Question

我正在建立一个查询，以“宽”格式（长为CSV）返回高/窄表。

我提出的查询如下：

SELECT
  CONV(a.assetID, 10, 16) AS assetID,
  a.notes,
  a.budgetReplaceDate,
  p.propertyName AS property,
  b.buildingName AS building,
  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, c.choice, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?
FROM
  asset_details AS ad
  INNER JOIN assets AS a ON a.assetID = ad.assetID
  LEFT JOIN asset_choices AS c ON c.choiceID = ad.choiceID
  LEFT JOIN property AS p ON a.propertyID = p.propertyID
  LEFT JOIN property_buildings AS b ON a.buildingID = b.buildingID
WHERE
  a.clientID IN(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
  AND a.formID = ?
  AND (
    a.deletedID IS NULL
    OR a.deletedID = 0
  )
GROUP BY
  assetID
ORDER BY
  assetID
LIMIT
  10 OFFSET 0

我要输入的参数如下：

array(29) {
    [0]=> string(3) "385"
    [1]=> string(4) "f385"
    [2]=> string(3) "386"
    [3]=> string(4) "f386"
    [4]=> string(3) "387"
    [5]=> string(4) "f387"
    [6]=> string(3) "388"
    [7]=> string(4) "f388"
    [8]=> string(3) "389"
    [9]=> string(4) "f389"
    [10]=> string(3) "390"
    [11]=> string(4) "f390"
    [12]=> string(2) "14"
    [13]=> string(2) "15"
    [14]=> string(2) "26"
    [15]=> string(2) "29"
    [16]=> string(2) "30"
    [17]=> string(2) "31"
    [18]=> string(2) "32"
    [19]=> string(2) "34"
    [20]=> string(2) "35"
    [21]=> string(2) "36"
    [22]=> string(2) "37"
    [23]=> string(2) "38"
    [24]=> string(2) "39"
    [25]=> string(2) "40"
    [26]=> string(2) "41"
    [27]=> string(2) "42"
    [28]=> string(2) "41"
}

当尝试通过PHP中的PDO执行它时，出现以下错误：

警告：PDOStatement :: execute（）：SQLSTATE [42000]：语法错误或访问冲突：1064您的SQL语法有错误；您可能会发现错误。在网上的report.php的第1行中，检查与您的MySQL服务器版本相对应的手册以获取正确的语法，以在'SELECT CONV（a.assetID，10，16）AS assetID，a.notes，a.budgetReplaceDate，'附近使用580

如果我去手动将数组中的每个值复制并粘贴到查询中的正确位置，然后通过phpMyAdmin（或adminer）运行它，则它将正常工作。

我认为可能是这部分：

  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, c.choice, NULL)) AS ?,
  MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS ?

