Question

我需要获得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UNTIL我尝试读取循环外的变量。

<?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工作..我根本没有wizz，我现在真的很困惑。任何人都可以对此有所了解吗？

提前致谢！

Answer 1

您在循环的后续迭代中覆盖$firstday变量，因此最后它包含null。

如果您启用错误报告，则会在访问$i > 1

时看到通过$temp[1]的迭代生成通知

您应该始终在开发和学习时启用错误报告，并且您应该（或者更好地）在发布代码之前解决所有警告。

以下是您的脚本中警告，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

Answer 2

$ firstDay正在被重复使用和覆盖：

<?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 -

希望这有帮助

变量在FOR循环外不可用

2 个答案: