通过命令行跳过输入的用户与PHP的交互

Question

我有3个相同的片段要求用户通过命令行输入。第一个查询被询问给用户和第三个查询，但由于某些原因，第二个代码段虽然代码相同，但输出不正确。下面是输出：

正如你可以看到它跳过content 2并直通内容3.任何想法是什么问题呢？

只要我点击内容1的y，代码就会直接运行到Would you like to print content 3！

<?php 

echo "\n\033[1;35m~~~~~~ CONTENT 1 ~~~~~~\033[0m\n\n"; 

echo "\033[1;37mWould you like to print content 1? (y/n) - "; 
$stdin = fopen('php://stdin', 'r'); 
$response = fgetc($stdin); 
    if ($response == 'y') { 
     echo "\033[0m"; 
     echo "Content 1 print success!"; 
    } 

echo "\n\033[1;35m~~~~~~ CONTENT 2 ~~~~~~\033[0m\n\n"; 

echo "\033[1;37mWould you like to print content 2? (y/n) - "; 
$stdin = fopen('php://stdin', 'r'); 
$response = fgetc($stdin); 
    if ($response == 'y') { 
     echo "\033[0m"; 
     echo "Content 2 print success!"; 

    } 

echo "\n\033[1;35m~~~~~~ CONTENT 3 ~~~~~~\033[0m\n\n"; 

echo "\033[1;37mWould you like to print content 3? (y/n) - "; 
$stdin = fopen('php://stdin', 'r'); 
$response = fgetc($stdin); 
      if ($response == 'y') { 
       echo "\033[0m"; 
       echo "Content 3 print success!\n"; 
      } 
echo "\033[0m"; 

?> 

使用DO /同时建议它可以让我输入了3次，但这次内容2和3输入请求被复制！

在这里看到：

<? 

    echo "\n\033[1;35m~~~~~~ CONTENT 1 ~~~~~~\033[0m\n\n"; 

    do { 
     echo "\033[1;37mWould you like to print content 1? (y/n) - "; 
     $stdin = fopen('php://stdin', 'r'); 

     $response = fgetc($stdin); 
    } while (!in_array($response, ['y','n'])); 
    if ($response == 'y') { 
     echo "\033[0m"; 
     echo "Content 1 print success!"; 
    } 

    echo "\n\033[1;35m~~~~~~ CONTENT 2 ~~~~~~\033[0m\n\n"; 

    do { 
     echo "\033[1;37mWould you like to print content 2? (y/n) - "; 
     $stdin = fopen('php://stdin', 'r'); 

     $response = fgetc($stdin); 
    } while (!in_array($response, ['y','n'])); 
    if ($response == 'y') { 
     echo "\033[0m"; 
     echo "Content 2 print success!"; 
    } 

    echo "\n\033[1;35m~~~~~~ CONTENT 3 ~~~~~~\033[0m\n\n"; 

    do { 
     echo "\033[1;37mWould you like to print content 3? (y/n) - "; 
     $stdin = fopen('php://stdin', 'r'); 

     $response = fgetc($stdin); 
    } while (!in_array($response, ['y','n'])); 
    if ($response == 'y') { 
     echo "\033[0m"; 
     echo "Content 3 print success!\n"; 
    } 


    ?> 

莫非因为输入一些空格或者其捕获新行？对不起，我只是一味地猜测说实话......

----- ----- UPDATE

如果statment在迫使回音，我可以得到我的代码，通过把一个额外的工作只打印一次。它是一个黑客，但如果任何人都可以想出更好的解决方案，请让我知道！

echo "\n\033[1;35m~~~~~~ CONTENT 3 ~~~~~~\n\n"; 
    $x = "1"; 
    do { 
     if ($x==1){ 
      echo "\033[1;37mWould you like to print content 2? (y/n) - \033[0m\n"; 
      $x = $x+1; 

     } 
     $stdin = fopen('php://stdin', 'r'); 

     $response = fgetc($stdin); 
    } while (!in_array($response, ['y','n'])); 
    if ($response == 'y') { 
     echo "\033[0m"; 
     echo "Content 3 print success!\n"; 
    } 

Answer 1

你的代码在这里工作。无论如何，如果你想获得有效的输入，使用该do/while，只接受Y或N作为输入：

echo "\n\033[1;35m~~~~~~ CONTENT 1 ~~~~~~\033[0m\n\n"; 

do { 
    echo "\033[1;37mWould you like to print content 1? (y/n) - "; 
    $stdin = fopen('php://stdin', 'r'); 

    $response = fgetc($stdin); 
} while (!in_array($response, ['y','n'])); 
if ($response == 'y') { 
    echo "\033[0m"; 
    echo "Content 1 print success!"; 
}