我有一些缺勤,每次缺勤都有一个员工的 FK,我想将其显示在生成的 HTML 表中。但是,如果我删除一名员工,顺序将从 0, 1, 2 更改为 1, 2(例如,如果第二名员工被删除)。
这搞乱了我的代码,因为我需要 FK 来检查我必须在哪里插入休假(其中 <tr>
)。这里我统计一下员工:
$result = mysql_query("select count(1) FROM employee");
$row = mysql_fetch_array($result);
$count_user = $row[0];
稍后在我的代码中我在循环中进行查询。循环运行的次数与用户的数量一样多。问题来了:如果一名员工被删除,它将无法到达另一名员工(如果有 2 个用户,$i 会先变为 0,然后变为 1,但如果删除其中一个,则 FK 为 3,因此需要再进一步) 。 至
$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;
}
有没有人有不基于 FK 顺序的想法? 此外,该主题需要一个更好的标题,因此请随意编辑它。
<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>
最佳答案
如果employee_ID对应于employee_FK,那么您不应该迭代$count_user
,而应该迭代$array_user
- 那么您的选择将如下所示:
$result = mysql_query("select start, end, type_FK, employee_FK FROM absences where employee_FK = {$array_user[$i]['employee_ID']}");
而$i
仍然代表序列号,因为
$count_user == count($array_user)
关于php - 检查FK订单编号是否正确,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28585908/