首页 \ 问答 \ CSS first-letter伪元素适用于chrome,但不适用于firefox(CSS first-letter pseudo element works on chrome, but not on firefox)

CSS first-letter伪元素适用于chrome,但不适用于firefox(CSS first-letter pseudo element works on chrome, but not on firefox)

标题说明了一切。 Chrome完全符合我的要求,但Firefox不会将任何样式规则应用于我指定的第一个字母。 这是代码,感兴趣的部分复制如下。

article::first-letter {
    font-size: 20pt;
    color: #990000;
    font-family: "Times New Roman";
}

http://codepen.io/patrickfbray/pen/QNdoJq


Title says it all. Chrome does exactly what I want it to do, but Firefox does not apply any of the style rules to the first letter that I specified. Here's the code, and the section of interest is copied below.

article::first-letter {
    font-size: 20pt;
    color: #990000;
    font-family: "Times New Roman";
}

http://codepen.io/patrickfbray/pen/QNdoJq


原文:https://stackoverflow.com/questions/36089192
更新时间:2022-04-29 17:04

最满意答案

如果你想获得每个连续的范围,你需要一个递归的cte。

SQL DEMO

WITH cte as (
    SELECT V_Key, B_Key, AssignStart, AssignEnd 
    FROM #temptable t1
    UNION ALL
    SELECT t1.V_Key, t1.B_Key, c.AssignStart, t1.AssignEnd 
    FROM #temptable t1
    JOIN cte c
       ON t1.AssignStart = c.AssignEnd
), create_ranges as (
    SELECT V_key, AssignStart, MAX(AssignEnd) as AssignEnd
    FROM cte
    GROUP BY V_key, AssignStart
)
SELECT C1.*
FROM create_ranges c1
LEFT JOIN create_ranges c2
  ON c1.AssignStart BETWEEN c2.AssignStart  AND C2.AssignEnd
 AND c1.AssignStart <> c2.AssignStart
 AND c1.V_key = c2.V_key
WHERE c2.V_key IS NULL 
ORDER BY c1.V_key 
;

OUTPUT

在此处输入图像描述


If you want get every consecutive range you need a recursive cte.

SQL DEMO

WITH cte as (
    SELECT V_Key, B_Key, AssignStart, AssignEnd 
    FROM #temptable t1
    UNION ALL
    SELECT t1.V_Key, t1.B_Key, c.AssignStart, t1.AssignEnd 
    FROM #temptable t1
    JOIN cte c
       ON t1.AssignStart = c.AssignEnd
), create_ranges as (
    SELECT V_key, AssignStart, MAX(AssignEnd) as AssignEnd
    FROM cte
    GROUP BY V_key, AssignStart
)
SELECT C1.*
FROM create_ranges c1
LEFT JOIN create_ranges c2
  ON c1.AssignStart BETWEEN c2.AssignStart  AND C2.AssignEnd
 AND c1.AssignStart <> c2.AssignStart
 AND c1.V_key = c2.V_key
WHERE c2.V_key IS NULL 
ORDER BY c1.V_key 
;

OUTPUT

enter image description here

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