Php 带有日期范围问题的周号下拉列表

我想用日期范围显示一年中当前周数和下一周数的下拉列表

我用过这个代码

  for ($i = $j; ($i <= $j + 1) && ($i <= 52); $i++) {
    $str = '';
    if (in_array($i, $week_no_array)) {
        $str ='selected = "selected"';
    }
    ?>

        <option value="<?php echo $i; ?>" <?php echo $str; ?> >
        <?php
    $year = date('Y');

    $week_start_date = date('d/m/Y', strtotime($year . "W" . $i . '1'));
    $week_end_date = date('d/m/Y', strtotime($year . "W" . $i . '7'));

    echo $i . ' (' . $week_start_date . ' - ' . $week_end_date . ')'
    ?>
        </option>

对于（$i=$j；（$i更改行：

  for ($i = $j; ($i <= $j + 1) && ($i <= 52); $i++) {
    $str = '';
    if (in_array($i, $week_no_array)) {
        $str ='selected = "selected"';
    }
    ?>

        <option value="<?php echo $i; ?>" <?php echo $str; ?> >
        <?php
    $year = date('Y');

    $week_start_date = date('d/m/Y', strtotime($year . "W" . $i . '1'));
    $week_end_date = date('d/m/Y', strtotime($year . "W" . $i . '7'));

    echo $i . ' (' . $week_start_date . ' - ' . $week_end_date . ')'
    ?>
        </option>

$year=日期（'Y'）；
  for ($i = $j; ($i <= $j + 1) && ($i <= 52); $i++) {
    $str = '';
    if (in_array($i, $week_no_array)) {
        $str ='selected = "selected"';
    }
    ?>

        <option value="<?php echo $i; ?>" <?php echo $str; ?> >
        <?php
    $year = date('Y');

    $week_start_date = date('d/m/Y', strtotime($year . "W" . $i . '1'));
    $week_end_date = date('d/m/Y', strtotime($year . "W" . $i . '7'));

    echo $i . ' (' . $week_start_date . ' - ' . $week_end_date . ')'
    ?>
        </option>


  for ($i = $j; ($i <= $j + 1) && ($i <= 52); $i++) {
    $str = '';
    if (in_array($i, $week_no_array)) {
        $str ='selected = "selected"';
    }
    ?>

        <option value="<?php echo $i; ?>" <?php echo $str; ?> >
        <?php
    $year = date('Y');

    $week_start_date = date('d/m/Y', strtotime($year . "W" . $i . '1'));
    $week_end_date = date('d/m/Y', strtotime($year . "W" . $i . '7'));

    echo $i . ' (' . $week_start_date . ' - ' . $week_end_date . ')'
    ?>
        </option>

与：

$year=日期（'o'）；
  for ($i = $j; ($i <= $j + 1) && ($i <= 52); $i++) {
    $str = '';
    if (in_array($i, $week_no_array)) {
        $str ='selected = "selected"';
    }
    ?>

        <option value="<?php echo $i; ?>" <?php echo $str; ?> >
        <?php
    $year = date('Y');

    $week_start_date = date('d/m/Y', strtotime($year . "W" . $i . '1'));
    $week_end_date = date('d/m/Y', strtotime($year . "W" . $i . '7'));

    echo $i . ' (' . $week_start_date . ' - ' . $week_end_date . ')'
    ?>
        </option>

$i=sprintf（'%02d'，$i）；

  for ($i = $j; ($i <= $j + 1) && ($i <= 52); $i++) {
    $str = '';
    if (in_array($i, $week_no_array)) {
        $str ='selected = "selected"';
    }
    ?>

        <option value="<?php echo $i; ?>" <?php echo $str; ?> >
        <?php
    $year = date('Y');

    $week_start_date = date('d/m/Y', strtotime($year . "W" . $i . '1'));
    $week_end_date = date('d/m/Y', strtotime($year . "W" . $i . '7'));

    echo $i . ' (' . $week_start_date . ' - ' . $week_end_date . ')'
    ?>
        </option>



在您的foreach循环中，您有条件
$i本周是2014年的第1周。不清楚您是否认为不应该，或者您是否知道不应该，但您的代码返回2013年。请您澄清一下，好吗？谢谢，但我还想展示前3周的情况，因此我必须在代码中进行更改。
  for ($i = $j; ($i <= $j + 1) && ($i <= 52); $i++) {
    $str = '';
    if (in_array($i, $week_no_array)) {
        $str ='selected = "selected"';
    }
    ?>

        <option value="<?php echo $i; ?>" <?php echo $str; ?> >
        <?php
    $year = date('Y');

    $week_start_date = date('d/m/Y', strtotime($year . "W" . $i . '1'));
    $week_end_date = date('d/m/Y', strtotime($year . "W" . $i . '7'));

    echo $i . ' (' . $week_start_date . ' - ' . $week_end_date . ')'
    ?>
        </option>