Question

我想要一个带有php的表单。但是好几个小时我都遇到了错误，错误是当$ _POST = [＆＃39; name＆＃39;]想要被检查为空或不是空的时候。当我检查数据库时，行是白色的，没有任何东西。为了检查$ _POST是否为空，我打印出单词＆＃39; empty＆＃39;要确定它是空的，它将被打印出来并且是空的＆＃39 ;; 我的错误在哪里，它与数据库mysql有关，还是只与代码有关？请帮助我，我感到困惑和无聊。这是我的全部代码：

<!doctype html>
<html lang="fa">
<head>
<meta charset="utf-8">
<title>form</title>
<link href="addContact.css" rel="stylesheet"/>
<link href="main.css" rel="stylesheet"/>
<link href="table.css" rel="stylesheet"/>
</head>
<body>

<?php
$name = "";
$nameErr = "";

if ($_SERVER["REQUEST_METHOD"] == "POST") {
 if (empty($_POST["name"])) {
    echo 'empty';
    $nameErr = "Name is required";
} else {
    echo 'full';
    $name = test_input($_POST["name"]);
    // check if name only contains letters and whitespace
    if (!preg_match("/^[a-zA-Z ]*$/", $name)) {
        $nameErr = "Only letters and white space allowed";
    }
  }


  $servername = "localhost";
  $username = "abc";
  $password = "abc";
  $dbname = "abc";

  try {
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, 
  $password);
    // set the PDO error mode to exception
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $sql = "INSERT INTO abc (firstname)
    VALUES ('$name')";
    // use exec() because no results are returned
    $conn->exec($sql);
    $last_id = $conn->lastInsertId();
    echo "New record created successfully. Last inserted ID is: " . $last_id;
} catch (PDOException $e) {
    echo $sql . "<br>" . $e->getMessage();
}

$conn = null;

}

function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}

?>


<form method="post" action="<?php echo 
htmlspecialchars($_SERVER["PHP_SELF"]);?>">
Name: <input type="text" name="name" value="<?php echo $name;?>">
<span class="error">* <?php echo $nameErr;?></span>
<br><br>
<input type="submit" name="submit" value="Submit">
</form>

</body>
</html>

提前谢谢

Answer 1

问题在于使用empty()

您可以在变量上使用它，而不是在值上使用它。

请参阅此处查看PHP documentation page。

要检查此项，请先将其保存在另一个变量中，然后检查：

$tempVal = $_POST["name"];
if (empty($tempVal))

Answer 2

你可以使用这个简单的例子

{{1}}

我建议这样： Understanding $_SERVER['PHP_SELF']

如何修复空$ _POST错误[＆＃39;名称＆＃39;]

2 个答案: