Question

我的查询有什么问题？我确定它的工作几天前，但今天我不知道它为什么会发生。我怎么能解决这个问题？抱歉，我的英文不好

错误：

查询：

function getUserGroupList() {
    $rulesQuery = "SELECT * FROM t_rules";
    $rulesResult = pg_query($this->DB, $rulesQuery);

    $nextQuery = '';

    while ($data = pg_fetch_array($rulesResult)) {
        $nextQuery .= ",(SELECT COUNT(*) FROM t_user_group_rules ugp WHERE ugp.ref_user_group = ug.id AND ugp.ref_rules = ".$data['id'].") as ".str_replace(' ', '_', $data['c_name'])."_count,(SELECT c_status FROM t_user_group_rules ugp WHERE ugp.ref_user_group = ug.id AND ugp.ref_rules = ".$data['id'].") as ".str_replace(' ', '_', $data['c_name']);
    }

    $query = "SELECT ug.id, ug.c_name $nextQuery
        FROM t_user_group ug
        WHERE ug.c_active = 1";

    $result = pg_query($this->DB, $query);
    // echo $query;
    // exit();
    $all = array();

    while ($data = pg_fetch_assoc($result)) {
        array_push($all, $data);
    }

    echo json_encode($all);
}

如果我回复$query，它就会变成这样

SELECT
   ug.id,
   ug.c_name,
   (SELECT count(*)
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 1
   ) AS command_count,
   (SELECT c_status
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 1
   ) AS command,
   (SELECT count(*)
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 2
   ) AS user_count,
   (SELECT c_status
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 2
   ) AS USER,
   (SELECT count(*)
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 3
   ) AS executive_count,
   (SELECT c_status
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 3
   ) AS executive,
   (SELECT count(*)
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 4
   ) AS rekon_count,
   (SELECT c_status
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 4
   ) AS rekon,
   (SELECT count(*)
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 5
   ) AS ej_count,
   (SELECT c_status
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 5
   ) AS ej,
   (SELECT count(*)
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 12
   ) AS tes_count,
   (SELECT c_status
    FROM t_user_group_rules ugp
    WHERE ugp.ref_user_group = ug.id
       AND ugp.ref_rules = 12
   ) AS tes
FROM t_user_group ug
WHERE ug.c_active = 1;

Answer 1

如果查询中的任何count(*)返回的值大于1，您将收到此错误。

这是因为SELECT列表中的子选择可能不会返回多个值，因为它将导致单个值（如果没有结果，则会得到SQL NULL）。

pg_query（）：查询失败：错误：用作表达式的子查询返回的多行

1 个答案: