Question

$query = "SELECT users FROM tabl ORDER BY RAND()";
$result = mysql_query ($query)
    or die ("Query '$query' failed with error message: \"" . mysql_error () . '"');

while ($row = mysql_fetch_array($result)) {
    $users[] = $row[0];
}

$current = end($users);
$partners = array();
foreach ($users as $user)
{
        $partners[$user] = $current;
        $current = $user;
}
print_r($partners);

$query2 = "UPDATE tabl SET partner = {$partners[0]} WHERE users = '$users'";
mysql_query ($query2)
    or die ("<br>Query '$query2' failed with error message: \"" . mysql_error () . '"');

这就是我正在使用的代码。查询2之前一切都很好。我已经尝试了我能想到的每一种变化，但没有任何作用。

表上有两个字段：用户和合作伙伴。代码以随机顺序拉动用户，然后将它们分配给一个圆圈中的另一个。我需要使用分配填充合作伙伴字段。

Answer 1

将更新查询放在foreach循环中，然后您可以提供合作伙伴和用户变量，而不会在以后潜入数组：

foreach ($users as $user)
{
        $partners[$user] = $current;
        $current = $user;
        $query2 = "UPDATE tabl SET partner = '{$partners[$user]}' WHERE users = '{$user}'";
        mysql_query ($query2)
          or die ("<br>Query '$query2' failed with error message: \"" . mysql_error ()
}

Answer 2

使用WHERE users IN ('.implode(',',$users).')'

Answer 3

我会将代码更改为：

$current = end($users);
$partners = array();
foreach ($users as $user)
{
    $partners[$user] = $current;
    $current = $user;
    $query2 = "UPDATE tabl SET partner = {$partners[$user]} WHERE users = '$user'";
    mysql_query ($query2)
        or die ("<br>Query '$query2' failed with error message: \"" . mysql_error () . '"');
}
print_r($partners);

但是你也可以根据你想要的结果做以下事情：

$userList = join(',', $users);
$query2 = "UPDATE tabl SET partner = {$partners[0]} WHERE users IN ($userList)";
mysql_query ($query2)
    or die ("<br>Query '$query2' failed with error message: \"" . mysql_error () . '"');

如何从阵列更新表？

3 个答案: