如何在php中证明参数

嘿

我有一个用户注册php与一些参数，他们都应该证明。我的格式很难看。我将所有的验证都打包在复杂的if条款中

看一看:

include '../db_connect.php';
$arr = array('Data' => null,'Code' => null);

$birthdate = mysql_real_escape_string($_POST['birth']);
$gender = mysql_real_escape_string($_POST['gender']);
$uname = mysql_real_escape_string($_POST['username']);
$password = mysql_real_escape_string($_POST['pw']);
$email = mysql_real_escape_string($_POST['email']);
$lang = mysql_real_escape_string($_POST['lang']);

if (!proof_value($birthdate) && !proof_value($gender) && !proof_value($uname) && !proof_value($password) && !proof_value($email))
{
    if (!user_exist($uname))
    {
        if(!email_exist($email))
        {
            if (count($pw)==32)
            {
                    if(count($gender)==1 && ($gender=='m' ||$gender =='f'))
                    {
                        $code = genverification();
                        $sql = "Insert into USER (DATE_BIRTH,GENDER,USER_NAME,PASSWORD,EMAIL,VERIFICATION) VALUES ('$birthdate','$gender','$uname','$password','$email','$code')";
                        $result = mysql_query("Insert into USER (DATE_BIRTH,GENDER,USER_NAME,PASSWORD,EMAIL,VERIFICATION) VALUES ('$birthdate','$gender','$uname','$password','$email','$code')");
                        if ($result)
                        {
                            require_once("mailer.php");
                            if (sendmail($email,$link, $lang))
                            {
                                $arr['Code'] = 200;
                            }
                            else
                            {
                                $arr['Code'] = 422;
                            }


                        }
                        else
                        {
                            $arr['Code'] = 421;
                            //$arr['Date'] = $sql;
                        }

                    }
                    else
                    {
                        $arr['Code'] = 420;
                    }

            }
            else
            {
                $arr['Code']=423;
            }


        }
        else
        {
            $arr['Code']=419;
        }


    }
    else
    {
        $arr['Code']=418;
    }

}
else
{
$arr['Code']=400;
}

mysql_close($db);
echo json_encode($arr);

如您所见，如果验证失败，我的脚本将返回一个错误代码。我想把我的实际格式改成可读性更好的格式，但我现在知道如何解决这个问题了

---

Thx

有很多可能性，如果你敢尝试：-

PHP中的其他

---

我的格式很难看。

是的，是的。如果这些用户函数只是返回

isset（）

，那么为什么不直接使用

isset（）

，

count

应该被

strlen

替换，而且

$pw

永远不会被替换set@MA42 :) . 如果你写了这样的评论，你也应该发布一个答案。我一直听说每个人都应该避免使用breaksyeah-thx作为你的答案，但是代码对我来说太难了（我习惯用java编写）np，尽管我很惊讶你说java更严格

$success = null;
while (empty($error))
{
  if (...)
  {
    $error = xxx;
    break;
  }
  // repeat other checking
  // lastly
  $sql = ...;
  if ( $insert_ok )
  {
    $success = ...;
    break;
  }
}

if ( ! empty($error))
{
   //  something error
}
if ( ! empty($success))
{
  // something right
}

<?php 
//Allowed array of parameters
$allowed = array('birth','gender','username','passwordw','email','lang');

$cont=true;
//Loop through the post and check & assign the variables
foreach($_POST as $key=>$value){
    if(in_array($key,$allowed) && $value!=''){
        //m or f set $cont to false if not
        if($key=='gender' && ($value!='m' || $value!='f')){$arr['Code'] = 420; $cont=false;}
        //chek if email set $cont to false if not
        if($key=='email' && filter_var($value, FILTER_VALIDATE_URL, FILTER_FLAG_SCHEME_REQUIRED)==true){$arr['Code'] = 419; $cont=false;}
        //check pass len set $cont to false if not
        if($key=='password' && strlen($value)==32){$arr['Code'] = 423; $cont=false;}
        //Assign the variable
        $$key=$value;
    }else{
        //Rouge key in post or value blank
        $cont=false;
    }
}

//if alls ok
if($cont===true){

    $code = genverification();

    //PDO connect to database
    try {
        $dbh = new PDO("mysql:host=localhost;dbname=YOURDB", $dbusername, $dbpassword);
    }
    catch(PDOException $e)
    {
        echo $e->getMessage();
    }

    //Use prepared statement to avoid sql injections
    $sth = $dbh->prepare('INSERT into USER
            (DATE_BIRTH,GENDER,USER_NAME,PASSWORD,EMAIL,VERIFICATION) 
             VALUES (:birth,:gender,:username,:password,:email,:code)');

    //bind the variables to the parameters
    $sth->bindParam(':birth', $birth, PDO::PARAM_STR, strlen($birth));
    $sth->bindParam(':gender', $gender, PDO::PARAM_STR, 1);
    $sth->bindParam(':username', $username, PDO::PARAM_STR, strlen($username));
    $sth->bindParam(':password', $password, PDO::PARAM_STR, strlen($password));
    $sth->bindParam(':email', $email, PDO::PARAM_STR, strlen($email));
    $sth->bindParam(':code', $code, PDO::PARAM_STR, strlen($code));
    $sth->execute();

    //Do your mail
    require_once("mailer.php");
    if (sendmail($email,$link, $lang)){
        $arr['Code'] = 200;
    }else{
        $arr['Code'] = 422;
    }
}else{
    #Show your errors
}
?>