我从数据库中选择了一些日期,并按月显示这些日期。我使用以下代码
$work_res = mysql_query("(SELECT DISTINCT date FROM `work_details` WHERE employee_id='" . $emp_id . "' and date between '" . $qsdate . "' and '" . $qedate . "') UNION (SELECT holy_date from holiday where holy_date between '" . $qsdate . "' and '" . $qedate . "')");
while ($row = mysql_fetch_array($work_res)) {
echo date("F", $test_date).'<br>';
while ((date("Y-m-d", $test_date) < $row['date']) && ($flag = 0)) {
if (!(date('N', strtotime(date("Y-m-d", $test_date))) >= 6)) {
echo "<tr ><td align=center class=fontclass style=color:FF0000>" . date("Y-m-d F", $test_date) . "</td></tr>";
}
$test_date = $test_date + ($day_incrementer * 60 * 60 * 24);
}
$flag = 1;
while ((date("Y-m-d", $test_date) != $row['date'])) {
if (!(date('N', strtotime(date("Y-m-d", $test_date))) >= 6)) {
echo "<tr><td align=center class=fontclass style=color:FF0000>" . date("Y-m-d F", $test_date) . "</td></tr>";
}
$test_date = $test_date + ($day_incrementer * 60 * 60 * 24);
}
$test_date = $test_date + ($day_incrementer * 60 * 60 * 24);
}
while (date("Y-m-d", $test_date) <= date("Y-m-d", $end_date)) {
if (!(date('N', strtotime(date("Y-m-d", $test_date))) >= 6)) {
echo "<tr><td align=center class=fontclass style=color:FF0000>" . date("Y-m-d F", $test_date) . "</td></tr>";
}
$test_date = $test_date + ($day_incrementer * 60 * 60 * 24);
}
return;
}
得到的结果像
2012-01-16 January
2012-01-26 January
2012-01-27 January
2012-02-02 February
2012-03-21 March
2012-03-22 March
我想显示这些日期,例如
January (3)
2012-01-16
2012-01-26
2012-01-27
February (1)
2012-02-02
March(2)
2012-03-21
2012-03-22
这可能吗?请帮忙
这是可以工作但需要测试的东西。您需要在 dateIsValid() 函数中插入逻辑。
<?php
$work_res = mysql_query("(SELECT DISTINCT date FROM `work_details` WHERE employee_id='" . $emp_id . "' and date between '" . $qsdate . "' and '" . $qedate . "') UNION (SELECT holy_date from holiday where holy_date between '" . $qsdate . "' and '" . $qedate . "')");
//Group all dates into their Months
while($result = @mysql_fetch_array($work_res))
{
$date = $result['date'];
$month = date("F",$date );
//Your complex logic in between...
if(dateIsValid())
$output[$month][] = date("Y-m-d", $date);
}
//Display as required
foreach($output as $month => $dates)
{
echo $month." (".count($dates).")"; //Echo the month heading
foreach($dates as $date) echo $date; //Echo the date
}
笔记:
- 这种分组在 SQL 中是可能的,并且很可能应该完成那边。
- 将所有逻辑移动到单独的函数
- 对于数据库访问,请使用PHP的PDO或像redBean这样的ORM。
- 在代码中编写注释 注释应解释已实现的逻辑。
- 使用更好的变量名称。
- 使用 DateTime 而不是 date()(请参阅下面的注释)。
你想要一个数组,其中键是完整的月份名称。你会想要做这样的事情...
下面是一个示例脚本:http://codepad.org/pJOpDr17
$dateArray=array();
while (($row = mysql_fetch_array($work_res)){
$d = new DateTime($row['date']);
$monthName = $d->format('F');
//$monthName = $d->format('F_Y'); (If you want month_year)
$dateArray[$monthName][] = $row['date'];
}
示例输出
array(3) {
["January"]=>
array(3) {
[0]=>
string(10) "2012-01-16"
[1]=>
string(10) "2012-01-26"
[2]=>
string(10) "2012-01-27"
}
["February"]=>
array(1) {
[0]=>
string(10) "2012-02-02"
}
["March"]=>
array(2) {
[0]=>
string(10) "2012-03-21"
[1]=>
string(10) "2012-03-22"
}
}
而不是
在将日期添加到数组时打印日期,例如
无论您在哪里使用 echo $test_date 附加到数组 say
$leavesTakes[]=$date("Y-m-d F", $test_date)
现在使用循环显示日期
for($i=0;$i<count($leavesTakes);$i++)
{
$cnt=0;
for($j=$i;$j<count($leavestaken);$j++)
{
if($leavesTaken[$i]!=$leavesTaken[$j])
break;
$cnt++;
}
//print month name
echo '<tr><td> month name (' + $cnt + ')'; //print month name
for(;$i<$j;$i++)
echo '<br/>$date("Y-m-d F", $leavesTaken[$i]);
echo '</td></tr>';
}
我不知道从日期打印月份名称的格式。所以请忽略打印月份名称,并替换为实际的php函数来打印月份名称
你可以做这样的事情来得到你想要的东西
像荷兰语432建议的那样使用日期时间
result = array();
while ($row = mysql_fetch_array($work_res))
{
$tmp_time = strtotime(row['date']);
$tmp_month = date('F',$tmp_time);
$tmp_date = date('Y-m-d', $tmp_time);
if(!is_array($result[$tmp_month]))
{
$result[$tmp_month] = array();
}
array_push(result[$tmp_month], $tmp_date);
}
# this will print something similar to your req
foreach($result as $month => $dates)
{
echo $month . " (" . $result[$month].length . ")";
foreach($dates as $date)
{
echo $date;
}
}