Question

DOH！它现在已经解决了：

人为错误：如果我按照以下所有人的建议迭代数据，请帮助...添加以下代码以循环结果...

while ($rowprevprev = mysql_fetch_assoc($resultprevprev)) {
            print_r($rowprevprev);
        }

[羞愧地抱头......适当地尴尬......]

为每个人欢呼！

小时。

您好，

只需一段代码（感谢之前的帖子）：

SELECT
  AVG (q1) AS q1, AVG (q2) AS q2, AVG (q3) AS q3, AVG (q4) AS q4,
  AVG (q5) AS q5, AVG (q6) AS q6, AVG (q7) AS q7, AVG (q8) AS q8,
  AVG (q9) AS q9, AVG (q10) AS q10, AVG (q11) AS q11, AVG (q12) AS q12,
  AVG (q13) AS q13, AVG (q14) AS q14, AVG (q15) AS q15, AVG (q16) AS q16,
  AVG (q17) AS q17, AVG (q18) AS q18, AVG (q19) AS q19, AVG (q20) AS q20,
  AVG (q21) AS q21, AVG (q22) AS q22
FROM thotels_results
WHERE brand IN ('XYZ','ABC','EFG')
AND date = 'NOV2010' GROUP BY brand;

使用查询浏览器的输出是：

q1       q2       q3       q4       q5       q6       q7       q8       q9       q10      q11      q12      q13      q14      q15      q16      q17      q18      q19      q20      q21      q22
8.1724   8.2414   8.2414   7.8966   8.5862   8.5517   9.0000   8.5862   8.1724   7.9655   8.8966   8.6207   8.2414   8.3793   7.8276   8.3793   7.9310   8.4138   8.6897   8.3448   8.8621   8.5172
8.7714   8.8429   8.1643   8.7500   8.7571   8.9000   9.4071   9.1214   8.5714   8.7643   9.5143   8.9429   9.1643   8.9857   7.9500   8.9286   8.7000   9.0429   9.0143   8.7214   9.1214   9.3071
8.6009   8.5686   7.8528   8.3133   8.3423   8.6410   9.0301   8.6912   8.3233   8.3389   9.2029   8.3969   8.6856   8.5017   7.8071   8.4816   8.3512   8.6789   8.6789   8.3913   8.6388   8.8986

但是当我 VAR_DUMP 时，我得到以下内容，看起来并非所有数据都存在：

array
  0 => string '8.1724' (length=6)
  'q1' => string '8.1724' (length=6)
  1 => string '8.2414' (length=6)
  'q2' => string '8.2414' (length=6)
  2 => string '8.2414' (length=6)
  'q3' => string '8.2414' (length=6)
  3 => string '7.8966' (length=6)
  'q4' => string '7.8966' (length=6)
  4 => string '8.5862' (length=6)
  'q5' => string '8.5862' (length=6)
  5 => string '8.5517' (length=6)
  'q6' => string '8.5517' (length=6)
  6 => string '9.0000' (length=6)
  'q7' => string '9.0000' (length=6)
  7 => string '8.5862' (length=6)
  'q8' => string '8.5862' (length=6)
  8 => string '8.1724' (length=6)
  'q9' => string '8.1724' (length=6)
  9 => string '7.9655' (length=6)
  'q10' => string '7.9655' (length=6)
  10 => string '8.8966' (length=6)
  'q11' => string '8.8966' (length=6)
  11 => string '8.6207' (length=6)
  'q12' => string '8.6207' (length=6)
  12 => string '8.2414' (length=6)
  'q13' => string '8.2414' (length=6)
  13 => string '8.3793' (length=6)
  'q14' => string '8.3793' (length=6)
  14 => string '7.8276' (length=6)
  'q15' => string '7.8276' (length=6)
  15 => string '8.3793' (length=6)
  'q16' => string '8.3793' (length=6)
  16 => string '7.9310' (length=6)
  'q17' => string '7.9310' (length=6)
  17 => string '8.4138' (length=6)
  'q18' => string '8.4138' (length=6)
  18 => string '8.6897' (length=6)
  'q19' => string '8.6897' (length=6)
  19 => string '8.3448' (length=6)
  'q20' => string '8.3448' (length=6)
  20 => string '8.8621' (length=6)
  'q21' => string '8.8621' (length=6)
  21 => string '8.5172' (length=6)
  'q22' => string '8.5172' (length=6)

VAR_DUMP 似乎只会带回查询的第一行而不是所有三行。我需要访问所有三行数据，以便我可以执行计算。

有任何想法和/或建议的人吗？

谢谢，

荷马。

Answer 1

问题不在于var_dump，而是用于从MySQL获取数据的代码。从昨天开始看this question - 我认为你犯了同样的错误。

Answer 2

如果您使用的是php，

    $query = mysql_query("do your select query here");

while ($row = mysql_fetch_assoc($query)) {
    var_dump($row);
}

MySQL WHERE IN问题

2 个答案: