使用isset()函数的PHP表单验证问题
我一直在学习php表单验证,似乎对以下内容感到困惑使用isset()函数的PHP表单验证问题,php,Php,我一直在学习php表单验证,似乎对以下内容感到困惑 <html> <head> <title> learning </title> </head> <?php function test_input($data) { $data = trim($data); $data = stripslashes($data); $data = htmlspecialchars($data); return
<html>
<head>
<title>
learning
</title>
</head>
<?php
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
?>
<?php
$name = $email = $gender = $comment = $website = "";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$name = test_input($_POST["name"]);
$email = test_input($_POST["email"]);
$website = test_input($_POST["website"]);
$comment = test_input($_POST["comment"]);
$gender = test_input($_POST["gender"]);
}
?>
<?php
// define variables and set to empty values
$nameErr = $emailErr = $genderErr = $websiteErr = "";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["name"])) {
$nameErr = "Name is required";
} else {
$name = test_input($_POST["name"]);
}
if (empty($_POST["email"])) {
$emailErr = "Email is required";
} else {
$email = test_input($_POST["email"]);
}
if (empty($_POST["website"])) {
$website = "";
} else {
$website = test_input($_POST["website"]);
}
if (empty($_POST["comment"])) {
$comment = "";
} else {
$comment = test_input($_POST["comment"]);
}
if (empty($_POST["gender"])) {
$genderErr = "Gender is required";
} else {
$gender = test_input($_POST["gender"]);
}
}
?>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "test";
$conn = mysqli_connect($servername,$username,$password,$database);
if(isset($name) && isset($email) && isset($gender))
{
$sql = "insert into learn values('$name','$email','$website','$comment','$gender')";
if(mysqli_query($conn,$sql))
{
echo 'Entry to database successful';
}
else {
echo 'Error!';
}
}
?>
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
Name: <input type="text" name="name">
<span class="error">* <?php echo $nameErr;?></span>
<br><br>
E-mail:
<input type="text" name="email">
<span class="error">* <?php echo $emailErr;?></span>
<br><br>
Website:
<input type="text" name="website">
<span class="error"><?php echo $websiteErr;?></span>
<br><br>
Comment: <textarea name="comment" rows="5" cols="40"></textarea>
<br><br>
Gender:
<input type="radio" name="gender" value="female">Female
<input type="radio" name="gender" value="male">Male
<span class="error">* <?php echo $genderErr;?></span>
<br><br>
<input type="submit" name="submit" value="Submit">
</form>
它似乎起作用了
编辑1:好的,我已经设法解决了这个问题。感谢帮助过的人。检查此情况,而不是现有情况
if(isset($name) && $name != NULL && isset($email) && $email != NULL && isset($gender) && $gender != NULL)
检查此条件而不是现有条件
if(isset($name) && $name != NULL && isset($email) && $email != NULL && isset($gender) && $gender != NULL)
您可以在代码中显式设置所有这些变量。它们总是被设置为结果。是的,变量总是被设置的。验证是否设置了
$\u POST
项。@johncode,以便在表单未填写时取消设置它们?或者我应该这样做!empty()?检查是否设置了一条错误消息。或者更好的方法是,将错误添加到空数组中,并在将数据添加到数据库之前检查该数组是否仍然为空。您可以在代码中显式设置所有这些变量。它们总是被设置为结果。是的,变量总是被设置的。验证是否设置了$\u POST
项。@johncode,以便在表单未填写时取消设置它们?或者我应该这样做!empty()?检查是否设置了一条错误消息。或者更好的方法是,将错误添加到空数组中,并在将数据添加到数据库之前检查该数组是否仍然为空。