Question

我是一个完整的新手（在上周开始）

问题是这样的：

基本上，我试图确保一旦填写了子表单，就不会改变它。所以，我使用了！isset来显示子表单（即如果！isset为true），如果！isset为false，则隐藏该子表单并显示下一个子表单（个人表单仅设计）

        <?php include($_SERVER['DOCUMENT_ROOT'].'/officespace/includes/functions.php');

    echo'<html>
        <head>
            <title> Create </title>
        </head>

    <body>';

    if(!isset($_POST["Category"])){
    /* if no category is selected, then this code will display the form to select the category*/

    Echo "Pls Select Category before clicking on Submit Category";
    /* Breaking out of PHP here, to make the form sticky by using a php code inside form action*/
    ?>
    <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
    <fieldset>
        <legend>Select the Category of Person: </legend><br />
            <input type="radio" name="Category" value="Individual" /> Individual<br /><br />
            <input type="radio" name="Category" value="Company, Pvt Ltd" /> Company, Pvt Ltd<br /><br />
            <input type="radio" name="Category" value="Company, Ltd" /> Company, Ltd<br /><br />
            <input type="radio" name="Category" value="Partnership Firm" /> Partnership Firm<br /><br />
            <input type="radio" name="Category" value="LLP Firm" /> LLP Firm<br /><br />
            <input type="radio" name="Category" value="HUF" /> HUF<br /><br />
            <input type="submit" name='Submit Category' value="Submit Category" /><br />
    </fieldset>
    </form>
    <?php
    } Else {

    $Category = $_POST["Category"];


        Echo "$Category";
        Echo "<br />";
    /* Using swich statement to test the value of Category, and accordingly echo appropriate forms*/
        switch ($Category) {
            case "Individual":
            if(!isset($_POST['Submit_Data_for_Individual'])){

            //if no data for individual is submitted,
            //then this code will display the form to enter and submit data for Individual

                Echo  /*displays message*/
                "Pls Enter the Data for the Individual";
    ?>
                <form action="<?php Echo $_SERVER['PHP_SELF']; ?>" method="post"> 
                <fieldset>
                <br />
                    First Namee: <input type="text" name="Individual_First_Name" />
                <br />
                    Middle Name: <input type="text" name="Individual_Middle_Name" />
                <br />
                    Last Name: <input type="text" name="Individual_Last_Name" />
                <br />
                    Date of Birth: <input type="text" name="date_of_birth/incorporation" />
                <br />
                    Gender: 
                <br />
                    <input type="radio" name="Gender" value="male" /> Male
                <br />
                    <input type="radio" name="Gender" value="female" /> Female
                <br />
                    Email 1: <input type="text" name="email_1" />
                <br />
                <input type="submit" name="Submit_Data_for_Individual" value="Submit Data for Individual" />
                </fieldset>
                </form>
    <?php                               
                }Else
                {
                $email_1 = $_POST["email_1"];
                $Gender = $_POST["Gender"];

                validate_email($email_1); // this is a custom function which i made
                Echo $Gender; // just to see if value has been passes. problem lies here because its not showing anything

                // run other validations here
                // and if all valid then run mysqli insert query for individuals record
                }

                break;
            case "Company, Pvt Ltd":
                echo "Company, Pvt Ltd";
                break;
            case "Company, Ltd":
                echo "Company, Ltd";
                break;
            case "Company, Ltd":
                echo "Company, Ltd";
                break;
            case "Partnership Firm":
                echo "Partnership Firm";
                break;
            case "LLP Firm":
                echo "LLP Firm";
                break;
            case "HUF":
                echo "HUF";
                break;
            case NULL:
                echo "Error: nothing selected";
                break;

        }
    }

    echo '</body>


    </html>';

    ?>

Answer 1

立即看到一个问题。

您正在检查名为Submit Data for Individual的表单输入，但这是一个没有value属性的提交按钮的name。在提交按钮上设置name='submit-data'属性并更改条件以检查名称而不是其值：

// This will never match.
if(!isset($_POST["Submit Data for Individual"])){

// Change it to
if(!isset($_POST["submit-data"])){

// Then change this
<input type="submit" value="Submit Data for Individual" />

// To this:
<input type="submit" name='submit-data' value="Submit Data for Individual" />

此外，switch语句中的默认大小写使用default关键字：

    // You may safely change this:
    case NULL:
        echo "Error: nothing selected";
        break;

    // To this:
    default:
        echo "Error: nothing selected";
        break;

<强>附录：

以下代码永远不可访问，因为表单会发布到另一个脚本create.php。如果您将<form>操作属性更改为回发到<?php $_SERVER['PHP_SELF'];?>而不是create.php，则应该看到else案例。现在，它不起作用，因为您设置了if的{{1}}测试。只有在提交表单时才能设置它，但表单提交到外部脚本。

$_POST["submit-data"]

要解决此问题并看到您的// This else case can never be reached... }Else { validate_email($_POST["email_1"]); // this is a custom function which i made Echo $_POST["Gender"]; // just to see if value has been passes. problem lies here because its not showing anything // run other validations here // and if all valid then run mysqli insert query for individuals record }已回显，请暂时更改

Gender

附录2

您正在检查是否已设置类别，但在发布用户表单后，它不会是：

<form action="create.php" method="post"> 
// change to
<form action="' . $_SERVER['PHP_SELF'] . '" method="post">

然后，您需要测试用户表单是否已提交。在// Change if(!isset($_POST["Category"])){ // To check that the user form was not submitted if (!isset($_POST["Category"]) && !isset($_POST['submit-data'])) {部分之前，添加Else { $Category = $_POST['Category'];以处理用户表单。

else if

最后，从if (!isset($_POST["Category"]) && !isset($_POST['submit-data'])) { // Show the Category form... } // Process the user form... else if (isset($_POST['submit-data'])) { validate_email($_POST["email_1"]); // this is a custom function which i made Echo $_POST["Gender"]; } // Now process the categories or show the user form... else { $Category = $_POST['Category']; // etc... }案例中移除整个Else块，因为它不能在那里使用。

PHP表单（值消失，不知道如何处理）

1 个答案: