將表或數組添加到每個用戶的行中

Question

首先，我很抱歉發佈如此龐大的代碼塊。這可能與問題甚至不相關，但以防萬一......代碼保留了一個簡單的ToDo列表，我希望將其納入現有的PHP網站，該網站爲每個用戶存儲大量信息。換句話說，我想將它添加到mySQL DB中的用戶行信息中。

我是PHP的新手，但通過提出想法並弄清楚如何使它們工作，已經走過了很長的路。你能否指出我加入這樣一個功能的方向，即通過將刪除信息行添加到分配給用戶的字段列表中來存儲信息？

另一種說法：我想給我的用戶一種維護自己的待辦事項列表的方法。

<?php 
$conn = mysql_connect('server, 'db', 'password') or die(mysql_error()); 
$db = mysql_select_db('db',$conn) or die(mysql_error()); 
// if an arrow link was clicked... 
if ($_GET['dir'] && $_GET['id']) { 
    // make GET vars easier to handle 
    $dir = $_GET['dir']; 
    // cast as int and couple with switch for sql injection prevention for $id 
    $id = (int) $_GET['id']; 
    // decide what row we're swapping based on $dir 
    switch ($dir) { 
     // if we're going up, swap is 1 less than id 
     case 'up': 
     // make sure that there's a row above to swap 
     $swap = ($id > 1)? $id-- : 1; 
     break; 
     // if we're going down, swap is 1 more than id 
     case 'down': 
     // find out what the highest row is 
     $sql = "SELECT count(*) FROM info"; 
     $result = mysql_query($sql, $conn) or die(mysql_error()); 
     $r = mysql_fetch_row($result); 
     $max = $r[0]; 
     // make sure that there's a row below to swap with 
     $swap = ($id < $max)? $id++ : $max; 
     break; 
     // default value (sql injection prevention for $dir) 
     default: 
     $swap = $id; 
    } // end switch $dir 
    // swap the rows. Basic idea is to make $id=$swap and $swap=$id 
    $sql = "UPDATE info SET usort = CASE usort WHEN $id THEN $swap WHEN $swap THEN $id END WHERE usort IN ($id, $swap)"; 
    $result = mysql_query($sql, $conn) or die(mysql_error()); 
} // end if GET 
// set a result order with a default (sql infection prevention for $sortby) 
$sortby = ($_GET['sortby'] == 'name')? $_GET['sortby'] : 'usort'; 
// pull the info from the table 
$sql = "SELECT usort, name FROM info ORDER BY $sortby"; 
$result = mysql_query($sql, $conn) or die(mysql_error()); 
// display table 
echo "<table border = '1'>"; 
echo "<tr>"; 
// make column names links, passing sortby 
echo "<td><a href='{$_SERVER['PHP_SELF']}?sortby=usort'>usort</a></td>"; 
echo "<td><a href='{$_SERVER['PHP_SELF']}?sortby=name'>name</a></td>"; 
echo "</tr>"; 
// delete from table 
if ($_GET['del'] == 'true') { 
    // cast id as int for security 
    $id = (int) $_GET['id']; 
    // delete row from table 
    $sql = "DELETE FROM info WHERE usort = '$id'"; 
    $result = mysql_query($sql, $conn) or die(mysql_error()); 
    // select the info, ordering by usort 
    $sql = "SELECT usort, name FROM info ORDER BY usort"; 
    $result = mysql_query($sql, $conn) or die(mysql_error()); 
    // initialize a counter for rewriting usort 
    $usort = 1; 
    // while there is info to be fetched... 
    while ($r = mysql_fetch_assoc($result)) { 
     $name = $r['name']; 
     // update the usort number to the one in the next number 
     $sql = "UPDATE info SET usort = '$usort' WHERE name = '$name'"; 
     $update = mysql_query($sql, $conn) or die(mysql_error()); 
     // inc to next avail number 
     $usort++; 
    } // end while 
} // end if del 
// display data 1 row at a time 
while ($r = mysql_fetch_assoc($result)) { 
    echo "<tr>"; 
    // make the links to change custom order, passing direction and the custom sort id 
    echo "<td align = 'center'><a href='{$_SERVER['PHP_SELF']}?dir=up&id={$r['usort']}'>/\</a> "; 
    echo "<a href='{$_SERVER['PHP_SELF']}?dir=down&id={$r['usort']}'>\/</a></td>"; 
    echo "<td>{$r['name']}</td>"; 
    echo "<td><a href='{$_SERVER['PHP_SELF']}?del=true&id={$r['usort']}'>delete</a></td>"; 
    echo "</tr>"; 
} // end while $r 
echo "</table>"; 
// end display table 
?> 

Answer 1

用戶待辦事項列表似乎是另一個表給我。你不需要改變用戶信息的任何價值。只需添加，刪除或更改另一個表中的任務順序。像圖像下面的東西