Question

我只是无法正确解决这个问题，我有两个数组。第一个数组$ attendance，其中包含以下键和值

Array
(
[attendance] => Array
    (
        [1] => 1
        [2] => 1
        [3] => 0
        [4] => 1
        [5] => 1
        [6] => 0
        [7] => 0
    )

如果选中的值为1，则来自复选框。否则值为0 第二个数组是$ remark

[remark] => Array
    (
        [1] =>
        [2] => 
        [3] =>  'sick'
        [4] => 
        [5] => 
        [6] =>  'leave'
        [7] =>  'On assignment'

    )

现在这就是密钥1- 7所代表的内容，该脚本用于员工的出勤，密钥1-7是我数据库中employee表中的employeeID。

现在我想要实现的是以这样的方式连接数组

Array
(
[0] => Array
    (
        [employeeID] => 7
        [attendance] => 0
        [remark] => 'On assignment'
    )

[1] => Array
    (
        [employeeID] => 6
        [attendance] => 0
        [remark] => 'leave'
    )

[2] => Array
    (
        [employeeID] => 5
        [attendance] => 1
        [remark] => 
    )

//and so on
)

我正在使用Codeigniter如果我能够连接它，我也很想知道如何将多个数据插入到员工表中，如下所示，

employee's table
employeeID | date | status | remarks

我计划使用CURDATE（）的日期然后状态将从出勤中保持0或1 再次：备注和出勤数组中1-7的密钥都是employeeID

更新!! 这是我试过但没有用的。

$att = $_POST['attendance'];
$remarks = $_POST['remark'];

foreach($att as $employee_id=>$status)
        {
            $x=0;
            $employee[$x]['employee_id']=$employee_id;
            $employee[$x]['status']=$status;


            foreach($remarks as $employee_id=>$remark)
            {

                $employee[$x]['remark']=$remark;
                $x++;
            }
       }

Answer 1

$attendance = $_POST['attendance'];
$remarks = $_POST['remark'];

$collect = array();
foreach ($attendance as $employeeID => $attending) {
    $collect[] = array(
        "employeeID" => $employeeID,
        "attendance" => $attending,
        "remark" => isset($remarks[$employeeID]) ? $remarks[$employeeID] : ""
    );
}

$values = array();
foreach ($collect as $entry) {
    $values[] = "(" . ((int) $entry["employeeID"]) . ",
        CURDATE(),
        " . ((int) $entry["attendance"]) . ",
        '" . sql_escape($entry["remark"]) . "'
    )";
}
$query = "INSERT INTO `attendance`
    (`employeeID`, `date`, `status`, `remarks`)
    VALUES
    (" . implode(",", $values) . ")";

确保用适当的转义函数替换sql_escape。如果你正在使用PDO那么。

Answer 2

很简单，说实话......

$numOfEmployees = 8;

$concatArray = array();

foreach($attendance as $k => $v)
{
     $concatArray[] = array("employeeID" => $k, "attendance" => $v, "remark" => $remark[$k];
}

如果您的考勤表与SQL类似，您可以在上面提到的concatArray中使用foreach进行处理：

foreach($concatArray as $key => $value)
{
     $remark = $value['remark'];
     // sanitize remark for database
     $query = "INSERT INTO employees (ID, Date, Status, Remarks) VALUES ({$value['employeeID']}, NOW(), {$value['attendance']}, '{$value['remark']}');";
     // perform the actual query
}

请记住，这是一个非常一般的例子，对数据库做了很多假设。

正如Frits所说，应该对备注字段进行消毒，可能是对MySQLi使用mysqli_real_escape_string或对PDO使用quote - 取决于您使用的是什么。

LAMP考勤系统

2 个答案: