使用isset（）函數進行PHP表單驗證問題

Question

我一直在學習php表單驗證，並且似乎在以下內容中存在混淆

<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>

好吧所以我試圖用PHP進行表單驗證。 如果其中一個必填字段為空但頁面確實顯示錯誤，但仍然輸入數據庫中的值而不管錯誤。 我懷疑它是因為isset因為如果我更換線

if(!empty($name) && !empty($email) && !empty($gender))

它似乎工作。

編輯1：好的，我已設法解決問題。 感謝幫助過的人們。

Answer 1

檢查此條件而不是現有條件

if(isset($name) && $name != NULL && isset($email) && $email != NULL  && isset($gender) && $gender != NULL)

使用isset（）函數進行PHP表單驗證問題

問題描述

1 個解決方案

解決方案1
0 已采納 2016-09-15 12:10:22

使用isset（）函數進行PHP表單驗證問題

問題描述

1 個解決方案

解決方案1 0 已采納 2016-09-15 12:10:22

解決方案1
0 已采納 2016-09-15 12:10:22