MySQL PHP查询优化-反馈

Question

我需要每天（cron）运行此脚本以更新1个月的反馈总数（至少这是我现在设计的）。 这是我的代码。 有人对我应该如何解决这个问题有更好的主意吗？ 也许改变了我的方式或优化了updateMonthlyFeedback.php脚本？

updateMonthlyFeedback.php

session_start();
include("db.php");

$sql="SELECT MAX(uid) as maxUID FROM users";
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($result);
$maxUID = $row['maxUID'];

for($i=0;$i<$maxUID;$i++){
    $sql="SELECT COUNT(*) as negativeCount FROM feedbacks WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY ) AND type = -1 AND uid = '$i'";
    $result = mysql_query($sql) or die(mysql_error());
    $row = mysql_fetch_array($result);
    $negativeCount = $row['negativeCount'];
    $sql="SELECT COUNT(*) as neutralCount FROM feedbacks WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY ) AND type = 0 AND uid = '$i'";
    $result = mysql_query($sql) or die(mysql_error());
    $row = mysql_fetch_array($result);
    $neutralCount = $row['neutralCount'];
    $sql="SELECT COUNT(*) as positiveCount FROM feedbacks WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY ) AND type = 1 AND uid = '$i'";
    $result = mysql_query($sql) or die(mysql_error());
    $row = mysql_fetch_array($result);
    $positiveCount = $row['positiveCount'];
    $sql = "UPDATE feedback_totals SET negativeCount = '$negativeCount', neutralCount = '$neutralCount', positiveCount = '$positiveCount' WHERE uid = '$i'";
    $result=mysql_query($sql) or die(mysql_error());
}

MySQL表

CREATE TABLE feedback_totals (
    uid                     VARCHAR(40),
    negativeCount           int,
    neutralCount            int,                
    positiveCount           int,
    halfStarCount           int,
    oneStarCount            int,
    oneHalfStarCount        int,
    twoStarCount            int,
    twoHalfStarCount        int,
    threeStarCount          int,
    threeHalfStarCount      int,
    fourStarCount           int,
    fourHalfStarCount       int,
    fiveStarCount           int,
    PRIMARY KEY             (uid)
    #FOREIGN KEY            (uid) REFERENCES users(uid) ON DELETE CASCADE
);

CREATE TABLE feedback_last_month (
    uid                     VARCHAR(40),
    negativeCount           int,
    neutralCount            int,                
    positiveCount           int,
    halfStarCount           int,
    oneStarCount            int,
    oneHalfStarCount        int,
    twoStarCount            int,
    twoHalfStarCount        int,
    threeStarCount          int,
    threeHalfStarCount      int,
    fourStarCount           int,
    fourHalfStarCount       int,
    fiveStarCount           int,
    PRIMARY KEY             (uid)
    #FOREIGN KEY            (uid) REFERENCES users(uid) ON DELETE CASCADE
);

CREATE TABLE feedback (
    feedback_id             INT NOT NULL AUTO_INCREMENT,
    uid                     VARCHAR(40),INDEX (uid),
    sender_id               VARCHAR(40),
    type                    int,                #-1 = neg, 0 = neutral, 1 = positive
    starCount               VARCHAR(40),
    description             VARCHAR(80),
    date_created            timestamp DEFAULT CURRENT_TIMESTAMP, 
    fromType                VARCHAR(40), # buyer or seller
    fromUsername            VARCHAR(40),
    PRIMARY KEY             (feedback_id)
    #FOREIGN KEY            (uid) REFERENCES users(uid) ON DELETE CASCADE
);

Answer 1

显然这里有很多重复。 大部分代码都可以通过重构代码来删除，但是即使从当前流程开始，也可以通过使用更好的数据库API使其性能更好。

因此，我建议的第一件事是停止使用mysql_xxx()函数，而改用PDO库。 无论如何，旧的mysql函数已被弃用，因此不建议尽可能使用它们，但是在这种情况下，使用PDO是一个特定的原因，因为它比旧的函数具有明显的性能优势。

