Question

我不知道变量是否适合在这里使用。因此，我在Concrete5中设置了一个块，允许列出几个小时的某些部门。

这是我的示例脚本：

<script type="text/javascript">
    $(function () {
        $("#Saturdayopen").datetimepicker({
"pickDate":false,
"pickTime":true,
"useMinutes":true,
"useSeconds":false,
"useCurrent":false,
"showToday":false,
"useStrict":false,
"sideBySide":false,
"minuteStepping":1,
"minDate":"",
"defaultDate":"",
"icons ":{
"time":"glyphicon glyphicon-time",
"date":"glyphicon glyphicon-calendar",
"up":"glyphicon glyphicon-chevron-up",
"down":"glyphicon glyphicon-chevron-down"},
"language":"en",
"disabledDates":[],
"enabledDates":[],
"daysOfWeekDisabled":[]});
    });
</script>

问题是我需要每次为每周的每一天使用相同的脚本两次。所以我在同一个文件中有14个这样的脚本。有什么办法可以将代码压缩成一个脚本吗？

例如，在这一行：

$("#Saturdayopen").datetimepicker({

有没有办法添加#Saturdayclosed，＃Sundayopen等等？

Answer 1

这实际上不需要对Concrete5做任何事情，但它与Javascript有关。 <击> 变量在这里不是正确的术语，您正在寻找一种向jQuery选择器添加多个元素的方法。 jQuery使用CSS选择器，所以你可以像在CSS中一样：string'em和逗号：

$("#Saturdayopen, #Sundayopen, #etcetera").datetimepicker({...

<击>

对it0ken评论的回复：

这可能是由于您正在使用的.datetimepicker()插件的限制。您可以使用for循环轻松修复：

var days = ["#Saturdayopen", "#Sundayopen", "#etc"]; for(var i = 0; i < days.length; ++i){ $(days[i]).datetimepicker({ //options }); }

您可能还可以在代码示例中删除一些默认参数，这些参数应该将其整理得很多。

Answer 2

您正在使用daysOfWeekDisabled和datetimepicker来控制哪些日期已启用？

如果你的目标是压缩代码，你可以声明一些可以重用的默认值而不是创建14个匿名对象。这可能比批处理更灵活，除非每个var DateTimePicker = { //return a new copy of the defaults Defaults: function() { return { "pickDate": false, "pickTime": true, "useMinutes": true, "useSeconds": false, "useCurrent": false, "showToday": false, "useStrict": false, "sideBySide": false, "minuteStepping": 1, "minDate": "", "defaultDate": "", "icons ": { "time": "glyphicon glyphicon-time", "date": "glyphicon glyphicon-calendar", "up": "glyphicon glyphicon-chevron-up", "down": "glyphicon glyphicon-chevron-down" }, "language": "en", "disabledDates": [], "enabledDates": [], "daysOfWeekDisabled": [] } } } /* JSON.stringify [https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/JSON/stringify] prints the results of a JSON object so we can view the results $.extend [https://api.jquery.com/jquery.extend/] will merge any override settings into the defaults */ //lets set Saturday as open $("#Saturdayopen").html(JSON.stringify($.extend(DateTimePicker.Defaults(), { enabledDates: ["Saturday"] }))); //now lets try setting Saturday as closed $("#Saturdayclosed").html(JSON.stringify($.extend(DateTimePicker.Defaults(), { daysOfWeekDisabled: ["Saturday"] }))); /*daysOfWeekDisabled using this in your code would look like $("#Saturdayopen").datetimepicker($.extend(DateTimePicker.Defaults(), {enabledDates:["Saturday"]})); $("#Saturdayclosed").datetimepicker($.extend(DateTimePicker.Defaults(), {daysOfWeekDisabled:["Saturday"]})); */的属性始终相同。

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>Saturday Open:
  <div id="Saturdayopen"></div>
</div>
<br />
<div>Saturday Closed:
  <div id="Saturdayclosed"></div>
</div>

max-height: 100vh

将多个变量添加到一个函数中

2 个答案:

对it0ken评论的回复：