我正在制作一个ajax登录脚本,我的html代码如下:
<form class="form-container" action="login.php" method="POST">
<div class="form-title"><h2>Sign up</h2></div>
<div class="form-title">Name</div>
<input class="form-field" type="text" name="username" /><br />
<div class="form-title">Password</div>
<input class="form-field" type="password" name="pass" /><br />
<div class="submit-container">
<input class="submit-button" type="submit" value="Submit" />
</div>
</form>
我的php代码如下:
$client_username = $_POST["username"];
$client_password = $_POST["pass"]
当我运行代码时,我在wamp服务器中出现错误,在我的php代码中显示Unidentified Index。有什么问题,我该如何解决?
编辑:
我尝试过但却徒劳无功:
if(isset($_POST["username"]))
{
$client_username = $_POST["username"];
}
if(isset($_POST["pass"]))
{
$client_password = $_POST["pass"];
}
echo $client_username;
echo $client_password;
答案 0 :(得分:1)
试试这个,只需使用isset()来检查POST变量
$client_username = "";
$client_password = "";
if(isset($_POST["username"]));
{
$client_username = $_POST["username"];
}
if(isset($_POST["pass"]))
{
$client_password = $_POST["pass"];
}