PDO允许您使用一种称为“准备查询”的功能，如果您反复调用类似的查询，则数据库可以更有效地缓存查询。

其次，查询本身。 是的，这些绝对可以简化。 可以使用GROUP BY将循环中的三个查询合并为一个查询。 查询如下所示：

SELECT COUNT(*) FROM users
WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY )
AND uid = :uid
AND type = -1 OR type = 1 OR type = 1
GROUP BY type

您应该从此查询中获得与要提取的三个记录相同的三个值。

您还可以做很多事情，但这是一个好的开始。 我相信您还会得到其他答案，以进一步帮助您。

希望能有所帮助。

Answer 2

好的，就像其他人说的那样，请使用PDO / MYSQLI。 但是，使用已有的代码，这里有两种方法可能会起作用并且性能更好。

首先是使用相关子查询来获取负/正/中性值。 这很不错，因为它很短，但绝不是理想的选择。 您仍在数据库上执行大量查询（每个uid 3个查询+初始更新）。 但是，您仅从php向服务器发送一个查询，并让数据库完成所有其余工作。 这对于一些用户来说可能效果不错，但是过一会儿就会出现性能问题。 此查询将更新feedback_totals中的所有行。 但是，如果feedback_totals中没有一行，则不会为任何新的uid插入新行。

//one query, this is it. updates it all.
$sql = "UPDATE `feedback_totals`
        SET
            `negativeCount`=(SELECT COUNT(*) FROM `users` WHERE `uid`=`feedback_totals`.`uid` AND `date_created` >= (CURDATE() - INTERVAL 30 DAY) AND `type`=-1),
            `positiveCount`=(SELECT COUNT(*) FROM `users` WHERE `uid`=`feedback_totals`.`uid` AND `date_created` >= (CURDATE() - INTERVAL 30 DAY) AND `type`=1),
            `neutralCount`=(SELECT COUNT(*) FROM `users` WHERE `uid`=`feedback_totals`.`uid` AND `date_created` >= (CURDATE() - INTERVAL 30 DAY) AND `type`=0)";
$result=mysql_query($sql) or die(mysql_error());

---

从长远来看，第二可能更好。 查询一次查询中需要的所有数据。 遍历该结果，将其格式化为php，遍历该遍并进行更新。 这可能会更好，因为您正在运行的查询数量要少得多（1个用于获取数据+ 1个每个uid）。

//query for all the data
$sql="SELECT
            `uid`,
            `type`,
            COUNT(*) AS cnt
        FROM `users`
        WHERE `date_created` >= (CURDATE() - INTERVAL 30 DAY)
        GROUP BY `uid`,`type`";
$result=mysql_query($sql) or die(mysql_error());

$data = array();
//loop through the result
while($row=mysql_fetch_assoc($result)){
    //if the uid is not in $data
    if(!isset($data[$row['uid']])){
        //add it with a blank array
        $data[$row['uid']] = array('negativeCount'=>0,'neutralCount'=>0,'positiveCount'=>0);
    }
    //add to the data for this uid depending on type
    if($row['type']==-1){
        $data[$row['uid']]['negativeCount']=$row['cnt'];
    } elseif($row['type']==1){
        $data[$row['uid']]['positiveCount']=$row['cnt'];
    } else {
        $data[$row['uid']]['neutralCount']=$row['cnt'];
    }
}

//now loop through the data and update the table
foreach($data as $uid=>$cnt){
    $sql = "UPDATE `feedback_totals`
            SET
                `negativeCount`={$cnt['negativeCount']},
                `positiveCount`={$cnt['positiveCount']},
                `neutralCount`={$cnt['neutralCount']}
            WHERE `uid`=$uid";
    $result=mysql_query($sql) or die(mysql_error());
}

Answer 3

我将运行三个查询来获取您需要的信息：

SELECT uid, COUNT(*) as negativeCount FROM users
WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY ) AND type = -1 
GROUP BY uid ORDER BY uid ASC";

SELECT uid, COUNT(*) as neutralCount FROM users
WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY ) AND type = 0
GROUP BY uid ORDER BY uid ASC";

