[英]How can I pass 18 variables into one function to validate my PHP/MySQL Interaction
我在我的网站上有一个个人资料页面,您可以在其中登录和更新您的帐户详细信息,例如:名字,姓氏,用户名,公司,地址行1等。
看到代码:
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$AccountID = $row["AccountID"];
$FName = $row["FName"];
$LName = $row["LName"];
$Username = $row["Username"];
$Company = $row["Company"];
$AddressL1 = $row["AddressL1"];
$AddressL2 = $row["AddressL2"];
$Town = $row["Town"];
$County = $row["County"];
$PostCode = $row["PostCode"];
$Password = $row["Password"];
$DFName = $row["DFName"];
$DLName = $row["DLName"];
$DAddressL1 = $row["DAddressL1"];
$DAddressL2 = $row["DAddressL2"];
$DTown = $row["DTown"];
$DCounty = $row["DCounty"];
$DPostCode = $row["DPostCode"];
}
if ($_SESSION['login_user']) {
如果它们已登录,则它将在输入字段中回显每个值,您可以按需更改和更新它们。
而不是写出来
$login_session = stripslashes($login_session);
$login_session = mysql_real_escape_string($login_session);
$login_session = trim($login_session);
对于每个变量,最佳选择是什么?
您可以执行以下操作:
function makeSafe($var)
{
$return = stripslashes($var);
$return = mysql_real_escape_string($return);
$return = trim($return);
return $return;
}
$login_session = makeSafe($login_session);
这样可以避免一页上多行代码,并且避免重复代码
就像我在评论中说的那样,请尝试避免mysql_ *功能,并用mysqli或PDO替换它
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