，但是有一个简化的查询版本就可以了。我一生无法弄清楚发生了什么。我想念什么？

下面是常规日志：

Tcp port: 3306  Unix socket: /var/run/mysqld/mysqld.sock
Time                 Id Command    Argument
2019-04-11T08:54:26.499114Z 376072 Connect  *USERNAME*@localhost on *DATABASE* using TCP/IP
2019-04-11T08:54:26.499374Z 376072 Query    SELECT sessionID, uid, loginDate, lastActive, ipAddress FROM session WHERE loggedOut = 0 AND token = '*REDACTED*' AND lastActive >= '2019-04-10 15:54:26' - INTERVAL 27 HOUR
2019-04-11T08:54:26.499889Z 376073 Connect  *USERNAME*@localhost on *DATABASE* using TCP/IP
2019-04-11T08:54:26.500007Z 376073 Query    SELECT userID, userEmail, userLevel, userGroupID, joinDate, firstname, lastname, subscriptionEndDate, contactNo, contractorID, homePropertyID, timezone, uuid FROM users WHERE userID = '2'
2019-04-11T08:54:26.500487Z 376074 Connect  *USERNAME*@localhost on *DATABASE* using TCP/IP
2019-04-11T08:54:26.500595Z 376074 Query    SELECT * FROM userGroups WHERE userGroupID = '1'
2019-04-11T08:54:26.500891Z 376073 Query    SELECT ClientID FROM client
2019-04-11T08:54:26.501095Z 376073 Quit 
2019-04-11T08:54:26.501180Z 376072 Query    UPDATE session SET lastActive = '2019-04-11 18:54:26', browser = NULL WHERE sessionID = '2691'
2019-04-11T08:54:26.503173Z 376074 Query    SELECT * FROM markers
2019-04-11T08:54:26.503416Z 376072 Quit 
2019-04-11T08:54:26.503954Z 376074 Query    SELECT f.formID, f.formName, f.formTable, f.category, f.budget, s.formName AS srvcName, s.formID AS srvcFormID, s.formTable AS srvcTable, s.srvcFreq AS freq FROM forms as f
            LEFT JOIN srvcForms AS s ON f.formID = s.assetTable WHERE f.formID = '41' LIMIT 1
2019-04-11T08:54:26.504237Z 376074 Query    SELECT * FROM formField WHERE formID = '41' ORDER BY position, fieldID
2019-04-11T08:54:26.504948Z 376074 Query    SELECT 
    CONV(a.assetID, 10, 16) AS assetID,
    a.notes,
    a.budgetReplaceDate,
    p.propertyName AS property,
    b.buildingName AS building
    ,
    MAX(IF(ad.fieldID = '385', ad.text, NULL)) AS 'f385',
    MAX(IF(ad.fieldID = '386', ad.text, NULL)) AS 'f386',
    MAX(IF(ad.fieldID = '387', ad.text, NULL)) AS 'f387',
    MAX(IF(ad.fieldID = '388', ad.text, NULL)) AS 'f388',
    MAX(IF(ad.fieldID = '389', c.choice, NULL)) AS 'f389',
    MAX(IF(ad.fieldID = '390', ad.text, NULL)) AS 'f390'
FROM asset_details AS ad
INNER JOIN assets AS a ON a.assetID = ad.assetID
LEFT JOIN asset_choices AS c ON c.choiceID = ad.choiceID
LEFT JOIN property AS p ON a.propertyID = p.propertyID
LEFT JOIN property_buildings AS b ON a.buildingID = b.buildingID

WHERE a.clientID IN('14', '15', '26', '29', '30', '31', '32', '34', '35', '36', '37', '38', '39', '40', '41', '42') AND a.formID = '41' AND (a.deletedID IS NULL OR a.deletedID = 0)
GROUP BY assetID ORDER BY assetID LIMIT 10 OFFSET 0
2019-04-11T08:54:26.512918Z 376074 Quit 
2019-04-11T08:54:32.938734Z 376067 Query    set global general_log = "OFF"

Answer 1

您不能使用?占位符来指定MySQL准备语句中列的别名。因此，您应该只使用硬编码的别名：

SELECT
    CONV(a.assetID, 10, 16) AS assetID,
    a.notes,
    a.budgetReplaceDate,
    p.propertyName AS property,
    b.buildingName AS building,
    MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS max_one,
    MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS max_two,
    MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS max_three,
    MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS max_four,
    MAX(IF(ad.fieldID = ?, c.choice, NULL)) AS max_five,
    MAX(IF(ad.fieldID = ?, ad.text, NULL)) AS max_six
  FROM asset_details AS ad
  -- the rest of your query the same

但是，我认为没有充分的理由来控制别名。您的PHP脚本可能还需要一个硬编码的已知别名才能提取该值。

Answer 2

结果证明它确实在工作，这是一个进一步的错误。

我正在重写一个现有函数，但是没有注意到该函数进一步重复使用SQL查询来执行总记录数（以显示“查看1-10，共1000个结果”）。该错误实际上就是此错误，并且可以快速轻松地解决。

经验教训，请实际检查错误所在的行号！

说句公道话，我确实有一些错别字，只是没有注意到行号何时更改。

MySQL查询通过PHP PDO失败，通过phpMyAdmin成功

2 个答案: