.length) {
        var id = $tr.removeClass(this.options.cssSelectedClass).find(':checkbox').attr('checked', false).val();
        return this._removeId(id);
    }
},
uncheckAllRows: function () {
    var self = this;
    this.options.$mainTable.find('tr.' + this.options.cssSelectedClass).each(function () {
        self._uncheckRow($(this));
    });
    return self;
},
_removeId: function (id) {
    this._checkedIds.splice(this._checkedIds.indexOf(id), 1);
    return this._checkedIds;
},
_addId: function (id) {
    if (this._checkedIds.indexOf(id) == -1)
        this._checkedIds.push(id);
    return this._checkedIds;
},
_onClick: function (event, $tr, redrawTables) {
    if ($tr.hasClass(this.options.cssSelectedClass) && this._checkedIds.length > this.options.minCount) {
        this._uncheckRow($tr);
    } else {
        if (!this.options.canMultiSelect) {
            this.uncheckAllRows();
        }
        this.checkRow($tr);
    }

    this.redrawTables();
    this._trigger('click');

},
redrawTables: function () {
    $.each(this.options.childTables, function () {
        this.fnDraw();
    });
},
checkedIds: function () {
    return this._checkedIds;
}


});

And then the code to instantiate them.

tables['schedule'].selectAndFilter({canMultiSelect:selectMulti, childTables: redrawTables});
tables['start'].selectAndFilter({canMultiSelect: selectMulti})
tables['batch'].selectAndFilter({minCount:0});

Upvotes: 4

Views: 2008

Answers (1)

Brandon
Brandon

Reputation: 456

The problem was the _checkedIds array was global to the widget instead of in individual context.

Added this line to the _create method:

this._checkedIds = [];

And removed this line from the widget:

_checkedIds: [],

Upvotes: 6

Related Questions