谷歌大查詢行之間的時間差異

Question

我目前正在嘗試計算谷歌大查詢行之間的時間戳差異附加是我用來測試代碼的示例表。

我使用此代碼

SELECT 
    A.row, 
    A.issue.updated_at, 
(B.issue.updated_at - A.issue.updated_at) AS timedifference 
FROM [icxmedia-servers:icx_metrics.gh_zh_data_production] A 
INNER JOIN [icxmedia-servers:icx_metrics.gh_zh_data_production] B 
    ON B.row = (A.row + 1) 
WHERE issue.number==6 and issue.name=="archer" 
ORDER BY A.requestid ASC 

從這個問題Calculate the time difference between of two rows

Answer 1

比JOIN而是引用，這是更自然的使用解析函數來表示。 analytic functions with standard SQL in BigQuery的文檔解釋了分析函數的工作原理和語法。舉個例子，如果你想採取如命令是由y列決定在x值連續的差異，你可以這樣做：

WITH T AS (
    SELECT 
    x, 
    y 
    FROM UNNEST([9, 3, 4, 7]) AS x WITH OFFSET y) 
SELECT 
    x, 
    x - LAG(x) OVER (ORDER BY y) AS x_diff 
FROM T; 

注意，要BigQuery中運行這個，你需要取消選中「使用舊式SQL「框下的」顯示選項「來啓用標準SQL。 WITH T條款僅僅爲示例設置了一些數據。

針對您的特殊情況下，你可能會想要一個查詢，如：

SELECT 
    row, 
    issue.updated_at, 
    issue.updated_at - LAG(issue.updated_at) OVER (ORDER BY issue.updated_at) AS timedifference 
FROM `icxmedia-servers.icx_metrics.gh_zh_data_production` 
WHERE issue.number = 6 
    AND issue.name = "archer" 
ORDER BY requestid ASC; 

如果你想確定updated_at外面只是一個單一發行數量的差異，你可以使用一個PARTITION BY條款也是如此。例如：

SELECT 
    row, 
    issue.name, 
    issue.number, 
    issue.updated_at, 
    issue.updated_at - LAG(issue.updated_at) OVER (
     PARTITION BY issue.number 
     ORDER BY issue.updated_at) AS timedifference 
FROM `icxmedia-servers.icx_metrics.gh_zh_data_production` 
ORDER BY requestid ASC;