Question

我最终得到了PHP中最糟糕的编写代码（它有效，但它太可怕了！也许你可以告诉我一个更好的方法）。我想在我的网页上显示14天的统计信息。我需要每天显示4个值（文件计数已下载，上传;文件大小已下载和上传）。我所拥有的是2个表（files和downloads_log），结构是：

文件（file_name，file_size，upload_datetime） downloads_log（file_name，dl_datetime）

所以我需要以下内容：

Count all files uploaded from table `files` in specific day
Count all files downloaded from table `downloads_log` in specific day
SUM file_size in table `files` in specific day
SUM file_size in table `files` where files.file_name = downloads_log.file_name in specific day

这是我过去14天所需要的一切。

＆＃34;脚本＆＃34;我已经完成了：

  for($i = 13; $i >= 0; $i--) {
    $query_downloads  = mysqli_query($con, "SELECT files.file_name, files.file_size, downloads_log.dl_file_name, downloads_log.dl_datetime, COUNT(file_name), SUM(file_size) FROM files, downloads_log WHERE files.file_name = downloads_log.dl_file_name AND `dl_datetime` >= '".date("Y-m-d", strtotime("-$i day"))." 00:00:00' AND `dl_datetime` <= '".date("Y-m-d", strtotime("-$i day"))." 23:59:59'");
    $result_downloads = mysqli_fetch_assoc($query_downloads);

    $query_uploads  = mysqli_query($con, "SELECT file_name, file_size, COUNT(file_name), SUM(file_size) FROM files WHERE `upload_datetime` >= '".date("Y-m-d", strtotime("-$i day"))." 00:00:00' AND `upload_datetime` <= '".date("Y-m-d", strtotime("-$i day"))." 23:59:59'");
    $result_uploads = mysqli_fetch_assoc($query_uploads);
  }

建议我，如何以更简单的方式实现这个问题，不要给mysql带来巨大的负担，谢谢：）

Answer 1

$i = 13;

$query = mysqli_query($con, "SELECT files.file_name, files.file_size, downloads_log.dl_file_name, downloads_log.dl_datetime, DATE(dl_datetime) AS datum, COUNT(file_name), SUM(file_size) FROM files, downloads_log WHERE files.file_name = downloads_log.dl_file_name AND `dl_datetime` >= '".date("Y-m-d", strtotime("-$i day"))." 00:00:00' AND `dl_datetime` <= '".date("Y-m-d", strtotime("-$i day"))." 23:59:59' GROUP BY DATE_FORMAT(dl_datetime, '%Y%m%d')");
while ($result = mysqli_fetch_assoc($query))
    $result_downloads[$result['datum']] = $result;

$query = mysqli_query($con, "SELECT file_name, file_size, COUNT(file_name), SUM(file_size), DATE(upload_datetime) AS datum FROM files WHERE `upload_datetime` >= '".date("Y-m-d", strtotime("-$i day"))." 00:00:00' AND `upload_datetime` <= '".date("Y-m-d", strtotime("-$i day"))." 23:59:59' GROUP BY DATE_FORMAT(upload_datetime, '%Y%m%d')");
while ($result = mysqli_fetch_assoc($query))
    $result_uploads[$result['datum']] = $result;

这会按不同日期对查询进行分组：GROUP BY DATE_FORMAT(upload/dl_datetime, '%Y%m%d')创建一个唯一的日期时间戳（％Y = 4位数年份，％m = 2位数月份，％d = 2位数日）

结果填入数组$ result_uploads / downloads，以日为索引。

Answer 2

希望这会有所帮助：

SELECT files.file_name, files.file_size, downloads_log.dl_file_name, downloads_log.dl_datetime, COUNT(file_name), SUM(file_size) 
FROM files
join downloads_log on files.file_name = downloads_log.dl_file_name and to_days(now())-14<to_days(dl_datetime)
group by to_days(dl_datetime)

在其他查询中使用相同的想法。你没事。

它每天选择最近14天和每组。

PHP脚本错误，帮我减少28个mysql_queries

2 个答案: