根据所选年份显示选项

我想根据他选择的年份向用户显示选项。在我有一个函数来获取一年，发回这view2.php同一页面文件的顶部，但我敢肯定，我做错了：根据所选年份显示选项

if(!empty($_POST['year'])){ 
    $year = htmlentities($_POST['year']); 
    header('Location: view2.php'); 
    //exit(); 
}

后来，我有一个表格提交，看起来像这样

<form method="post" action=""> 

    Select year: 
    <?php 
     $myOptions = array(
      'All' => 'All', 
      '2013' => '2013', 
      '2014' => '2014', 
      '2015' => '2015', 
      '2016' => '2016', 
      '2017' => '2017', 
      '2018' => '2018', 
      '2019' => '2019', 
      '2020' => '2020' 

     ); 
    ?> 
    <select name="year"> 
     <?php 
     foreach($myOptions as $key => $opt) { 
      $selected = null; 

     if(isset($_POST['View All']) && $key === $_POST['View All']) 
      $selected = ' selected'; 

     echo '<option value="' . $key . '"' . $selected. '>' . $opt . '</option>'; 
     } 
     ?> 
    </select>  
    <input type="submit" name="submit" value="GO"/> 
    </form>

如果我手动设置$year变量，我的SQL查询工作正常年份，因此函数（$users->expensePerYear($username, $year, $i);）是好的。这是我的功能：

<?php 
    $i=0; 
    $year = 2013; 
     //if (isset($year)){ 
     $users->expensePerYear($username, $year, $i); 
     $userQueryResult = $users->queryResult; 
     $count = $users->i; 
     $totalSpent = $users->totalSpent; 
     $remaining = $budget-$totalSpent;

问题是，我试图发送年份变量到同一页，我做错了。我希望用户在select中指定年份，并将其发送到同一页面并根据该变量显示内容。

来源

2013-11-15 Cataneo

没关系我，我只需要对header（）行进行注释并且它可以工作。

if (isset($_POST['submit2'])) { 
    $year = htmlentities($_POST['year']); 
    //header('Location: view2.php?$year=".$year."'); 
    //echo $year; 
}

来源

2013-11-15 09:58:23 Cataneo

根据所选年份显示选项

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