我已经创建了一个注册页面,它正在运行,但我有一个问题。
该页面包含用户名,密码,电子邮件,地址和电话。 根据我所达到的目的,用户可以添加所有这些要求,并且只能添加“用户名”以获得成功注册 我需要一种方法让用户添加所有这些要求(用户名,密码,电子邮件,地址和电话),而不仅仅是添加1个要求。
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("theater", $con);
$sql="INSERT INTO member (username, password, email, telephone, address)
VALUES
('$_POST[username]','$_POST[password]','$_POST[email]','$_POST[telephone]','$_POST[address]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "records added";
mysql_close($con);
?>
signup.php
页面是一个普通表格,其中包含输入用户名,电子邮件,密码,地址和带注册按钮的电话。我不认为我们应该在这里使用它,不是吗?
答案 0 :(得分:1)
您可以检查特定项目是否已设置,如果已设置,请继续。
$errors = array(); //Initialize an empty errors array.
if(!isset($_POST["username"]){
$errors[] = "Username not set; please try again.";
}
if(!isset($_POST["email"]){
$errors[] = "Email address not set; please try again.";
}
//If there are no errors in the $errors array (aka: every field is set):
if(empty($errors)) {
//Query the database and whatnot.
} else {
//Echo out the errors and demand proper values.
}
答案 1 :(得分:0)
您应该尝试这种方式来检查是否所有表单字段都已填充。
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("theater", $con);
if($_POST['username'] == '' || $_POST['password'] == '' || $_POST['email'] == '' || $_POST['telephone'] == '' || $_POST['address'] == '') $filled = false;
if($filled) {
$sql="INSERT INTO member (username, password, email, telephone, address)
VALUES
('$_POST[username]','$_POST[password]','$_POST[email]','$_POST[telephone]','$_POST[address]')"; } else echo 'Please fill all the fields!';
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "records added";
mysql_close($con);
?>