用PHP连接到FTP，密码为@ [英] Connect to FTP with PHP with a password that has @

问题描述

我遇到以下问题：

我需要连接到FTP并读取一个CSV文件。它的主要问题是密码有@，$，％...我如何连接especials字符？我尝试了以下方式进行连接：

文件打开

  $ filename ='ftp：// user：p @ s（$[e-mail]/file.csv'; 
 $ handle = fopen（$ filename，r）或die（Error） ; 
  

FTP登录

  $ ftp_server =ftp.myftp.url / file.csv; 
 $ ftp_user =user; 
 $ ftp_pass =p @ s（$ word; 
 $ conn_id = ftp_connect（$ ftp_server）或死（无法连接到$ ftp_server）; 
 $ login_result = ftp_login（$ ftp_server，$ ftp_user，$ ftp_pass）或死（无法连接到2 ）; 
 $ data = file_get_contents（'ftp.myftp.url / file.csv'）; 
  

感谢所有！

解决方案

您的两段代码有两个不同的问题。 b
$ b

---

文件打开

  $ filename ='ftp：// user：p @ s（$[e-mail]/file.csv'; 
 $ handle = fopen（$ filename，r）或死（错误）; 
  

问题在于 @ ，正如您已经正确猜测的那样，因为它在 URL语法中有含义（作为证书和主机名之间的分隔符）。

您必须 URL编码 @ 到％40 ：

  $ filename ='ftp：// user：p％40s（$[e-mail]/file.csv' ; 
  

您提到实际密码也有％。那必须以URL编码为 ％25 。

---

FTP登录

  $ ftp_server =ftp.myftp.url / file.csv; 
 $ ftp_user =user; 
 $ ftp_pass =p @ s（$ word; 
 $ conn_id = ftp_connect（$ ftp_server）或die（无法连接到$ ftp_server）; 
 $ login_result = ftp_login（ $ ftp_server，$ ftp_user，$ ftp_pass）或die（无法连接到2）; 
  

在这里，问题不是 @ ，因为不包含任何URL（％会是一个问题）在这里，问题是 $ ，因为你是使用双引号，所以 $ word 被替换为一个值（可能是未定义的）变量 word ，有效地使密码为 p @ s（ only。

使用单引号，以避免将 $ 解释为变量：

  $ ftp_pass ='p @ s（$ word'; 
  

也可以用<$ c转义 $
$ b $ $ p $ code $ $ ftp_pass =p @ s（\ $ word ;

I have the following problem:

I need connect to FTP and read one CSV file. The main problem it's password has @, $, %... How can I connect with especials characters? I tried the following ways to connect:

FILE OPEN

$filename = 'ftp://user:p@s($[e-mail]/file.csv'; 
$handle = fopen($filename, "r") or die("Error");

FTP LOGIN

$ftp_server = "ftp.myftp.url/file.csv";
$ftp_user = "user";
$ftp_pass = "p@s($word";
$conn_id = ftp_connect($ftp_server) or die("Could not connect to $ftp_server");
$login_result = ftp_login($ftp_server, $ftp_user, $ftp_pass) or die("Could not connect to 2");
$data = file_get_contents('ftp.myftp.url/file.csv');

Thanks for all!

解决方案

You have two distinct problems with your two pieces of code.

---

FILE OPEN

$filename = 'ftp://user:p@s($[e-mail]/file.csv'; 
$handle = fopen($filename, "r") or die("Error"); 

Here, the problem is the @, as you have correctly guessed, as it has a meaning in the URL syntax (as a separator between credentials and hostname).

You have to URL-encode the @ to %40:

$filename = 'ftp://user:p%40s($[e-mail]/file.csv'; 

You mentioned that the actual password also has %. That has to be URL-encoded to %25.

---

FTP LOGIN

$ftp_server = "ftp.myftp.url/file.csv";
$ftp_user = "user";
$ftp_pass = "p@s($word";
$conn_id = ftp_connect($ftp_server) or die("Could not connect to $ftp_server");
$login_result = ftp_login($ftp_server, $ftp_user, $ftp_pass) or die("Could not connect to 2");

Here, the problem is not @, as no URL is involved (neither % would be a problem). Here, the problem is the $, as you are using double-quotes, so $word is replaced with a value of (probably undefined) variable word, effectively making the password be p@s( only.

Use single quotes to avoid the $ being interpreted as a variable:

 $ftp_pass = 'p@s($word';

Or escape the $ with \:

 $ftp_pass = "p@s(\$word";

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