我有一个选择标签,当它的值发生变化时,我想根据该新值从数据库中获取一些数据,然后将这些新数据放在一个新的 div 上,当用户单击它时,这个 div 包含一个关闭符号,新的 div 将隐藏,我使用 jquery 做到了,除了隐藏新 div 外,一切都运行良好,我的 jQuery 用于关闭新 div 运行良好,但现在不,我不认为我在 jQuery 中做错了,但我没有知道我的错误在哪里,你能帮我吗?
jQuery代码
$(document).ready(function(){
$("#aqIoQuesSelector").on("change",function(){
var IO = $("#aqIoQuesSelector").val();
$("#aqSugInfo").html('<div class="closeSign1"></div>');
$.getJSON("http://localhost/Mar7ba/InformationObject/giveContenfForIO/"+IO+"/TRUE",function(data){
if(data.length>0){
$("#aqSugInfo").css("text-align","right");
for(var i=0;i<data.length;i++){
$("#aqSugInfo").append('<p><span class="aqoneContenSug">'+data[i]+'</span></p>');
}
}else{
$("#aqSugInfo").append('<span class="successMessage">no suggesiont</span>');
}
$("#aqSugInfo").css("display","block");
});
});
});
$(document).ready(function(){
$('.closeSign1').on('click', function() {
$(this).parent().hide();
});
});
$html代码#
<div id="addQuestion1" class="container">
<ul>
<li>
<label class="Paragraph">Question</label>
<p>
<label>Text</label>
<input id="aqQuestionText"type="text" class="longInput1"/>
</p>
<p>
<label>Answer</label>
<input id="aqQuestionAnswer"type="text" class="longInput1"/>
</p>
<p>
<label>Is Existed ?</label>
<input type="button" value="check" class="button1" id="aqQuestionSug"/>
</p>
<div id="aqSugQues" class="SuggestionsContainer">
<div class="closeSign1"></div>
</div>
</li>
<li>
<p><label class="Paragraph">Choices</label></p>
<p>
<label>First Choice</label>
<input type="text" class="longInput1" name="choice1"/>
</p>
<p>
<label>Second Choice</label>
<input type="text" class="longInput1" name="choice2"/>
</p>
<p>
<label>Third Choice</label>
<input type="text" class="longInput1" name="choice3"/>
</p>
</li>
<li id="aqQuestionIoli">
<label class="Paragraph">Question IO</label>
<p>
<label>Concept</label>
<select class="ConceptSelector1"></select>
</p>
<p>
<label>IO</label>
<select id="aqIoQuesSelector"></select>
</p>
<p>
<label>Info</label>
<input type="text" class="longInput1"/>
</p>
<div id="aqSugInfo" class="SuggestionsContainer">
<div class="closeSign1"></div>
</div>
</li>
<li id="aqAnswerIoli">
<label class="Paragraph">Answer IO</label>
<p>
<label>Concept</label>
<select class="ConceptSelector1"></select>
</p>
<p>
<label>IO</label>
<select id="aqIoAnswerSelector"></select>
</p>
</li>
<li>
<label class="Paragraph">Hints</label>
<p>
<label>First Hint</label>
<input type="text" class="longInput1"/>
</p>
<p>
<label>Second Hint</label>
<input type="text" class="longInput1"/>
</p>
<div id="aqSugHints" class="SuggestionsContainer">
<div class="closeSign1"></div>
</div>
</li>
<li>
<label>Type</label>
<select class="TypeSelector"></select>
</li>
<li>
<input type="submit" value="save" class="button1"/>
</li>
</ul>
</div>
注意这里用了三次 close 标志,第一次在 divaqSugQues
有效,但在 divaqSugInfo
和 div都无效aqSugHints
ajax 运行良好,我打印结果
谢谢你的帮助