Question

Echo不打印fname。请让我知道为什么回声不起作用。 Echo不打印fname。

html表单

<!DOCTYPE html>
<html lang="en">
    <head>
        <meta charset="UTF-8">
        <title>PHP Form</title>
        <link rel="stylesheet" href="style.css">
    </head>
    <body>
        <form action="submit.php" method="post">
            <h1>Sign Up</h1>
            <fieldset>
                <legend>New Users</legend>

                <label for="First Name">First Name</label>
                <input type="text" name="fname" placeholder="First Name" required>

                <label for="Last Name">Last Name</label>
                <input type="text" name="lname" placeholder="Last Name" required>

                <label for="Username">Username</label>
                <input type="text" id="username" placeholder="Username"requiredrequiredrequired>
                <br >

                <input type="Submit" required name="submit"value="Submit">
            </fieldset>
        </form>
    </body>
</html>

submit.php

<!DOCTYPE html>
<html lang="en">
    <head>
        <meta charset="UTF-8">
        <title>Document</title>
    </head>
    <body>
        <?php $fname = $_post["fname"];  echo  $fname; ?>    
        <?php $lname= $_post["lname"]; ?>     
    </body>
</html>

Answer 1

变化

<!DOCTYPE html>
<html lang="en">
    <head>
        <meta charset="UTF-8">
        <title>Document</title>
    </head>
    <body>
        <?php $fname = $_post["fname"];  echo  $fname; ?>    
        <?php $lname= $_post["lname"]; ?>     
    </body>
</html>

到

<!DOCTYPE html>
<html lang="en">
    <head>
        <meta charset="UTF-8">
        <title>Document</title>
    </head>
    <body>
        <?php $fname = $_POST["fname"];  echo  $fname; ?>    
        <?php $lname= $_POST["lname"]; ?>     
    </body>
</html>

我认为这是因为post变量需要大写

(PHP 4 >= 4.1.0, PHP 5)
$_POST -- $HTTP_POST_VARS [deprecated] — HTTP POST variables

http://php.net/manual/en/reserved.variables.post.php

Answer 2

先修复html 你是这样写的吗？

<input type="text" id="username" placeholder="Username"requiredrequiredrequired>

我试一试，它有效。我创建了echo.php。测试一下。

<!DOCTYPE html>
<html lang="en">
  <head>
    <meta charset="UTF-8">
    <title>PHP Form</title>
    <link rel="stylesheet" href="style.css">
  </head>
  <body>
    <?php
      $fname = $_POST["fname"];  echo  $fname;     
      $lname= $_POST["lname"]; echo $lname;
    ?>

    <form action="echo.php" method="post">
      <h1>Sign Up</h1>
      <fieldset>
      <legend>New Users</legend>

      <label for="First Name">First Name</label>
      <input type="text" name="fname" placeholder="First Name" />

      <label for="Last Name">Last Name</label>
      <input type="text" name="lname" placeholder="Last Name" />

      <label for="Username">Username</label>
      <input type="text" id="username" placeholder="Username" />
      <br >

      <input type="Submit" required name="submit"value="Submit">
      </fieldset>
    </form>
  </body>
</html>

你可以制作更漂亮的代码。

Answer 3

试试这个注意：为了方便起见，我将两个页面一起加入并命名为submit.php。你可以分开它

<?php
        if(isset($_POST["fname"])){
    ?>
            <!DOCTYPE html>
                <html lang="en">
                    <head>
                        <meta charset="UTF-8">
                        <title>Document</title>
                    </head>
                    <body>
                        <?php $fname = $_POST["fname"];
                        echo  $fname;
                        $lname= $_POST["lname"];
                        echo $lname?>     
                    </body>
                </html>
    <?php
        }
        else{
    ?>
    <!DOCTYPE html>
    <html lang="en">
        <head>
            <meta charset="UTF-8">
            <title>PHP Form</title>
            <link rel="stylesheet" href="style.css">
        </head>
        <body>
            <form action="submit.php" method="post">
                <h1>Sign Up</h1>
                <fieldset>
                    <legend>New Users</legend>

                    <label for="First Name">First Name</label>
                    <input type="text" name="fname" placeholder="First Name" required>

                    <label for="Last Name">Last Name</label>
                    <input type="text" name="lname" placeholder="Last Name" required>

                    <label for="Username">Username</label>
                    <input type="text" id="username" placeholder="Username"requiredrequiredrequired>
                    <br >

                    <input type="Submit" required name="submit"value="Submit">
                </fieldset>
            </form>
        </body>
    </html>
    <?php }?>

echo不能以PHP形式工作

3 个答案: