即使滿足條件，也執行其他語句

Question

這是提交按鈕被觸發時的2個表單和2個操作。第一種形式要求帳戶名稱並檢查驗證碼是否正確。當兩者都沒有問題時，用戶就會轉到下一個表格，詢問他2個問題，以便他可以找回他的密碼。代碼執行正常，但我有一個問題。當一切正常時：帳戶存在，驗證碼正常，兩個問題都可以，當用戶點擊第二個表單提交按鈕時，第一個動作的else語句再次執行，但表單連續正常，他得到了他的密碼e郵件。任何幫助將appriciated :)

if ('POST' === $_SERVER['REQUEST_METHOD']){ 
if($_POST['lostpassword']=='account' AND ($_POST["captcha"])&&$_POST["captcha"]!=""&&$_SESSION["code"]==$_POST["captcha"]) { 

connectdb($CONFIG['dbdbname'], $CONFIG['dbaddress'], $CONFIG['dbuser'], $CONFIG['dbpass']); 

$postusername = $_POST['account']; 
$postusername = antiinjection($postusername); 

$result = mssql_query (sprintf(SELECT_USER_FULLINFO, $postusername)); 
$rows=mssql_num_rows($result); 

if($rows>0) { 
    $rows=mssql_fetch_assoc($result); 
    extract($rows); 
    $error = 2; 
} else { 
    echo "Account doesn't exist.<br>"; 
    $error = 1; 
}  
} 
else { 
     echo '<script language="JavaScript"> 
      alert("Wrong Verification code. Please try again."); 
      </script>'; 
    } 
} 

形式：

<form name='lostpassword' action='index.php?page=lostpassword' method='post'  onsubmit='return checkform1()' autocomplete='off'> 
    <table CELLSPACING=0 BORDER=0 CELLPADDING=0 align=CENTER> 
     <tr> 
      <td width=200> 
       Account 
      </td> 
      <td> 
       <div align=right> 
        <input type=text maxlength=14 name=account> 
       </div> 
      </td> 
     </tr> 
     <tr> 
      <td valign=middle> 
       Verification Image <img src=\"captcha.php\"> 
      </td> 
      <td> 
       <div align=right> 
        <input type=text maxlength=4 name=captcha> 
       </div> 
      </td> 
     </tr> 

    </table> 
    <div align=center> 
     <BR> 
     <input type=hidden name=lostpassword value='account'> 
     <input type=submit name=Login value=' Submit '> 
    </div> 
</form> 

Nexti操作：

if($_POST['lostpassword']=='email') { 

$error = 3; 
$postusername = $_POST['account']; 
$postanswer1 = $_POST['answer1']; 
$postanswer2 = $_POST['answer2']; 
$postusername = antiinjection($postusername); 
$postanswer1 = antiinjection($postanswer1); 
$postanswer2 = antiinjection($postanswer2); 

connectdb($CONFIG['dbdbname'], $CONFIG['dbaddress'], $CONFIG['dbuser'], $CONFIG['dbpass']); 

$result = mssql_query (sprintf(SELECT_USER_FULLINFO, $postusername)); 
$rows=mssql_num_rows($result); 

if($rows>0) { 
    $rows=mssql_fetch_assoc($result); 
    extract($rows); 

    $postanswer1 = encrypt($postanswer1); 
    $postanswer2 = encrypt($postanswer2); 
    $answer1 = '0x' . substr(bin2hex($answer1), 0, 32); 
    $answer2 = '0x' . substr(bin2hex($answer2), 0, 32); 

    if($answer1!=$postanswer1) { 
     echo "Answer to security question #1 is incorrect.<br>"; 
     $error = 2; 
    } 
    if($answer2!=$postanswer2) { 
     echo "Answer to security question #2 is incorrect.<br>"; 
     $error = 2; 
    } 
} else { 
    echo "Account doesn't exist.<br>"; 
    $error = 1; 
} 

} 

構成了本次行動：

<form name='lostpassword' action='index.php?page=lostpassword' method='post'  onsubmit='return checkform2()' autocomplete='off'> 
<table CELLSPACING=0 BORDER=0 CELLPADDING=0 align=CENTER> 
<tr> 
    <td width=200> 
     Security Question #1 
    </td> 
    <td> 
     <div align=right> 
      {$quiz1} 
     </div> 
    </td> 
</tr> 
<tr> 
    <td> 
     Security Answer #1 
    </td> 
    <td> 
     <div align=right> 
      <input type=text maxlength=32 name=answer1> 
     </div> 
    </td> 
</tr> 
<tr> 
    <td> 
     Security Question #2 
    </td> 
    <td> 
     <div align=right> 
      {$quiz2} 
     </div> 
    </td> 
</tr> 
<tr> 
    <td> 
     Security Answer #2 
    </td> 
    <td> 
     <div align=right> 
      <input type=text maxlength=32 name=answer2> 
     </div> 
    </td> 
</tr> 
</table> 
<div align=center> 
    <BR> 
    <input type=hidden name=lostpassword value='email'> 
    <input type=hidden name=account value='{$postusername}'> 
    <input type=submit name=Login value=' Submit '> 
</div> 
</form> 

電子郵件代碼：

if($error==3) { 

$newpassword = mt_rand(1000000,9999999); 
$newpassword = md5($newpassword); 
$newpassword = substr($newpassword, 0, 15); 
$encnewpassword = encrypt($newpassword); 
echo '<br>'; 

mssql_query(sprintf(UPDATE_PASSWORD, $encnewpassword, $account)); 

if($CONFIG['email']==0) { 
    echo "<strong>Your password has been reseted to...</strong><br>{$newpassword}<br><br>"; 
} elseif($CONFIG['email']==1) { 
    sendemail($CONFIG['emailsmtp'], $CONFIG['emailuser'], $CONFIG['emailpass'], $CONFIG['emailaddress'], $CONFIG['servername'], "Lost Password", $email, $account, $newpassword, $ssn, "<strong>Your password has been reseted and sent to your email.</strong>"); 
} 

} 

Answer 1

從第一條件當然else語句執行的，因爲else語句驗證第一的否定，如果這是：

if ('POST' === $_SERVER['REQUEST_METHOD']){ 

} 
else { 

} 

當用戶提交的第二種形式，在第一種情況明顯進入。

您可以通過解決這個問題，例如添加另一名值提交輸入按鈕，如：

<input type='submit' name='LoginAct1' value='Submit' /> 
<input type='submit' name='LoginAct2' value='Submit' /> 

，然後您將驗證何種形式的數據被髮送。

if ('POST' === $_SERVER['REQUEST_METHOD'] && isset($_POST['LoginAct1'])){ 
    //blah blah 
} 

if ('POST' === $_SERVER['REQUEST_METHOD'] && isset($_POST['LoginAct2'])){ 
    //blah blah 
} 

Answer 2

你需要寫的第一個動作的「如果條件」如下：

if ('POST' === $_SERVER['REQUEST_METHOD'] && $_POST['lostpassword']=='account'){ 
    //rest of the code 
} 

那麼你的代碼將是工作的罰款。