2015-04-01 121 views

回答

3

試試這個。

function disabledWeekdays(date) { 
    var day = date.getDay(); 
    //0 is Sunday, 1 is Monday, 2 is Tuesday , 3 is Wednesday, 4 is Thursday, 5 is Friday and 6 is Saturday 
    if (day == 1 || day == 2 ||day == 3 ||day == 4 ||day == 5 ||day == 6) { 
     return [false] ; 
    } else { 
     return [true] ; 
    } 
} 

$('#datetimepicker').datetimepicker({ 

beforeShowDay: disabledWeekdays 

});  
+0

非常感謝!僅在今天(超過一年之後),我成功地嘗試了您的代碼。 – ozzmosis 2016-08-20 21:22:10

2

在jquery.datetimepicker.js文件中創建與'xdsoft_weekend'相同的if語句,如下所示。

if (start.getDay() === 0 || start.getDay() === 6 || options.weekends.indexOf(start.dateFormat(options.formatDate)) !== -1) { 
    classes.push('xdsoft_weekend'); 
} 

if (start.getDay() === 0 || options.weekends.indexOf(start.dateFormat(options.formatDate)) !== -1) { 
    classes.push('xdsoft_sunday'); 
} 

然後,創建一個jQuery如下。

$('#datetimepicker').datetimepicker({ 
    onGenerate:function(ct) { 
     jQuery(this).find('.xdsoft_date.xdsoft_sunday') 
      .addClass('xdsoft_disabled'); 
    }, 
    timepicker: false 
}); 
0

很容易:

onGenerate:功能(HU){ jQuery的(本).find( 'xdsoft_date.xdsoft_day_of_week0。 ')addClass(' xdsoft_disabled');。 },

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