Question

我尝试过使用jquery函数和触发器收音机，复选框和日期选择器。样本低于

单选按钮

$("input[name=receivedFluStatus]:radio").change(function () {
	reqdEntered = true;
	//alert('radio selected');
	if (this.id=='receivedFluOutside') {
  	alert(this.id);
		//alert('receivedFluOutside selected');
		//enableLocation();
		locationEntered = false;
    	dateReceivedEntered = false;
    }
    else {
    alert(this.id);
    	//alert('receivedFluOutside not selected');
    	//disableLocation();
    	locationEntered = true;
    	dateReceivedEntered = true;
    }

  
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
    <div style='display: block;'>
                <label>
                    <input type='radio' name='receivedFluStatus' id='receivedFluFrom'>I have received a flu vaccination
                </label>
            </div>
    <div style='display: block;'>
                <label>
                     <input type='radio' name='receivedFluStatus' id='receivedFluOutside'>I have received a flu vaccination
                </label>
            </div>

复选框

$("input[name=acknowledge]:checkbox").change(function() {
  //alert('acknowledge selected');
  var acknowledged = $('#acknowledge').is(':checked');
  alert(acknowledged);

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<div style='display: block;padding-left: 10px;'>
  <label for='acknowledge' style="font-weight: 400;">
<input type='checkbox' name='acknowledge' id='acknowledge' value='acknowledge'> By checking this box I acknowledge
  </label>
</div>

日期选择器

 

   $("#outsideDate").datepicker({
	changeMonth: true,
  changeYear: true,
  showButtonPanel: true,
  dateFormat: "mm/dd/yy",
    onSelect: function(dateText, inst) {
        console.log('date Entered'+dateText);
        dateReceivedEntered = true;
		$('#outsideDateErrorLabel').attr('style','display:none;color: red;font-weight: 200;');
    },
    onClose: function(dateText, inst) {
    outsideDateFocus();
  }
});

function outsideDateFocus() {
	if($('#outsideDate').val()=='MM/DD/YYYY' || $('#outsideDate').val()==''){
		dateReceivedEntered = false;
		$('#outsideDateErrorLabel').attr('style','display:initial;color: red;font-weight: 200;');
	}else{
		dateReceivedEntered = true;
		$('#outsideDateErrorLabel').attr('style','display:none;color: red;font-weight: 200;');		
	}
  alert(dateReceivedEntered);
};

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.ui/1.9.2/jquery-ui.min.js"></script>
<div style="float:left;margin-right:20px;padding-left: 30px;">
  <label id="outsideDateLabel" for="outsideDate" style="display:block;color: lightgrey;font-weight: 400;">Date Received:</label>
  <input id="outsideDate" type="text" name="outsideDate" value="MM/DD/YYYY" style="display:block;border-style: solid;border-width: 1px;border-color: lightgrey;color: lightgrey;" readonly="readonly">
  <label id="outsideDateErrorLabel" for="outsideDate" style="display:block;color: red;font-weight: 200;">This is a required field, please enter date received</label>
</div>

jquery版本是1.9.1，加载类型是onLoad。所有上述工作在jsfiddle中运行得很好，我尝试了所有浏览器和所有设备。虽然上述方法不适用于负载类型头。

但是当我在我的应用程序中添加上述函数时，这是一个在Jquery中使用ui的NodeJS应用程序，它们都不能在ios safari中工作并且间歇性地在chrome中工作

我研究了jsfiddle如何呈现onLoad或body load类型，我也尝试了以下内容

在index.html

上的脚本标记下添加了整个jquery脚本

 //<![CDATA[
   window.onload = function () {
 all the scripts here
  //]]>
 }

但接下来没有调用radio / checkbox，datepickers的处理程序

如果我们添加head标签，处理程序会在chrome中间歇性地工作，但不会在iOS safari中工作

任何人都可以帮我看一下

JQuery函数/触发器无法正常工作

0 个答案: