$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 '$query2' failed with error message: \"" . mysql_error () . '"');
这就是我正在使用的代码。在 query2 之前一切都很好。我已经尝试了所有我能想到的变体,但没有任何效果。
该表有两个字段:用户和合作伙伴。该代码以随机顺序拉出用户,然后将他们分配给彼此,形成一个圆圈。我需要用任务填充合作伙伴字段。
最佳答案
将更新查询放在 foreach 循环中,然后您就可以使用合作伙伴和用户变量,而无需稍后深入数组:
foreach ($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 ()
}
关于php - 如何从数组更新表?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3141919/