定义PHP变量

时间:2019-02-13 14:25:26

标签: php mysql variables undefined

我的PHP代码似乎还不错,但会产生错误。我已经注意到:未定义变量:C:\ xampp2 \ htdocs \ datamanager1.php在第137行上的showhaddress

注意:未定义的变量:C:\ xampp2 \ htdocs \ datamanager1.php中的showppmail行138 为什么会这样呢?这是我的代码:

 $stmt3 = mysqli_stmt_init($connect);

  $query4 = "  
  SELECT haddress FROM tabela 
  WHERE password = ?  
  "; 
if(!mysqli_stmt_prepare($stmt3, $query4)){
echo "Failed to authenticate";
} else {
    mysqli_stmt_bind_param($stmt3, "s", $_SESSION['password']);
    mysqli_stmt_execute($stmt3);
$result = mysqli_stmt_get_result($stmt3); } 
while($row2 = mysqli_fetch_object($result)){
    $showhaddress = $row2->haddress;
}

 $stmt4 = mysqli_stmt_init($connect);

  $query5 = "  
      SELECT ppname FROM tabela 
      WHERE password = ?  
      "; 
if(!mysqli_stmt_prepare($stmt4, $query5)){
echo "Failed to authenticate";
} else {
    mysqli_stmt_bind_param($stmt4, "s", $_SESSION['password']);
    mysqli_stmt_execute($stmt4);
$result = mysqli_stmt_get_result($stmt4); } 
while($row3 = mysqli_fetch_object($result)){
    $showppmail = $row3->ppname;
}

$showhaddress = decrypt($showhaddress); // here is the error
$showppmail = decrypt($showppmail); //here is the error

0 个答案:

没有答案