Question

我有一个字符串列表，我想逐个读取字符串并将其转换为整数列表，有没有办法将每个字符转换为新的列表？

public static function sendEmail($data) {

        $r_error = 1;
        $r_message = "";
        $r_data = array();

        $q = "select * from config where type='email_detail'";
        $r = self::DBrunQuery($q);
        $row = self::DBfetchRow($r);
        $detail = json_decode($row['value'], true);

        include "phpmailer/PHPMailerAutoload.php";


        if (!empty($data['email'])) {


            foreach ($data as $var) {
                $work_email = 'fahadansari12feb@gmail.com'; //$var['email_id']; 
                $name = 'fahad'; //$var['name'];
                $subject = $var['subject'];
                $body = $var['body'];
                $cc = $var['cc_detail'];
                $bcc = $var['bcc_detail'];
                $file_upload = $var['upload_file'];

                $mail = new PHPMailer;
                $mail->isSMTP();
                $mail->SMTPDebug = 0;
                $mail->Debugoutput = 'html';
                $mail->Host = '5.9.144.226'; //$detail['host'];
                $mail->Port = '2222'; //$detail['port'];
                $mail->SMTPSecure = 'tls';
                $mail->SMTPAuth = true;
                $mail->Username = 'fahadansari12feb@gmail.com'; //$detail['username']; //sender email address 
                $mail->Password = 'mypassword'; //$detail['password']; // sender email password
                $mail->setFrom('hr@excellencetechnologies.in', 'Excellence Technologies'); // name and email address from which email is send
                $mail->addReplyTo('hr@excellencetechnologies.in', 'Excellence Technologies'); // reply email address with name 
                $mail->addAddress($work_email, $name); // name and address to whome mail is to send
                if (sizeof($cc) > 0) {
                    foreach ($cc as $d) {
                        $mail->addCC($d[0], $d[1]);
                    }
                }
                if (sizeof($bcc) > 0) {
                    foreach ($bcc as $d2) {
                        $mail->addBCC($d2[0], $d2[1]);
                    }
                }
                $mail->Subject = $subject; // subject of email message 
                $mail->msgHTML($body); // main message 
                // $mail->AltBody = 'This is a plain-text message body';
                //Attach an image file
                if (sizeof($file_upload) > 0) {
                    foreach ($file_upload as $d3) {
                        $mail->addAttachment($d3);
                    }
                }
//send the message, check for errors
                if (!$mail->send()) {
                    $row3 = $mail->ErrorInfo;
                } else {
                    $row3 = "Message sent";
                }
            }
        }


        if ($row3 != "Message sent") {

            $r_error = 1;
            $r_message = $row3;
            $r_data['message'] = $r_message;
        } else {
            $r_error = 0;
            $r_message = "Message Sent";
            $r_data['message'] = $r_message;
        }
        $return = array();
        $return['error'] = $r_error;
        $return['data'] = $r_data;
        return $return;
    }

Answer 1

函数式编程的一个支柱是map，它具有以下特征：

map :: (a -> b) -> [a] -> [b]

换句话说，它需要一个函数，它需要a类型的值并返回类型b的值（注意这些类型可能是相同的类型，但不要be）以及a值列表，并返回b值列表。举个例子：

double :: Int -> Int
double = (*2)
{- equivalently written as:
   double x = x * 2 -}

as = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
bs = map double as

在此示例中，地图专用于a ~ Int和b ~ Int，解析为

map :: (Int -> Int) -> [Int] -> [Int]

然后bs为[2, 4, 6, 8, 10, 12, 14, 16, 18, 20]。

为什么map上有这么长的引物？因为它是函数式编程中许多答案的框架，包括你的问题。请注意，String只是[Char]的类型同义词，因此您尝试从[[Char]] -> [[Int]]开始。哎呀，这肯定看起来像地图类型签名的最后两个术语，不是吗？让我们专注map以这些术语运作。

map :: ([Char] -> [Int]) -> [[Char]] -> [[Int]]

但是等等，map期待的函数看起来像是映射的结果。我们也写一下。

map :: (Char -> Int) -> [Char] -> [Int]

所以我们想要的是 double 地图，应用于某个函数f，以便：

map (map f) :: [[Char]] -> [[Int]]
{- more idiomatically written as
   (map . map) f :: [[Char]] -> [[Int]] -}

这意味着我们需要一个f :: Char -> Int - 一些从单个字符到整数的函数。该操作的定义输入数量相当少，所以我只是写它。

digitToInt :: Char -> Int
digitToInt '0' = 0
digitToInt '1' = 1
digitToInt '2' = 2
digitToInt '3' = 3
digitToInt '4' = 4
digitToInt '5' = 5
digitToInt '6' = 6
digitToInt '7' = 7
digitToInt '8' = 8
digitToInt '9' = 9
digitToInt 'a' = 10
digitToInt 'A' = 10
digitToInt 'b' = 11
digitToInt 'B' = 11
digitToInt 'c' = 12
digitToInt 'C' = 12
digitToInt 'd' = 13
digitToInt 'D' = 13
digitToInt 'e' = 14
digitToInt 'E' = 14
digitToInt 'f' = 15
digitToInt 'F' = 15
digitToInt _   = error "Invalid digit"

但是N.B.此功能是Data.Char

中的标准功能

import Data.Char (digitToInt)

您的结果是：

result = (map.map) digitToInt ["123","346","789"]

Answer 2

import Data.List (intersperse)

f :: String :: [Int]
f stringnum = read $ "[" ++ intersperse  ',' stringnum ++ "]" :: [Int]

>>> map f ["123", "456"]
[[1,2,3],[4,5,6]]

Haskell字符串列表到Int列表

2 个答案: