[英]Sending the next element of the clicked button into jQuery UI dialog box
尝试为元素设置动画后,使用此脚本单击按钮时:
$(document).ready(function()
{
$("#oone, #otwo, #othree, #ofour, #ofive, #osix, #oseven, #oeight, #onine, #oten, #otwelve, #otwenty, #othirteen, #ofourteen, #ofifteen, #osixteen, #loone, #lotwo, #lothree, #lofour, #lofive, #losix, #loseven, #loeight, #lonine, #laaten, #lotwelve, #lotwenty, #lothirteen, #lofourteen, #lofifteen, #losixteen").click(function()
{
// $(this).css('border', "solid 2px red");
// var divs = $("#div_one, #div_two, #div_three, #div_four, #div_five, #div_six, #div_seven, #div_eight, #div_nine, #dive_ten, #div_eleven, #div_twelve, #div_thirteen, #div_fourteen, #div_fifteen, #div_sixteen, #div_lone, #div_ltwo, #div_lthree, #div_lfour, #div_lfive, #div_lsix, #div_lseven, #div_leight, #div_lnine, #dive_lten, #div_leleven, #div_ltwelve, #div_lthirteen, #div_lfourteen, #div_lfifteen, #div_lsixteen");
// $(divs).siblings().slideToggle();
$(this).next().slideToggle();
$(this).siblings().slideToggle();
});
});
我有一些不需要的动画效果。 所以我决定,为什么不使用jQuery UI
插件将结果发送到对话框bix中,而不是对单击按钮的下一个元素进行动画处理。 所以我尝试了以下方法:
<script src="../include/jquery-1.12.1.min.js"></script>
<script src="../include/jquery-ui.min.js"></script>
<script src="../include/bootstrap.min.js" type="text/javascript"></script>
$(document).ready(function() {
$("#oone, #otwo, #othree, #ofour, #ofive, #osix, #oseven, #oeight, #onine, #oten, #otwelve, #otwenty, #othirteen, #ofourteen, #ofifteen, #osixteen, #loone, #lotwo, #lothree, #lofour, #lofive, #losix, #loseven, #loeight, #lonine, #laaten, #lotwelve, #lotwenty, #lothirteen, #lofourteen, #lofifteen, #losixteen").click(function() {
$(this).next().dialog({
autoOpen: false,
hide: "puff",
show: "slide",
width: 800,
modal: true
});
//$(this).dialog("open");
});
});
这是仅前两个按钮的html代码,因为cod太长:
<div class="form-group col-md-12">
<input class="img1" type="image" style="width:60px;height:60px" src="../images/molar_left_t.png" id="oone" name="one" alt="button" />
<div id="div_one" class="collapse">3rd Molar:
<?php echo $resTeeth['one'] ?>
</div>
<input class="img1" type="image" style="width:60px;height:60px" src="../images/molar_left_t.png" id="otwo" name="two" alt="button" />
<div id="div_two" class="collapse">
<?php echo $resTeeth['two'] ?>
</div>
所以我有这个错误:
jquery-1.12.1.min.js:2 Uncaught Error: cannot call methods on dialog prior to initialization; attempted to call method 'open'
如何解决此错误,是否可以使用jQuery UI
将next()
div元素发送到dialog
中,还是应该为每个div指定ID,并为每个div创建一个对话脚本?
我假设您希望对话框包含下一个(以下)div的内容。
这样可以做到:
$(document).ready(function() {
var dialog = '<div class="mydialog" title="Basic dialog"><p class="dialogcontent">me</p></div>';
var newDiv = $(dialog);
newDiv.dialog({
autoOpen: false,
hide: "puff",
show: "slide",
modal: true,
buttons: {
Ok: function() {
$(this).dialog("close");
}
}
});
$(".form-group").on('click', ".img1", function() {
var me = $(this);
newDiv.find('.dialogcontent').html(me.next('.collapse').html())
newDiv.dialog("open");
});
});
实际示例: https : //jsfiddle.net/89pyhsuj/
您的问题是,您正在从下一个div创建对话框,但是您试图在CURRENT div上打开对话框。 您可以解决此问题:
$(document).ready(function() {
$("#oone, #otwo, #othree, #ofour, #ofive, #osix, #oseven, #oeight, #onine, #oten, #otwelve, #otwenty, #othirteen, #ofourteen, #ofifteen, #osixteen, #loone, #lotwo, #lothree, #lofour, #lofive, #losix, #loseven, #loeight, #lonine, #laaten, #lotwelve, #lotwenty, #lothirteen, #lofourteen, #lofifteen, #losixteen").click(function() {
var dialog = $(this).next();
dialog.dialog({
autoOpen: false,
hide: "puff",
show: "slide",
width: 800,
modal: true
});
dialog.dialog("open");
});
});
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