Question

**我正在使用php创建表单应用程序。此代码无法正常运行。我想在它为空时显示错误Student ID is required。并且该值必须仅包含4位数。不能输入字符串值和特殊字符。以上所有都应该得到验证。如何解决这个问题呢。任何人都可以帮助我。谢谢... ！！！ **

<!DOCTYPE html>
    <html>
        <head>
    <title> Enter Details </title>
        </head>
    <body>


<?php
    $id = "";
    $idErr = ""; 

if ($_SERVER["REQUEST_METHOD"] == "POST"){
        if (empty($id)){
            $idErr = "Student ID is required";
        }elseif(preg_match('/^\d+$/',$_GET['id'])){
            $idErr = "Only numbers allowed";
        }elseif(strlen($id) != 4) {
            $idErr = "Please enter  4 numbers";
        }
    }

    ?>

<form action = <?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?> method = "post">
    Student ID: <input placeholder="Student ID" type = "text" name= "id" value = ""/> 
    <span class="error">* <?php echo $idErr;?></span></br></br>
        <input type = "submit" name = "submit" value = "Register"/> 
</form>

</body>
</html>

Answer 1

您定义一个空变量，如果帖子提交，请检查它是否为空。这将始终为true，因为此变量的空字符串为值。您应该尝试使用$_POST。您还应将$_GET['id']更改为$_POST['id']：

<?php

$idErr = ""; 

if ($_SERVER["REQUEST_METHOD"] == "POST") {
    $id = $_POST['id'];

    if (empty($id)){
         $idErr = "Student ID is required";
    } elseif(preg_match('/^\d+$/',$_POST['id'])) {
         $idErr = "Only numbers allowed";
    } elseif(strlen($id) != 4) {
         $idErr = "Please enter  4 numbers";
    }
}

?>

如果你尝试这个，你应该能得到不同的结果。

Answer 2

    <?php
        $id = "";
        $idErr = ""; 
    if ($_SERVER["REQUEST_METHOD"] == "POST"){
            $idVal = $_POST['id'];
            if (empty($idVal)){
                $idErr = "Student ID is required";
            }elseif(!is_numeric($idVal)){
                $idErr = "Only numbers allowed";
            }elseif(strlen($idVal) != 4) {
                $idErr = "Please enter  4 numbers";
            }
        }
    ?>


//it check like below
   empty->number->exact 4 digit-> success.....

如何在php中验证整数

2 个答案: