Question

我有3个选择框，根据每个框中的选择加载动态。在我尝试使用jqtransform插件设置页面样式之前，这些工作正常。我试图弄清楚这一点，并搜索了这个网站和网页，但没有找到解决方案，为什么如果我使用jqtransform插件，选择菜单根本不会填充。任何帮助将不胜感激。

原始工作代码：

    `<script language="JavaScript" src="myminiAJAX.js"></script>
<script language="JavaScript" src="functions.js"></script>

<script language="JavaScript">
function init() {
    doAjax('man_list.php', '', 'populateComp', 'post', '1');

}

function showOutput(){
alert("This Is Your Model Id: "+getValue('manufacturer'));
}


    $(function(){
        $('form').jqTransform({imgPath:'jqtransformplugin/img/'});
    });

</script>
<style>
#loading{
background:url('loader64.gif') no-repeat;
height: 63px;
}
</style>
    </head>

<body onLoad="init();">
<p>
<form method="post" action="<?php echo $PHP_SELF;?>">
<b>System Type:</b>&nbsp;<select name="manufacturer" id="manufacturer"       onChange="resetValues();doAjax('type_list.php', 'man='+getValue('manufacturer'), 'populateType', 'post', '1')">
<option value="">Please select:</option></select>&nbsp;

<br /><b>Type of Equipment: </b>&nbsp;<select name="printertype" id="printertype" disabled="disabled" onChange="doAjax('model_list.php', 'man='+getValue('manufacturer')+'&typ='+getValue('printertype'), 'populateModel', 'post', '1')">
<option value="">Please select:</option></select>&nbsp;

<br /><b>Equipment to be depoted out:</b>&nbsp;<select name="printermodel" id="printermodel" disabled="disabled" onChange="showOutput();">
<option value="">Please select:</option></select>

<br /><br /><input type="submit" name="action" value="Submit" />

代码添加Jqtransform插件后：

    <link rel="stylesheet" href="jqtransformplugin/jqtransform.css" type="text/css" media="all" />
<link rel="stylesheet" href="demo.css" type="text/css" media="all" />
    <script language="JavaScript" src="myminiAJAX.js"></script>
<script language="JavaScript" src="functions.js"></script>
<script type="text/javascript" src="requiered/jquery.js" ></script>
<script type="text/javascript" src="jqtransformplugin/jquery.jqtransform.js" >        </script>

删除了缩短此帖子的功能，因为它们没有从原始代码更改。

<body onLoad="init();">
<p>
<form method="post" action="<?php echo $PHP_SELF;?>">
<div class="rowElem"><b>System Type:</b>&nbsp;<select name="manufacturer" id="manufacturer" onChange="resetValues();doAjax('type_list.php', 'man='+getValue('manufacturer'), 'populateType', 'post', '1')">
<option value="">Please select:</option></select>&nbsp;
</div>
<div class="rowElem"><br /><b>Type of Equipment: </b>&nbsp;<select name="printertype" id="printertype" disabled="disabled" onChange="doAjax('model_list.php', 'man='+getValue('manufacturer')+'&typ='+getValue('printertype'), 'populateModel', 'post', '1')">
<option value="">Please select:</option></select>&nbsp;
</div>
<div class="rowElem"><br /><b>Equipment to be depoted out:</b>&nbsp;<select name="printermodel" id="printermodel" disabled="disabled" onChange="showOutput();">
 <option value="">Please select:</option></select>
</div>
<div class="rowElem"><br /><br /><input type="submit" name="action" value="Submit" />        </div>
</p>
<div id="loading" style="display: none;"></div>
<div id="output"></div>

无法将jqtransform插件与Cascade下拉集成

0 个答案: