为什么我的条件语句无法正常工作?我想同时显示错误消息。在表单验证中同时显示所有错误消息
function validate() {
if (firstName.value == "") {
document.getElementById('error').innerHTML = "*Field is empty";
return false;
} else if (lastName.value == "") {
document.getElementById('errorTwo').innerHTML = "*Field is empty";
return false;
} else {
return true;
}
}
<form name="form" action="action.php" method="post" onsubmit="return validate()">
<div class="wrapper">
<span>Name</span>
<br>
<input type="text" name="firstName" id="firstName" placeholder="First Name" onfocus="this.placeholder=''" onblur="this.placeholder='First Name'" />
<label id="error"></label>
<br>
<input type="text" name="middleName" id="middleName" placeholder="Middle Name (optional)" onfocus="this.placeholder=''" onblur="this.placeholder='Middle Name (optional)'" />
<input type="text" name="lastName" id="lastName" placeholder="Last Name" onfocus="this.placeholder=''" onblur="this.placeholder='Last Name'" />
<label id="errorTwo"></label>
<br>
<br>
</div>
你错过了关闭'}'。您是否在[JavaScript控制台]中看到任何错误(http://webmasters.stackexchange.com/questions/8525/how-to-open-the-javascript-console-in-different-browsers)? – 2015-03-31 14:49:36
您已将脚本标记粘贴到HTML标记范围之外,这是您的第一个问题,第二个问题是您尚未声明姓氏或姓氏 – Brad 2015-03-31 14:54:27
对不起,我只是没有将它包含在我的帖子中,但我已经关闭在我的实际代码..我没有任何错误,它只是不工作的方式,我想它是 – 2015-03-31 14:54:36