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我要自动编号,而是由用户组PHP的编号通过唯一的ID分组
查询:
$query = "select distinct(DATE_FORMAT(a.in_time, '%Y%m%d')) as in_time, a.user_id, a.notes, b.nama from attendance a left join dat_login b on(a.user_id=b.ID) where DATE_FORMAT(a.in_time, '%Y')='2017' and DATE_FORMAT(a.in_time, '%M')='April' and id_shop!=2 order by b.nama asc";
$querynm = "select distinct(b.nama) as nama from attendance a left join dat_login b on(a.user_id=b.ID) where DATE_FORMAT(a.in_time, '%Y')='2017' and DATE_FORMAT(a.in_time, '%M')='April' group by b.nama order by b.nama asc";
$stid = mysqli_query($conn, $query) or die (mysqli_error($conn));
$stnm = mysqli_query($conn, $querynm) or die (mysqli_error($conn));
结果:
$alphabet_start = 'A';
while ($data = mysqli_fetch_assoc($stid)) {
$data['user_id'] += 1;
$no_cell = 11;
echo $data['nama'] . " ";
echo $data['user_id'] . " ";
if ($data['notes']=="") {
echo chr(ord("A") + date('d', strtotime($data['in_time']))) . $no_cell . " - " . chr(ord("A") + date('d', strtotime($data['in_time']))) . ". " . "Masuk <br>";
} else if ($data['notes']=="S") {
echo chr(ord("A") + date('d', strtotime($data['in_time']))) . $no_cell . " - " . chr(ord("A") + date('d', strtotime($data['in_time']))) . ". " . "Sakit <br>";
} else if ($data['notes']=="L") {
echo chr(ord("A") + date('d', strtotime($data['in_time']))) . $no_cell . " - " . chr(ord("A") + date('d', strtotime($data['in_time']))) . ". " . "Libur <br>";
}
$alphabet_start++;
$no_cell++;
$data['user_id']++;
}
但是,该值显示是这样的:
我想要的值显示如下:
正如你所看到的数量在接下来的名称改变,该怎么办呢? 对不起,我的语法错误。以前感谢:)
谢谢你的答案先生。 styl3r,但user_id值是从数据库中获得的,第一个值并不总是52,而user_id值非常多。我不能一个一个地在开关盒里做。 –
你如何计划增加'$ no_cell'? – styl3r
$ no_cell在每个名称中递增,一个名称为一个名称 –