Php SQL到CodeIgniter阵列缺少数据问题

查询在phpMyAdmin中运行时运行良好，并在附加的图像中返回完整的结果。但是，在CodeIgniter中打印数组时，它在phpMyAdmin中运行的SQL查询中没有一个字段的值，七个利润，只是这一字段中的差异，从SQL到php数组。。。我只是不明白，为什么在打印数组时，一个字段（其值应为26）不包含任何内容？有什么想法吗？我将字段名从以数字开头改为试图修复它，但没有区别

我知道这很复杂，看起来很可怕，任何帮助或只是遇到类似情况的人都很想知道，谢谢

萨姆

编辑：查询的格式要更规范一些：

$query = $this->db->query("SELECT t1.numberofbets, t1.profit, t2.seven_profit, t3.28profit, user.user_id, username, password, email, balance, user.date_added, activation_code, activated FROM user LEFT JOIN (SELECT user_id, SUM(amount_won) AS profit, count(tip_id) AS numberofbets FROM tip GROUP BY user_id) as t1 ON user.user_id = t1.user_id LEFT JOIN (SELECT user_id, SUM(amount_won) AS seven_profit FROM tip WHERE date_settled > '$seven_daystime' GROUP BY user_id) as t2 ON user.user_id = t2.user_id LEFT JOIN (SELECT user_id, SUM(amount_won) AS 28profit FROM tip WHERE date_settled > '$twoeight_daystime' GROUP BY user_id) as t3 ON user.user_id = t3.user_id where activated = 1 GROUP BY user.user_id ORDER BY user.date_added DESC");

return $query->result_array();

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好的，问题是SQL查询包含PHP变量，这些变量必须被视为0或其他没有给出错误的变量，因此它在seven_price列中显示一个值

在CodeIgniter中运行查询时，它使用$seven_daystime变量的正确值，因此将数据放入数组时结果不同

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/归咎于疲劳

我建议您设置查询格式，以便其他人和您可以阅读。这样可能更容易发现问题。

$query = $this->db->query("
  SELECT
    t1.numberofbets, t1.profit, t2.seven_profit, t3.28profit,
    user.user_id, username, password, email,
    balance, user.date_added, activation_code, activated
  FROM
    user
  LEFT JOIN
    (
      SELECT
        user_id,
        SUM(amount_won) AS profit,
        count(tip_id) AS numberofbets
      FROM
        tip
      GROUP BY
        user_id
    ) as t1
  ON
    user.user_id = t1.user_id
  LEFT JOIN
    (
      SELECT
        user_id,
        SUM(amount_won) AS seven_profit
      FROM
        tip
      WHERE
        date_settled > '$seven_daystime'
      GROUP BY
        user_id
    ) as t2
  ON
    user.user_id = t2.user_id
  LEFT JOIN
    (
      SELECT
        user_id,
        SUM(amount_won) AS 28profit
      FROM
        tip
      WHERE
        date_settled > '$twoeight_daystime'
      GROUP BY
        user_id
    ) as t3
  ON
    user.user_id = t3.user_id
  WHERE
    activated = 1
  GROUP BY
    user.user_id
  ORDER BY
    user.date_added DESC
");