检查FK订单是否编号正确

时间:2015-02-18 14:20:54

标签: php html mysql arrays foreign-keys

我有一些缺席,每次缺席都有一个FK给员工,我想在生成的HTML表格中显示。但是,如果我删除员工,则订单从0,1,2变为1,2(例如,如果第二个员工被删除)。

这搞砸了我的代码,因为我需要FK来检查我必须在哪里插入假(<tr>)。在这里,我统计了员工:

$result = mysql_query("select count(1) FROM employee");
$row    = mysql_fetch_array($result);
$count_user = $row[0];

稍后在我的代码中我在循环中进行查询。循环运行的次数与用户一样多。并且问题是:如果一个员工被删除它将不会到达另一个($ i变为0然后如果有2个用户则为1,但如果删除一个,则FK为3,因此需要进一步) 。 到

 $result = mysql_query("select start, end, type_FK, employee_FK FROM absences where employee_FK = $i");
                            while ($row = mysql_fetch_assoc($result)) {
                            $array_absences[] = $row;
                            }

有没有人的想法不是基于FK订单? 此主题还需要更好的标题,因此可以随意编辑它。

<html>
    <head>
        <title>Absence System</title>
    </head>

    <body>
        <?php
            $con = mysql_connect("localhost", "root", "");
            if (!$con) {
                die('Could not connect: ' . mysql_error());
            }

            mysql_select_db("absence_system", $con);


            $result = mysql_query("select count(1) FROM employee");
            $row    = mysql_fetch_array($result);
            $count_user = $row[0];




            $result = mysql_query("select count(1) FROM absences");
            $row    = mysql_fetch_array($result);
            $count_absences = $row[0];



            $result = mysql_query("select name, surename, employee_ID FROM employee");
                while ($row = mysql_fetch_assoc($result)) {
                    $array_user[] = $row;
            }


            for($i = 0; $i < $count_user; $i++){
                echo '<table = border = 1px>';
                    echo '</tr>';

                        $result = mysql_query("select start, end, type_FK, employee_FK FROM absences where employee_FK = $i");
                        while ($row = mysql_fetch_assoc($result)) {
                        $array_absences[] = $row;
                        }

                        $count = count($array_absences);
                        echo $count;

                        print_r($array_absences);

                        for($j = 0; $j < 32; $j++){
                        $true = 0;
                        if($j == 0){
                        echo '<td>';
                        echo $array_user[$i]['name'], " ", $array_user[$i]['surename'] ;
                        echo '</td>';
                        }

                        for($k = 0; $k < $count; $k++)
                        {

                        $array_absences[$k]['start'] = substr($array_absences[$k]['start'], -2);
                        $array_absences[$k]['end']   = substr($array_absences[$k]['end'], -2);

                        $array_absences[$k]['start'] = ereg_replace("^0", "", $array_absences[$k]['start']);
                        $array_absences[$k]['end']   = ereg_replace("^0", "", $array_absences[$k]['end']);

                        if($j == $array_absences[$k]['start'] && $array_absences[$k]['employee_FK'] == $i){
                        $true = 1;
                        echo '<td>';
                        echo $array_absences[$k]['type_FK'];
                        echo '</td>';
                        }
                        }


                        if($j != 0 && $true == 0){
                        echo '<td>';
                        echo "$j";
                        echo '</td>';
                        }
                        }
                    echo '</tr>';
                echo '</table>';
            }
        ?>
    </body>

</html>

1 个答案:

答案 0 :(得分:1)

如果employee_ID对应employee_FK,那么你不应该迭代$count_user,而是迭过$array_user - 那么你的选择就会像这样:

$result = mysql_query("select start, end, type_FK, employee_FK FROM absences where employee_FK = {$array_user[$i]['employee_ID']}");

虽然$i仍然代表序列号,因为

$count_user == count($array_user)