How to remove a single widget from gridster.js by his dynamically created id

问题

I need to know how can I remove a single gridster.js widget by his dynamically created id with gridster.add_widget. Every new widget created has an own button to remove that single widget, but I can't make it work.

Here is my Code

$(document).ready(function(){

var count = 0;
var newid = count + 1;

    $(document).on("click", "#newGrid", function() {
        var gridster = $(".gridster ul").gridster().data('gridster');
        gridster.add_widget('<li id="block"'+newid'>Hello, now delete me <span id="remove"'+newid'>x</span></li>',2 ,1);
    });

    $(document).on('click', '#remove'+newid, function() {
    var gridster = $(".gridster ul").gridster().data('gridster');
        gridster.remove_widget( '#block'+newid );
    });

});

For adding widgets it works fine, but I can't remove widgets.

回答1:

Actually you don't need ID for widget removing. Please check my script.

    var gridster = $('.gridster ul').gridster().data('gridster');
    $(document).on( "click", ".gridster ul li", function() {
        $(this).addClass("activ");
        gridster.remove_widget($('.activ'));
    });

回答2:

In fact, you just need to wrap your '#block'+newid in a jQuery element, as stated in the function comment:

gridster.remove_widget( $( '#block'+newid ) );

Function comment (in jquery.gridster.js):

/**
 * Remove a widget from the grid.
 *
 * @method remove_widget

* @param {HTMLElement} el The jQuery wrapped HTMLElement you want to remove.

 * @param {Boolean|Function} silent If true, widgets below the removed one
 * will not move up. If a Function is passed it will be used as callback.
 * @param {Function} callback Function executed when the widget is removed.
 * @return {Class} Returns the instance of the Gridster Class.
 */
fn.remove_widget = function(el, silent, callback) {...}

回答3:

newid var should be increased after click add
li tag not had id correctly

please check code below, should work with you

$(document).ready(function(){

var count = 0;
var newid = count + 1;

$(document).on("click", "#newGrid", function() {
    var gridster = $(".gridster ul").gridster().data('gridster');
    gridster.add_widget('<li id="block'+newid+'">Hello, now delete me <span class="remove">x</span></li>',2 ,1);
    newid++;
});

    $(document).on('click', '.remove', function() {
        var gridster = $(".gridster ul").gridster().data('gridster');
        gridster.remove_widget( $(this).parent());
    });
});

回答4:

Since it works with jQuery objects, you don't need to add any classes or ids to identify the item you want to remove. Just create a click event and reference to $(this).

$('body').on( "click", ".gridster ul > li", function() {
     gridster.remove_widget($(this));
});

Or something more useful, where you have a "remove button" inside the li:

$('body').on( "click", ".gridster ul > li .remove", function() {
  gridster.remove_widget($(this).closest('li'));
});

回答5:

this works for me, i use a timestamp as unique id

ADD AN ITEM:

$( "#addMod").click(function(e) {
   gridster1.add_widget(
        '<li class="crItem" id="' + $.now() +  '">DESCRIPTION
         <i class="fi-wrench"></i>
         <i class="fi-x" id="delItem"></i></li>', 1, 1
      );  
 });

REMOVE AN ITEM:

$('#cashRegisterScreen').on('click', '#delItem', function() {          
    var id=$(this).parent().attr("id");             
    gridster.remove_widget($(".gridster ul").find("[id='" + id + "']"));

});

来源：https://stackoverflow.com/questions/20705110/how-to-remove-a-single-widget-from-gridster-js-by-his-dynamically-created-id

标签

jquery

widget

gridster