变量在 FOR 循环外不可用答案

【问题标题】：Variable not available outside FOR loop变量在 FOR 循环外不可用
【发布时间】：2011-09-22 22:09:59
【问题描述】：

我需要获取前一周的第一个和最后一个日期。我使用下面的代码，它工作正常，直到我尝试在循环外读取变量。

<?php
$current_week = date(W);
$previous_week = $current_week - 1;
$year = date(o);
for($day=1; $day<=7; $day++) {
  $temp = array ($day => date('Y-m-d', strtotime($year."W".$previous_week.$day)) . '   00:00:00');
  $firstday = $temp[1];
  // echo here works fine
  echo $firstday;
}

// echo here doesnt work anymore
echo $firstday;
?>

我真的不知道为什么这不起作用。我已经阅读了 PHP 文档，一切都告诉我它应该可以工作。我一点也不笨，现在我真的很困惑。任何人都可以对此有所了解吗？

提前致谢！

【问题讨论】：

标签： php variables date for-loop

【解决方案1】：

$firstDay 正在被重用和覆盖：

<?php
$current_week = date(W);
$previous_week = $current_week - 1;
$year = date(o);
for($day=1; $day<=7; $day++) {
  $temp = array ($day => date('Y-m-d', strtotime($year."W".$previous_week.$day)) . '   00:00:00');
  $firstday = $temp[1];
  // echo here works fine
  echo "Inloop - ".$firstday."\n";
}

// echo here doesnt work anymore
echo "After loop - ".$firstday."\n";
?>

Results:
Inloop - 2011-09-12   00:00:00
Inloop - 
Inloop - 
Inloop - 
Inloop - 
Inloop - 
Inloop - 
After loop -

希望对你有帮助

【讨论】：

谢谢！这帮助我解决了问题:)

【解决方案2】：

您在循环的后续迭代中覆盖了$firstday 变量，因此它最后包含null。

如果您打开错误报告，您将在访问$temp[1] 时看到由$i > 1 迭代生成的通知

您应该始终在开发和学习时打开错误报告，并且您应该（或最好必须）在发布代码之前解决所有警告。

以下是脚本中的警告，其中 error_reporting 设置为 E_ALL：

> php /tmp/foo.php 2>&1 | grep Notice
PHP Notice:  Use of undefined constant W - assumed 'W' in /private/tmp/foo.php on line 3
Notice: Use of undefined constant W - assumed 'W' in /private/tmp/foo.php on line 3
PHP Notice:  Use of undefined constant o - assumed 'o' in /private/tmp/foo.php on line 5
Notice: Use of undefined constant o - assumed 'o' in /private/tmp/foo.php on line 5
2011-09-12   00:00:00PHP Notice:  Undefined offset: 1 in /private/tmp/foo.php on line 8
Notice: Undefined offset: 1 in /private/tmp/foo.php on line 8
PHP Notice:  Undefined offset: 1 in /private/tmp/foo.php on line 8
Notice: Undefined offset: 1 in /private/tmp/foo.php on line 8
PHP Notice:  Undefined offset: 1 in /private/tmp/foo.php on line 8
Notice: Undefined offset: 1 in /private/tmp/foo.php on line 8
PHP Notice:  Undefined offset: 1 in /private/tmp/foo.php on line 8
Notice: Undefined offset: 1 in /private/tmp/foo.php on line 8
PHP Notice:  Undefined offset: 1 in /private/tmp/foo.php on line 8
Notice: Undefined offset: 1 in /private/tmp/foo.php on line 8
PHP Notice:  Undefined offset: 1 in /private/tmp/foo.php on line 8
Notice: Undefined offset: 1 in /private/tmp/foo.php on line 8

【讨论】：

谢谢！这帮助我解决了问题:)
@SjaakdeVries 查看我对答案的更新并应用我的建议以便从您的错误中学习。