Question

我刚刚开始学习并整理了以下表格。我的困惑是当我使用空来验证用户是否选择或输入了任何信息时，代码会生成以下错误。

我注意到我是否有以下代码行

if(empty($gender)) {
        $errormessage[2] = "Please select your gender";
    }

作为

if(empty($_POST["gender"])) {
        $errormessage[2] = "Please select your gender";
    }

和

if(empty($gender)) {
        $errormessage[2] = "Please select your gender";
    }

作为

if(empty($_POST["gender"])) {
        $errormessage[2] = "Please select your gender";
    }

我没有看到错误消息。我认为正在生成错误消息是因为性别和媒体表单元素的代码行$ _POST分别是单选按钮和复选框。但是，如果我想在顶部初始化所有变量，那么最好的方法是什么？

/* <?php

if(isset($_POST["submit"])) {

    $fname = $_POST["fname"];
    $lname = $_POST["lname"];
    $gender = $_POST["gender"];
    $age = $_POST["age"];
    $address = $_POST["address"];
    $media = $_POST['media'];

    $errormessage =  array();

    if(empty($fname)) {
        $errormessage[0] = "Please enter your first name";
    }

    if(empty($lname)) {
        $errormessage[1] = "Please enter your last name";
    }

    if(empty($gender)) {
        $errormessage[2] = "Please select your gender";
    }

    if(empty($age)) {
        $errormessage[3] = "Please select your age";
    }

    if(empty($address)) {
        $errormessage[4] = "Please enter your address";
    }

    if(empty($media)) {
        $errormessage = "Please select the type of media";

    }

}
?>
<html>
<head>
    <title>Sample Registration</title>
</head>
<body>
    <h2>Sample registration</h4>
        <form name="registration" method="post" action="registration.php">
            <div>
                First Name: <br />
                <input type="text" name="fname" value="">
            </div>

            <div>
                Last Name: <br />
                <input type="text" name="lname" value="">
            </div>

            <div>
                Gender: <br />
                male<input type="radio" name="gender" value="male">
                female<input type="radio" name="gender" value="female">
            </div>

            <div>
                Age: <br />
                <select name="age">
                    <option value="">Please select your age</option>
                    <option value="18-25">18-25</option>
                    <option value="26-33">26-33</option>
                </select>
            </div>

            <div>
                Address: <br />
                <textarea name="address" cols="10" rows="10"></textarea>
            </div>

            <div>
                Sign-me up: <br />
                <input type="checkbox" name="media['newsletter']" value="newsletter"> newsletter
                <input type="checkbox" name="media['specials']" value="specials"> specials
                <input type="checkbox" name="media['events']" value="events"> events

            <div>
                <input type="submit" name="submit" value="submit">
            </div>
        </form>
</body>
</html>

*/

/ *错误信息* /

! ) Notice: Undefined index: gender in C:\Program Files\EasyPHP-5.3.5.0\www\registration.php on line 7
Call Stack
#   Time    Memory  Function    Location
1   0.0004  341792  {main}( )   ..\registration.php:0
Dump $_SERVER

$_SERVER['REMOTE_ADDR'] =



string '127.0.0.1' (length=9)

$_SERVER['REQUEST_METHOD'] =



string 'POST' (length=4)

$_SERVER['REQUEST_URI'] =



string '/registration.php' (length=17)

Variables in local scope (#1)

$address =

    Undefined

$age =

    Undefined

$errormessage =

    Undefined

$errormsg =

    Undefined

$fname =



string '' (length=0)

$gender =

    Undefined

$lname =



string '' (length=0)

$media =

    Undefined

( ! ) Notice: Undefined index: media in C:\Program Files\EasyPHP-5.3.5.0\www\registration.php on line 10
Call Stack
#   Time    Memory  Function    Location
1   0.0004  341792  {main}( )   ..\registration.php:0
Variables in local scope (#1)

$address =



string '' (length=0)

$age =



string '' (length=0)

$errormessage =

    Undefined

$errormsg =

    Undefined

$fname =



string '' (length=0)

$gender =



null

$lname =



string '' (length=0)

$media =

    Undefined

Answer 1

在访问之前，您需要检查HTTP POST请求中是否实际设置了$_POST['gender']和$_POST['media']个变量;初始化每个$_POST var的更好的解决方案可能是这样的：

$fname = isset( $_POST['fname'] ) ? $_POST['fname'] : '';

上面的三元赋值相当于为您感兴趣的每个$_POST变量运行以下逻辑表达式：

if( isset( $_POST['fname'] ) ) {
    $fname = $_POST['fname'];
} else {
    $fname = '';
}

如果您实施其中任何一项，如果您在empty()上运行$gender，则不会收到通知;此外，如果未设置$_POST['gender']，empty()仍会按照您的预期行事。它增加了一些冗长，但要重写你的例子你可能会尝试：

if( isset( $_POST["submit"] ) ) {

    $fname = isset( $_POST['fname'] ) ? $_POST["fname"] : '';
    $lname = isset( $_POST['lname'] ) ? $_POST["lname"] : '';
    $gender = isset( $_POST['gender'] ) ? $_POST["gender"] : '';
    $age = isset( $_POST['age'] ) ? $_POST["age"] : '';
    $address = isset( $_POST['address'] ) ? $_POST["address"] : '';
    $media = isset( $_POST['media'] ) ? $_POST['media'] : '';

    $errormessage =  array();
    if( empty( $fname ) )
        $errormessage[] = "Please enter your first name";
    if( empty( $lname ) )
        $errormessage[] = "Please enter your last name";
    if( empty( $gender ) )
        $errormessage[] = "Please select your gender";
    if( empty( $age ) )
        $errormessage[] = "Please select your age";
    if( empty( $address ) )
        $errormessage[] = "Please enter your address";
    if( empty( $media ) )
        $errormessage[] = "Please select the type of media";
}

当然，如果您打算在SQL上下文中使用它，您需要清理数据，但这至少可以解决您所面临的错误。

如果你有很多这些变量 - 或者经常重复这个任务 - 你可以考虑将它变成一个函数！

Answer 2

尝试使用!isset()代替empty()。

作为旁注，如果warning值不存在，您分配变量的方式将创建$_POST通知。如果您想删除这些通知，请使用@ Ryan的解决方案。

PHP empty（）错误消息

2 个答案: