PHP在未調用時顯示“錯誤”

Question

我的PHP代碼有問題。

我不希望else echo "Check it again!"; 除非他們將一些數據輸入表格（輸入框）並且它無效，否則顯示。 但是當我加載頁面時，它會在框上方顯示錯誤。

<?PHP
if (filter_var($email, FILTER_VALIDATE_EMAIL)) {
    mysql_query("INSERT INTO newsletter (email) VALUES('$email')") or die(mysql_error());
        echo 'You have registered your E-Mail address to our database! You will now receive regular updates on the progess!';
    }else{
        echo "Check it again!";
}
?>

<form name="newsletter" method="post" action="<?PHP $_SERVER['PHP_SELF']?>">
<input type="text" name="newsletter" id="newsletter">
<input type="submit" value="SUBMIT!">
</form>

Answer 1

嘗試這個 ：）

<?PHP
if($_SERVER["REQUEST_METHOD"]=="POST"){
    if (filter_var($email, FILTER_VALIDATE_EMAIL)) {
        mysql_query("INSERT INTO newsletter (email) VALUES('$email')") or die(mysql_error());
            echo 'You have registered your E-Mail address to our database! You will now receive regular updates on the progess!';
        }else{
            echo "Check it again!";
    }
}
?>

代替

<?PHP

 if (filter_var($email, FILTER_VALIDATE_EMAIL)) {
    mysql_query("INSERT INTO newsletter (email) VALUES('$email')") or die(mysql_error());
    echo 'You have registered your E-Mail address to our database! You will now receive regular updates on the progess!';
    }else{
      echo "Check it again!";
    }
}

?>

編輯背后的邏輯：我們將檢查電子郵件或打印“再次檢查！” 只有提交表格。 現在，如果我們不檢查表單是否已提交或者這是第一次加載（或簡單刷新）頁面，則會發生插入或錯誤顯示。 我們不希望這樣:)

Answer 2

這只是我喜歡的樣式的@MuhammedHedayet答案的替代品，只在你需要的時候檢查它。 兩者都應該完美：

<?PHP
if (filter_var($email, FILTER_VALIDATE_EMAIL)) {
    mysql_query("INSERT INTO newsletter (email) VALUES('$email')") or die(mysql_error());
        echo 'You have registered your E-Mail address to our database! You will now receive regular updates on the progess!';
    } else if ($_SERVER["REQUEST_METHOD"]=="POST"){
        echo "Check it again!";
}
?>

<form name="newsletter" method="post" action="<?PHP $_SERVER['PHP_SELF']?>">
<input type="text" name="newsletter" id="newsletter">
<input type="submit" value="SUBMIT!">
</form>