为什么这些函数看不到我的变量？

Question

为什么会出现此错误：

Undefined variable key_2captcha

我运行此代码将验证码传递给 2captcha 服务器：

<?php
$id_Captcha=0;
$key_2captcha="key2captcha";
function send_captcha($base_file){

   $ch = curl_init("http://2captcha.com/in.php");
   curl_setopt($ch, CURLOPT_POSTFIELDS,
               array('method'=>"base64",
                     'key'=>$key_2captcha,
                     'numeric'=>1,
                     'max_len'=>1,
                     'body'=>$base_file,
                     'submit'=>'download and get the ID'));


   curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);


   $postResult = curl_exec($ch);


   curl_close($ch);

   return $postResult;
}

function getSolveCaptcha($id_captcha){
  $c = curl_init("http://2captcha.com/res.php?key=".$key_2captcha."&action=get&id=".$id_captcha);
  curl_setopt($c, CURLOPT_RETURNTRANSFER, 1);
  $postResult = curl_exec($c);
  curl_close($c);
  return $postResult;
}
?>

我运行这段代码在XAMPP.

Answer 1

使用下面的代码将 $key_2captcha 与全局一起使用。在这两个功能中。 read variable scope in PHP

function getSolveCaptcha($id_captcha){
  global $key_2captcha;

  $c = curl_init("http://2captcha.com/res.php?key=".$key_2captcha."&action=get&id=".$id_captcha);
  curl_setopt($c, CURLOPT_RETURNTRANSFER, 1);
  $postResult = curl_exec($c);
  curl_close($c);
  return $postResult;
}

Answer 2

将下面的代码与 $GLOBALS 一起使用 — 引用全局范围内可用的所有变量

<?php
    $id_Captcha=0;
    $key_2captcha="key2captcha";
    function send_captcha($base_file){

       $ch = curl_init("http://2captcha.com/in.php");
       curl_setopt($ch, CURLOPT_POSTFIELDS,
                   array('method'=>"base64",
                         'key'=>$GLOBALS['key_2captcha'],
                         'numeric'=>1,
                         'max_len'=>1,
                         'body'=>$base_file,
                         'submit'=>'download and get the ID'));


       curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);


       $postResult = curl_exec($ch);


       curl_close($ch);

       return $postResult;
    }

    function getSolveCaptcha($id_captcha){
      $c = curl_init("http://2captcha.com/res.php?key=".$GLOBALS['key_2captcha']."&action=get&id=".$id_captcha);
      curl_setopt($c, CURLOPT_RETURNTRANSFER, 1);
      $postResult = curl_exec($c);
      curl_close($c);
      return $postResult;
    }
    ?>

参考 PHP.net

Answer 3

我认为您遇到了变量作用域解析问题。

如果你想在泛型函数中使用这个变量，你必须在函数的签名中将这个变量作为参数传递。不要将变量用作全局变量，因为这是一种不好的做法，你必须制作通用函数，所以你必须使用通用参数。

试试这个代码：

<?php
$id_Captcha=0;
$key_2captcha="key2captcha";
function send_captcha($base_file, $key_2captcha){

   $ch = curl_init("http://2captcha.com/in.php");
   curl_setopt($ch, CURLOPT_POSTFIELDS,
               array('method'=>"base64",
                     'key'=>$key_2captcha,
                     'numeric'=>1,
                     'max_len'=>1,
                     'body'=>$base_file,
                     'submit'=>'download and get the ID'));


   curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);


   $postResult = curl_exec($ch);


   curl_close($ch);

   return $postResult;
}

function getSolveCaptcha($id_captcha, $key_2captcha){
  $c = curl_init("http://2captcha.com/res.php?key=".$key_2captcha."&action=get&id=".$id_captcha);
  curl_setopt($c, CURLOPT_RETURNTRANSFER, 1);
  $postResult = curl_exec($c);
  curl_close($c);
  return $postResult;
}

//Call Example
send_captcha($base_file, $key_2captcha);
?>

为什么这些函数看不到我的变量？

Why can't these functions see my variable?

php

2captcha