SELECT uid, COUNT(*) as positiveCount FROM users
WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY ) AND type = 1 
GROUP BY uid ORDER BY uid ASC";

然后，我将遍历结果，在与当前uid对齐的行中增加行。 返回结果的一个棘手的问题是，缺少uid表示计数为0。但是它们是有序的，因此您会知道要增加返回的结果（通过使用的fetch_row）或每个函数的单独索引结果。

它看起来像这样：

// Have to load up the first result
$p_row = mysql_fetch_array($positive_result);
$neu_row = mysql_fetch_array($neutral_result);
$neg_row = mysql_fetch_array($negative_result);

for($i = 0; $i < $maxUID; $i++){
  $positive = $neutral = $negative = 0;
  if($p_row[0] == $i){
    $positive = $p_row[1];
    $p_row = mysql_fetch_array($positive_result);
  }
  if($neu_row[0] == $i){
    $neutral = $neu_row[1];
    $neu_row = mysql_fetch_array($neutral_result);
  }
  if($neg_row[0] == $i){
    $negative = $neg_row[1];
    $neg_row = mysql_fetch_array($negative_result);
  }
  $sql = "UPDATE feedback_totals SET negativeCount = '$negative', neutralCount = '$neutral', positiveCount = '$positive' WHERE uid = '$i'";
  mysql_query($sql) or die(mysql_error());
}

Answer 4

所有这些都可以通过一个SQL查询来完成：

INSERT INTO `feedback_totals` (uid,negativeCount,positiveCount,neutralcount)
SELECT users.uid, 
  COUNT(neg.feedback_id) as negativeCount, 
  COUNT(pos.feedback_id) as positiveCount, 
  COUNT(neut.feedback_id) AS neutralCount
FROM users
LEFT JOIN feedback neg  ON neg.uid =users.uid AND neg.type=-1 AND neg.date_created >=(CURDATE() - INTERVAL 30 DAY)
LEFT JOIN feedback pos  ON pos.uid =users.uid AND pos.type=1  AND pos.date_created >=(CURDATE() - INTERVAL 30 DAY)
LEFT JOIN feedback neut ON neut.uid=users.uid AND neut.type=0 AND neut.date_created>=(CURDATE() - INTERVAL 30 DAY)
GROUP BY uid
ON DUPLICATE KEY UPDATE 
  negativeCount=VALUES(negativeCount), 
  positiveCount=VALUES(positiveCount), 
  neutralCount=VALUES(neutralCount);

ON DUPLICATE KEY UPDATE将允许查询添加新行以及更新现有行。 您可以在SQL Fiddle上玩这个游戏

Answer 5

首先尝试将您的Mysql更改为mysqli

只需将您的db.php更改为：

$mysqli = new mysqli("localhost", "XXX", "XXX", "XXX");

if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

并将您的代码更改为：

session_start();
include("db.php");
$sql="SELECT MAX(uid) as maxUID FROM users; ";

for($i=0;$i<$maxUID;$i++){
    $sql.="SELECT COUNT(*) as negativeCount FROM users WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY ) AND type = -1 AND uid = '$i'; ";
    $sql.="SELECT COUNT(*) as neutralCount FROM users WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY ) AND type = 0 AND uid = '$i'; ";
    $sql.="SELECT COUNT(*) as positiveCount FROM users WHERE date_created >= ( CURDATE() - INTERVAL 30 DAY ) AND type = 1 AND uid = '$i'; ";
    $sql.="UPDATE feedback_totals SET negativeCount = '$negativeCount', neutralCount = '$neutralCount', positiveCount = '$positiveCount' WHERE uid = '$i'; ";
}   

if ($mysqli->multi_query($sql)) {
    do {
        $rows=array();
        if ($result = $mysqli->store_result()) {
            while($rows[] = mysqli_fetch_assoc($result));
            array_pop($rows); 
            $result->free();
        }
        $data[]=$rows;
    } while ($mysqli->next_result());
    print_r($data);
} else
    echo "Error with SQL";

这将仅与数据库建立一个批处理连接，并将所有数据打印在一个阵列中。