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HTML表單顯示注釋

[英]HTML Forms to display comments

我正在HTML網頁上創建表單,以便訪問者可以添加有關我的頁面的評論。 這包括不同類型的表格,我正在努力為單選按鈕添加if else語句。 我的html代碼被發布了打擊,然后鏈接到它的php被發布了。 請幫忙!!!

<html>
<form action="process_comment.php"  method="post">
    Name:
    <input type="text" name="username" value=""/>
    <br>
</form>

<form action="process_comment.php" method="post" >
    Do you like this page?
    <input type = "radio" name = "like" value = "0"> Yes
    <input type = "radio" name = "like" value = "1"> No
    <br>
<input type="submit" value ="go" /><br>
</form>
</html>


<?php
$userName=$_POST["username"];
echo "Hello <b>$userName</b>!<br>";
?>   

<?php
$like = $_POST["like"];
if($like = "post") {
    echo "I am happy that you <b>like</b> this page :)";
}   else {
    echo "I am sorry that you <b>do not like</b> my page :(";
}
?>

您正在使用兩種形式,一種操作嘗試下面的代碼,並將兩種形式合並為一種。

<html>
    <form action="process_comment.php"  method="post">
        Name:
        <input type="text" name="username" value=""/>
        <br>

        Do you like this page?
        <input type = "radio" name = "like" value = "0"> Yes
        <input type = "radio" name = "like" value = "1"> No
        <br>
        <input type="submit" name="submit" value ="go" /><br>
    </form>
 </html>

並在process_comment.php中

<?php

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

      $userName=$_POST["username"];
      echo "Hello <b>$userName</b>!<br>";
      $like = $_POST["like"];
      if($like) {
        echo "I am happy that you <b>like</b> this page :)";
      }   
      else {
        echo "I am sorry that you <b>do not like</b> my page :(";
      }
    }
 ?>

如果value是1或0,則可以在if條件中使用變量,因為它可以是true或false。

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