duandu8202 2013-07-08 23:02 采纳率: 0%
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PHP在未调用时显示“错误”

I'm having a problem with my php code.

I don't want the else echo "Check it again!"; to show unless they input some data into the form(input box) and it's not valid. But when I load the page it shows the error above the box.

<?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>
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2条回答 默认 最新

  • dtrz99313 2013-07-08 23:06
    关注

    Try this :)

    <?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!";
        }
    }
    ?>
    

    instead of

    <?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!";
        }
    }
    
    ?>
    

    Logic behind the edit: We'll check for the email or print "Check it again!" only if the form is submitted. Now if we don't check whether the form is submitted or this is the first time loading (or simple refresh) of the page, either insertion or the error display will occur. And we don't want that :)

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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