Question

我有一个数据库表# This library is needed to extact zip archives. A .docx is a zip archive # .NET 4.5 or later is requried Add-Type -AssemblyName System.IO.Compression.FileSystem # This function gets plain text from a word document # adapted from http://stackoverflow.com/a/19503654/284111 # It is not ideal, but good enough function Extract-Text([string]$fileName) { #Generate random temporary file name for text extaction from .docx $tempFileName = [Guid]::NewGuid().Guid #Extract document xml into a variable ($text) $entry = [System.IO.Compression.ZipFile]::OpenRead($fileName).GetEntry("word/document.xml") [System.IO.Compression.ZipFileExtensions]::ExtractToFile($entry,$tempFileName) $text = [System.IO.File]::ReadAllText($tempFileName) Remove-Item $tempFileName #Remove actual xml tags and leave the text behind $text = $text -replace '</w:r></w:p></w:tc><w:tc>', " " $text = $text -replace '</w:r></w:p>', "`r`n" $text = $text -replace "<[^>]*>","" return $text } $fileList = Get-ChildItem "C:\Users\WP\Desktop\SearchFiles" -Include *.docx -Force -recurse # Adapted from http://stackoverflow.com/a/36023783/284111 $fileList | Foreach-Object {[regex]::matches((Extract-Text $_), '(?<=[A-Za-z]{3}\s*(?:-|–)\s*)\d{4}')} | Select-Object -ExpandProperty captures | Sort-Object value -Descending | Select-Object -First 1 -ExpandProperty value，其中包含以下列：

id | author_name | author_email | author_date（timestamp）| total_lines

示例内容为：

commits

我想得到如下结果：

1 | abc | abc@xyz.com | 2013-03-24 15:32:49 | 1234
2 | abc | abc@xyz.com | 2013-03-27 15:32:49 | 534
3 | abc | abc@xyz.com | 2014-05-24 15:32:49 | 2344
4 | abc | abc@xyz.com | 2014-05-28 15:32:49 | 7623

我在网上搜索类似的解决方案，但无法获得任何有用的解决方案。

我尝试了这个查询：

id | name | week | commits
1  | abc  | 1    | 2
2  | abc  | 2    | 0

但这不是正确的结果。

Answer 1

如果您有多年，您也应该考虑这一年。一种方法是：

SELECT date_part('year', author_date::date) as year,
       date_part('week', author_date::date) AS weekly,
       COUNT(author_email)           
FROM commits
GROUP BY year, weekly
ORDER BY year, weekly;

更自然的方法是使用date_trunc()：

SELECT date_trunc('week', author_date::date) AS weekly,
       COUNT(author_email)           
FROM commits
GROUP BY weekly
ORDER BY weekly;

Answer 2

问了这个问题已经很久了。
无论如何，只要有任何人通过这个。

如果您要计算所有中间周以及没有没有提交/记录的情况，则可以通过提供start_date end_date至generate_series()功能

SELECT t1.year_week week, 
       t2.commit_count 
FROM   (SELECT week, 
               To_char(week, 'IYYY-IW') year_week 
        FROM   generate_series('2020-02-01 06:06:51.25+00'::DATE, 
               '2020-04-05 12:12:33.25+00':: 
               DATE, '1 week'::interval) AS week) t1 
       LEFT OUTER JOIN (SELECT To_char(author_date, 'IYYY-IW') year_week, 
                               COUNT(author_email)             commit_count 
                        FROM   commits 
                        GROUP  BY year_week) t2 
                    ON t1.year_week = t2.year_week;

输出将是：

     week | commit_count  
----------+-------------
2020-05   | 2
2020-06   | NULL  
2020-07   | 1

如何在postgresql中按周分组

2 个答案: