Question

我现在已经搞乱了3天，研究和试验：是时候寻求帮助了。

我有一些代码用于调度数据库表的调度显示页面，每个约会类型一次。它收集本周所有类型的约会，并将它们连续返回，如下所示：

+------+------+-----+-----+-----+-----+-----+-----+-----+------+
| Name | Year | Mon | Tue | Wed | Thu | Fri | Sat | Sun | Purp |
+------+------+-----+-----+-----+-----+-----+-----+-----+------+
| IST  |  NA  | 9-4 |     |     |     |     |     |     | ABC  |
+------+------+-----+-----+-----+-----+-----+-----+-----+------+
| OT   |  NA  |     | 8-2 | 8-2 | 8-2 | 8-2 |     |     | DEF  |
+------+------+-----+-----+-----+-----+-----+-----+-----+------+

我的代码完全符合我的要求，但有一个致命的缺陷。如果＆＃34;名称＆＃34;有多行，我会将其作为我的行：

+------+------+-----+-----+-----+-----+-----+-----+-----+------+
| Name | Year | Mon | Tue | Wed | Thu | Fri | Sat | Sun | Purp |
+------+------+-----+-----+-----+-----+-----+-----+-----+------+
| IST  |  NA  | 9-4 |     |     |     |     |     |     | ABC  |
+------+------+-----+-----+-----+-----+-----+-----+-----+------+
| OT   |  NA  |     | 8-2 | 8-2 | 8-2 | 8-2 |     |     | DEF  |
+------+------+-----+-----+-----+-----+-----+-----+-----+------+-+-----+-+-+-+-+-+-----+
| MTG  |      | 1-2 |     |     |     |     |     |     |      | | 1-2 | | | | | | GHI |
|      |      |     |     |     |     |     |     |     |      | |     | | | | | | JKL |
+------+------+-----+-----+-----+-----+-----+-----+-----+------+-+-----+-+-+-+-+-+-----+

因此，它不是插入适当的＆lt; td＆gt;而是创建一组全新的。这真的令人沮丧，因为我确信它很简单，但是我看不到它...... :(

代码：

$apptnamestop = array("IST", "OT", "MTG", "TR-CN", "EVENT", "EN", "REC", "TO");
$daysofweek = array("1", "2", "3", "4", "5", "6", "0");
foreach ( $apptnamestop as $name) {
    print str_repeat($tab, 8) . "<tr>\n";
    print str_repeat($tab, 9) . "<td class=\"td1s\">$name</td>\n";
    print str_repeat($tab, 9) . "<td class=\"td1s\">N/A</td>\n";
    $getdataquery = "SELECT appointmentName as name
, appointmentType as type, appointmentStartDateTime as sdt
, appointmentEndDateTime as edt, appointmentPurpose as purp 
from tblAppointments 
where appointmentStaffIDsToAttend like '%{$_SESSION['user_id']}%' 
and appointmentName = '$name' 
and (appointmentStartDateTime >= DATE_ADD(CURDATE(), INTERVAL (9 - IF(DAYOFWEEK(CURDATE())=1, 8, DAYOFWEEK(CURDATE()))) DAY) 
and appointmentEndDateTime < DATE_ADD(CURDATE(), INTERVAL (16 - IF(DAYOFWEEK(CURDATE())=1, 8, DAYOFWEEK(CURDATE()))) DAY))
";
    $getdataqueryresults = $mysqli->query($getdataquery)  
       or trigger_error("<p class=\"error\">We're very sorry, but an error has occurred when interacting with the CHAIRS database.  Please try again and see if the error repeats.  If it does, please get the following information in its entirety to your database adminapptrator so the CHAIRS developer can get the error resolved.<br />Error Message: " . $mysqli->error, E_USER_ERROR);
    $datarowcnt = $getdataqueryresults->num_rows;
    if ($datarowcnt > 0) {
        while ($row = $getdataqueryresults->fetch_row()) {
            $rows[] = $row;
        }

        foreach ($rows as $row) {
            $title = $row[0];
            $type = $row[1];
            $sdt = $row[2];
            $edt = $row[3];
            $purp = $row[4];

            $c=7;

            if ($type == 1) {
                $typew = "Mandatory";
            } else {
                $typew = "Elective";
            }

            $sparts = explode(" ", $sdt);
            $eparts = explode(" ", $edt);
            $tdiff = getTimeDiff($sparts[1], $eparts[1]);

            foreach ($daysofweek as $day) {
                if ($title == $name) {
                    if ($day == date('w', strtotime("$sparts[0]"))) {
                        if ($sparts[0] == $eparts[0]) {
                            print str_repeat($tab, 9) . "<td class=\"td1s\">$sparts[1] - $eparts[1]<br />($tdiff) - $typew</td>\n";
                            $c--;
                        } else {
                            $s = strtotime("$sparts[0]");
                            $e = strtotime("$eparts[0]");

                            for ($i=$s; $i<=$e; $i+=86400) {
                                print str_repeat($tab, 9) . "<td class=\"td1s\">$sparts[1] - $eparts[1]<br />($tdiff) - $typew</td>\n";
                                $c--;
                            }
                        }
                    } 
                    if ( $c > 0) {
                            $c--;
                            print str_repeat($tab, 9) . "<td class=\"td1s\"></td>\n";
                    }
                }
            }
        }
        $rc++;
    } else {
        foreach ($daysofweek as $day) {
            print str_repeat($tab, 9) . "<td class=\"td1s\"></td>\n";
        }
    }
    print str_repeat($tab, 9) . "<td class=\"td1s\">$purp</td>\n";
    $purp = "";
    print "</tr>\n";
}

Answer 1

新行在此处开始：

foreach ( $apptnamestop as $name) {
    print str_repeat($tab, 8) . "<tr>\n";
    print str_repeat($tab, 9) . "<td class=\"td1s\">$name</td>\n";

如您所见，这个新行在$apptnamestop中每个元素启动一次，而不是每行从数据库启动一次。代码并不关心它在表中找到多少匹配的行，它只会在每个名称中插入一个新的tr。

根据您的需要，有几种方法可以改善这种情况。

如果您需要始终为每个名称插入一行，无论数据库中是否存在匹配的行，请保持第4,5和6行与现在相同。我们需要修改内部循环（foreach ($rows as $row)）以输出超出第一个循环的每个循环的新表行：

$rowCounter = 0;
foreach ($rows as $row) {
    if ($rowCounter > 0) {
        print "</tr>\n"; // Close the previous table row
        print str_repeat($tab, 8) . "<tr>\n";
        print str_repeat($tab, 9) . "<td class=\"td1s\">$name</td>\n";
        print str_repeat($tab, 9) . "<td class=\"td1s\">N/A</td>\n";
    }
    $rowCounter++;
    // Remainder of the loop code goes here
    ...
}

如果您希望在找不到匹配的数据库行时跳过表行的创建，请将行创建语句完全移动到内部循环中;这样它只有在找到数据库行时才会运行。

最后，请考虑实施视图模板系统;通过将逻辑代码与显示代码分开，您可以轻松地修改任何一个而不会破坏另一个。您的代码将来会更容易阅读和修改。

PHP mysqli fetch_row - 迭代多个记录

1 个答案: