PHP Dropdown 填充年份 - 如何选择当前年份答案

【问题标题】：PHP Dropdown populating years - how to select current yearPHP Dropdown 填充年份 - 如何选择当前年份
【发布时间】：2016-02-02 20:09:39
【问题描述】：

我有这个 PHP 代码在 SELECT 框中填充过去 10 年和未来 10 年的值。

<select name="fromYear"';
   $starting_year  =date('Y', strtotime('-10 year'));
   $ending_year = date('Y', strtotime('+10 year'));

    for($starting_year; $starting_year <= $ending_year; $starting_year++) {
 echo '<option value="'.$starting_year.'">'.$starting_year.'</option>';
  }             
 echo '<select>

如何让它自动选择当前年份？

【问题讨论】：

您可以使用$curyear = date('Y') 获取当前年份，因此只需检查循环内的内容即可。

标签： php

【解决方案1】：

<select name="fromYear"';
 $starting_year  =date('Y', strtotime('-10 year'));
 $ending_year = date('Y', strtotime('+10 year'));
 $current_year = date('Y');
 for($starting_year; $starting_year <= $ending_year; $starting_year++) {
     echo '<option value="'.$starting_year.'"';
     if( $starting_year ==  $current_year ) {
            echo ' selected="selected"';
     }
     echo ' >'.$starting_year.'</option>';
 }               
 echo '<select>';

【讨论】：

【解决方案2】：

<?php
//get the current year
$Startyear=date('Y');
$endYear=$Startyear-10;

// set start and end year range i.e the start year
$yearArray = range($Startyear,$endYear);
?>
<!-- here you displaying the dropdown list -->
<select name="year">
    <option value="">Select Year</option>
    <?php
    foreach ($yearArray as $year) {
        // this allows you to select a particular year
        $selected = ($year == $Startyear) ? 'selected' : '';
        echo '<option '.$selected.' value="'.$year.'">'.$year.'</option>';
    }
    ?>
</select>

【讨论】：

最好按照规范定义selected：$selected = ($year == $Startyear) ? 'selected="selected"' : '';，否则浏览器可能会忽略选中状态。

【解决方案3】：

用当前年份检查年份并选择它。

for($starting_year; $starting_year <= $ending_year; $starting_year++) {
    if($starting_year == date('Y')) {
        echo '<option value="'.$starting_year.'" selected="selected">'.$starting_year.'</option>';
    } else {
        echo '<option value="'.$starting_year.'">'.$starting_year.'</option>';
    }
}

【讨论】：

【解决方案4】：

将您的回声更改为：

 echo '<option'.($starting_year == date('Y')) ? "selected=\"selected\"" : "".' value="'.$starting_year.'">'.$starting_year.'</option>';

【讨论】：

【解决方案5】：

echo '<select name="fromYear">';
$cur_year = date('Y');
for($year = ($cur_year-10); $year <= ($cur_year+10); $year++) {
    if ($year == $cur_year) {
        echo '<option value="'.$year.'" selected="selected">'.$year.'</option>';
    } else {
        echo '<option value="'.$year.'">'.$year.'</option>';
    }
}               
echo '<select>';

【讨论】：

【解决方案6】：

您应该使用selected attributed

$starting_year  =date('Y', strtotime('-10 year'));
$ending_year = date('Y', strtotime('+10 year'));
for($starting_year; $starting_year <= $ending_year; $starting_year++) {
   if(date('Y')==$starting_year) { //is the loop currently processing this year?
      $selected='selected'; //if so, save the word "selected" into a variable
   } else {  
      $selected='' ; //otherwise, ensure the variable is empty
   }
   //then include the variable inside the option element
   echo '<option '.$selected.' value="'.$starting_year.'">'.$starting_year.'</option>';
}

【讨论】：

【解决方案7】：

for($starting_year; $starting_year <= $ending_year; $starting_year++) { 
 echo '<option value="'.$starting_year.'"'. if(starting_year == date('Y')) echo "selected"  .'>'.$starting_year.'</option>';   

}

【讨论】：

【解决方案8】：

for($starting_year; $starting_year <= $ending_year; $starting_year++) {
    if($starting_year == date('Y')){
        echo '<option selected=selected value="'.$starting_year.'">'.$starting_year.'</option>';
    }else{
        echo '<option value="'.$starting_year.'">'.$starting_year.'</option>';
    }
}

请使用上面的代码 for 循环，而不是你的。

【讨论】：