我试图在一个按钮上单击显示三个按钮.在按钮单击之前,所有三个按钮都被隐藏.我设置了隐藏属性,我也在点击按钮时调用了一个函数.问题是,当我加载页面时,所有按钮都是可见的.在我添加css之前,它工作得很好.我不明白问题出在哪里.这是HTML代码:
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<link rel="STYLESHEET" type="text/css" href="dba_style/buttons.css" />
<title>Untitled Document</title>
</head>
<script type="text/javascript">
function change(){
document.getElementById("save").hidden = "";
document.getElementById("change").hidden = "";
document.getElementById("cancel").hidden = "";
}
</script>
<body>
<form name="form1" method="post" action="">
<div class="buttons">
<button class="regular" name="edit" id="edit" onclick="change();">
<img src="dba_images/textfield_key.png" alt=""/>
Edit
</button>
<button type="submit" class="positive" name="save" id="save" hidden="hidden">
<img src="dba_images/apply2.png" alt=""/>
Save
</button>
<button class="regular" name="change" hidden="hidden" id="change">
<img src="dba_images/textfield_key.png" alt=""/>
change
</button>
<button class="negative" name="cancel" id="cancel" hidden="hidden">
<img src="dba_images/cross.png" alt=""/>
Cancel
</button> …我有这个数组从mysql填充.但它没有向我展示第一张唱片.我无法弄清楚问题是什么.应该很简单.这是代码.
$result=mysql_query("SELECT user_instance.instance_name, user_instance.host_name FROM
dba_account, user_instance WHERE dba_account.account_id = user_instance.account_id AND
dba_account.account_id = '1'");
echo mysql_error();
$nume = mysql_fetch_row($result);
while($note = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td><input type='text' name='instance_name' class='instance_name'
disabled='disabled' value='$note[instance_name]' size='25' /></td>";
echo "<td><input type='text' name='host_name' class='host_name' disabled='disabled'
value='$note[host_name]' size='25' /></td>";
echo "</tr>";
}