Question

我需要将名为TEAM的列中的值从一行复制到另一行。两行都需要具有相同的团队名称。这是我的查询不起作用：

$query = "UPDATE profiles SET team = (SELECT team FROM profiles WHERE id = '$coach_id') WHERE id = '$player_id'";

我尝试删除单引号，删除“FROM profiles”，将值更改为table.value，尝试提供newdata.clan别名，我甚至尝试将值更改为整数而不是参数。什么都行不通，这就是我得到的：

错误：您的SQL中有错误句法;检查手册对应于您的MySQL服务器用于正确语法的版本靠近'WHERE id =''）WHERE id ='''at 第3行

Answer 1

$query1 = "SELECT team FROM profiles WHERE id = '$coach_id'";

/* get the value of the first query and assign it to a variable like $team_name */

$query2 = "UPDATE profiles SET team = '$team_name' WHERE id = '$player_id'";

Answer 2

此外，您应该用花括号括起PHP变量：

$query = "UPDATE profiles SET team = \"(SELECT team FROM profiles WHERE id = '{$coach_id}')\" WHERE id = '{$player_id}'";

Answer 3

来自MySQL手册：

“目前，您无法更新表格并从a中的同一个表中选择子查询“。

来源：http://dev.mysql.com/doc/refman/5.0/en/update.html

使用FinalForm编写的方法：

<?
$coach_id = 2;
$player_id = 1;

$query1 = "SELECT team FROM profiles WHERE id = '$coach_id'";
$rs = mysql_query($query1);
if ($row = mysql_fetch_array($rs)) {
  $team_name = $row['team'];
  $query2 = "UPDATE profiles SET team = '$team_name' WHERE id = '$player_id'";
  mysql_query($query2);
  // Done, updated if there is an id = 1
} else {
  // No id with id = 2
}
?>

使用更新查询复制列数据

3 个答案: