php - 从 PHP 表中一次更新多个 mysql 条目

Question

我有一个表，其中包含来自 mysql 数据库的一些数据。我需要从表本身内部更新该信息，因此当我按下提交按钮(表位于表单内)时，我需要在数据库上更新所有数据。我可以毫无问题地从数据库获取信息，但无法更新它!这是代码:

<table border=1 cellpadding=4 cellspacing=0 width=960px style='font-size:10px'> 
<form id='form1' name='form1' method='post' action='itself.php'>
<thead>
<tr>
<th colspan=3> People on the list </th>
<th><input type='submit' name='filtrar' id='filtrar' value='Filtrar'/></th>
<th><input type='submit' name='modificar' id='modificar' value='Modificar'/></th>
</tr>
<tr>
<th><label for='id'>ID</label></th>
<th><label for='id'>Friends</label></th>
<th><label for='id'>On the list?</label></th>
</tr>


<?
while($row = mysqli_fetch_array($result_listas, MYSQLI_BOTH))
$id= $row['id'];
{
?>
<tr>
<td><? echo "$row[id]" ?>
</td>
<td><select name='friends[<? echo "$id" ?>]' size='1' id='friends[<? echo "$id" ?>]'>
<option selected='selected'><? echo "$row[friends]" ?></option>
<option>less than 10</option>
<option>more than 10</option>   
</select>
</td>
<td><select name='onlist[<? echo "$id" ?>]' size='1' id='onlist[<? echo "$id" ?>]'>
<option selected='selected'><? echo "$row[onlist]" ?></option>
<option>SI</option>
<option>NO</option> 
</select>
</td>

<? 
$ssql_min="select min(id) as id from listas_old";
$result_min=  mysqli_query($link, $ssql_min);
$resultado_min = mysqli_fetch_array($result_min, MYSQLI_BOTH);
$ssql_max="select max(id) as id from listas_old";
$result_max=  mysqli_query($link, $ssql_max);
$resultado_max = mysqli_fetch_array($result_max, MYSQLI_BOTH);      

if(isset($_POST[modificar]))

{
for($i=$resultado_min['id']; $i<$resultado_max['id']; $i++)
            {
$sql1="UPDATE listas_old SET friends='$friends[$i]', onlist='$onlist[$i]' WHERE id='$i'";
$result1=mysqli_query($link, $sql1);
}

}
?>

Answer 1

OP wrote in a comment :

Got the solution for myself! This is what I did...

Change the for sentence like this:

for($i=$resultado_min['id'];
    $i<=$resultado_max['id'];
    $i++) {
    $friendsfor= $_POST['friends('.$i.')'];
    $onlistfor= $_POST['onlist('.$i.')'];
    $sql1="UPDATE listas_old SET friends='$friendsfor', anotado= '$onlistfor' WHERE id='$i'";
    }