用户通过命令行跳过输入与 PHP 交互

Question

我有 3 个相同的 sn-ps，它们通过命令行要求用户输入。向用户询问第一个查询和第三个查询，但由于某种原因，第二个 sn-p 未正确输出，尽管其代码相同。下面是输出：

如您所见，它跳过 content 2 并直接运行到内容 3。您知道出了什么问题吗？

只要我点击 y 获取内容 1，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 / while 建议它允许我输入 3 次，但这次对于内容 2 和 3，输入请求是重复的！

看这里：

    <?

    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";
    }


    ?>

可能是因为输入有一些空白或捕获换行符吗？对不起，我只是盲目地猜测说实话......

----- 更新 -----

我可以通过添加一个额外的 if 语句来强制 echo 只打印一次来让我的代码工作。这是一个 hack，但如果有人能提出更好的解决方案，请告诉我！

    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

您的代码在这里有效。无论如何，如果您想获得有效的输入，请使用此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!";
}