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用于检索最早/最早节点的xpath表达式

msxml6 msxml xquery xpath xml

gkrogers · 技术社区 · 15 年前

我有一个XML片段,所以:

<STATES>
  <STATE>
    <NAME>Alabama</NAME>
    <ABBREVIATION>AL</ABBREVIATION>
    <CAPITAL>Montgomery</CAPITAL>
    <POPULATION>4661900</POPULATION>
    <AREA>52419</AREA>
    <DATEOFSTATEHOOD>14 December 1819</DATEOFSTATEHOOD>
  </STATE>
  <STATE>
    <NAME>Alaska</NAME>
    <ABBREVIATION>AK</ABBREVIATION>
    <CAPITAL>Juneau</CAPITAL>
    <POPULATION>698473</POPULATION>
    <AREA>663268</AREA>
    <DATEOFSTATEHOOD>1 January 1959</DATEOFSTATEHOOD>
  </STATE>
  <STATE>
    <NAME>Delaware</NAME>
    <ABBREVIATION>DE</ABBREVIATION>
    <CAPITAL>Dover</CAPITAL>
    <POPULATION>885122</POPULATION>
    <AREA>2490</AREA>
    <DATEOFSTATEHOOD>7 December 1787</DATEOFSTATEHOOD>
  </STATE>
</STATES>
<etc, etc.>

我想取回(例如)最古老的州的首府(即“多佛”)。我已经做到了:

//STATES/STATE[DATEOFSTATEHOOD='7 December 1787']/CAPITAL/text()

但不知道怎么说“国家日期”=最早的国家日期。

有人能给我指个方向吗?

解决方案: 马特的解决方案或多或少是恰到好处的。我必须重新格式化日期(我使用了YYYYMMDDD),因为正如前面指出的那样,xpath 1.0不支持我使用的日期格式。此外,Microsoft的XML库(4.0和6.0)返回了带有Matt表达式的整个节点列表。反向测试解决了这个问题,使它只返回最早的节点。

所以:

//STATES/STATE[(DATEOFSTATEHOOD < //STATES/STATE/DATEOFSTATEHOOD)]/CAPITAL/text()

3 回复 | 直到 15 年前

Matt Weldon 15 年前

xpath 1.0不支持您提供的格式的日期。如果您能够使用这些日期的数字表示,例如17871207,那么您可以很容易地这样做:

//STATES/STATE[not(DATEOFSTATEHOOD > //STATES/STATE/DATEOFSTATEHOOD)]/CAPITAL/text()

如果这不可行,那么尝试格式化 DATEOFSTATEHOOD 节点作为 xs:date 并执行同样的操作:

//STATES/STATE[not(xs:date(DATEOFSTATEHOOD) > xs:date(//STATES/STATE/DATEOFSTATEHOOD))]/CAPITAL/text()

语法可能不完全正确,但希望它能帮助您开始。

Dave Cassel 15 年前

你能把它们重新格式化为xs:dates吗?

let $dates := (xs:date('2000-10-23'), xs:date('1999-12-26'))
let $min := fn:min($dates)
let $max := fn:max($dates)
return $min

在MarkLogic服务器中完成,但我认为这都是标准配置。

Chris Wallace 15 年前

可以使用xquery重新格式化日期,并使用min()查找最早的日期:

declare variable $monthnames := ("January","February","March","April","May","June","July","August","September","October","November","December");

declare function local:pad-zero($s as xs:string) as xs:string {
  if (string-length($s) = 1) then concat("0",$s) else $s
};

declare function local:df ($d as xs:string) as xs:date {
  let $dp := tokenize($d," ")
  let $year := $dp[3]
  let $month := local:pad-zero(string(index-of($monthnames,$dp[2])))
  let $day := local:pad-zero($dp[1])
  return
    concat($year,"-",$month,"-",$day)


};

let $states := 
<STATES>
  <STATE>
    <NAME>Alabama</NAME>
    <ABBREVIATION>AL</ABBREVIATION>
    <CAPITAL>Montgomery</CAPITAL>
    <POPULATION>4661900</POPULATION>
    <AREA>52419</AREA>
    <DATEOFSTATEHOOD>14 December 1819</DATEOFSTATEHOOD>
  </STATE>
  <STATE>
    <NAME>Alaska</NAME>
    <ABBREVIATION>AK</ABBREVIATION>
    <CAPITAL>Juneau</CAPITAL>
    <POPULATION>698473</POPULATION>
    <AREA>663268</AREA>
    <DATEOFSTATEHOOD>1 January 1959</DATEOFSTATEHOOD>
  </STATE>
  <STATE>
    <NAME>Delaware</NAME>
    <ABBREVIATION>DE</ABBREVIATION>
    <CAPITAL>Dover</CAPITAL>
    <POPULATION>885122</POPULATION>
    <AREA>2490</AREA>
    <DATEOFSTATEHOOD>7 December 1787</DATEOFSTATEHOOD>
  </STATE>
</STATES>


return 
   $states//STATE
     [local:df(DATEOFSTATEHOOD) = 
      min($states//STATE/local:df(DATEOFSTATEHOOD))
     ]

你可以在 eXist sandbox