php - 基于当前日期php构建选择列表

Question

我创建了一个下拉选择列表，它根据当前日期生成一个日期列表，

4 天前和 7 天前，我尝试了几个较短的脚本，但都失败了，我一直在使用以下脚本，但它有一些限制......谢谢!

<option value=""> select </option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")-4),date("Y")));?>" style="color:red;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")-4),date("Y")));?></option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")-3),date("Y")));?>" style="color:red;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")-3),date("Y")));?></option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")-2),date("Y")));?>" style="color:red;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")-2),date("Y")));?></option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")-1),date("Y")));?>" style="color:red;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")-1),date("Y")));?></option>
<option value="<?php echo date ('m/d/Y');?>"><?php echo date ('m/d/Y');?> TODAY </option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+1),date("Y")));?>" style="color:green;font-weight:bold;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+1),date("Y")));?></option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+2),date("Y")));?>" style="color:green;font-weight:bold;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+2),date("Y")));?></option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+3),date("Y")));?>" style="color:green;font-weight:bold;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+3),date("Y")));?></option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+4),date("Y")));?>" style="color:green;font-weight:bold;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+4),date("Y")));?></option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+5),date("Y")));?>" style="color:green;font-weight:bold;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+5),date("Y")));?></option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+6),date("Y")));?>" style="color:green;font-weight:bold;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+6),date("Y")));?></option>
<option value="<?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+7),date("Y")));?>" style="color:green;font-weight:bold;"><?php echo date ("m/d/Y", mktime (0,0,0,date("m"),(date("d")+7),date("Y")));?></option>

Answer 1

这应该可以满足您的要求。我已经重写了您的代码，使其尽可能简单易读。

<?php
echo '<select>';
echo '<option value="">select</option>';

for($i = -4; $i <= 7; ++$i)
{
    $date = date('m/d/Y', strtotime($i.' days'));
    echo '<option value="'.$date.'" style="color:red;">'.$date.'</option>';
}

echo '</select>';
?>