Question

我有一个像这样的查询创建的表：

while ($row = mysqli_fetch_array($result))
    {       
        $fields[] = $row['unit']; $fields[] = $row['reservation']; $fields[] = $row['guest'];
        $fields[] = $row['arrival']; $fields[] = $row['departure']; $fields[] = $row['balance'];
        $fields[] = $row['checked_in'];

        //rowid == id of table (the row corresponds to the row in the table)
        $rowid = $row['reservation'];

        $count = count($fields);

        //make checkboxes
        echo "<tr class='edit_tr' id='".$rowid."'>";
        echo "<td><input type='checkbox' value='".$fields[1]."' name='checkins[]'/></td>"; 

        for ($j = 0; $j < $count; $j++)
        {
            echo "<td><span id='".$fields[1]."'>", ($j+1 == $count ? ($fields[$j] == 0 ? "Not Checked In" : "Checked In") : "".$fields[$j]."");
            echo "</span></td>"; 
        }

        echo "</tr>";

        //empty array
        $fields = "";
    }

我POST将PHP文件的所有复选框UPDATE发送到if(isset($_POST['checkins'])) { //open connection $con = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME); //create WHERE clause for new checkins $where = " WHERE id ="; foreach ($_POST['checkins'] as $id) { $where .= " $id OR id ="; } //remove last 8 characters from WHERE clause $where = substr($where, 0, -8); echo $query = "UPDATE reservations SET checked_in = CASE WHEN checked_in = 0 THEN 1 END $where)"; mysqli_query($con, $query); mysqli_close($con); }数据库。

POST

我知道我不能checked_in未经检查的框，但我需要一种方法来告诉PHP文件未选中的内容。这是我到目前为止所想到的：

1）查询所有checked_out = 1的数据库，其中checked_out = 0 （我只显示表中checked_in = 0的行）如果id不在POST结果中，则设置{{1}} = 0。

2）将ajax数组发送到另一个PHP文件，其中包含以前检查过的所有框的列表，现在不是。

我相信还有更多，但这就是我的想法。不过，它们似乎都是变通办法。有人有更好的解决方案吗？

POST之前选中的复选框到PHP

0 个答案: