如何按顺序替换字符串

问题描述

我想按顺序替换一些值

例如，下面是一个xpath的示例

/MCCI_IN200100UV01[@ITsversion='XML_1.0'][@xsi:schemaLocation='urn:hl7-org:v3 MCCI_IN200100UV01.xsd']
/PORR_IN049016UV[r]/controlActProcess[@classCode='CACT']
[@moodCode='EVN']/subject[@typeCode='SUBJ'][1]/investigationEvent[@classCode='INVSTG']
[@moodCode='EVN']/outboundRelationship[@typeCode='SPRT'][relatedInvestigation/code[@code='2']
[@codeSystem='2.16.840.1.113883.3.989.2.1.1.22']][r]/relatedInvestigation[@classCode='INVSTG']
[@moodCode='EVN']/subjectOf2[@typeCode='SUBJ']/controlActEvent[@classCode='CACT']
    [@moodCode='EVN']/author[@typeCode='AUT']/assignedEntity[@classCode='ASSIGNED']/assignedPerson[@classCode='PSN']
        [@determinerCode='INSTANCE']/name/prefix[1]/@nullFlavor",

并且，我想按顺序提取 [r] 并根据元素的数量从 [0] 替换为 [n]。

如何替换 [r] ？

解决方法

用str.replace()。例如：

>>> 'test[r]test'.replace('[r]','[0]')
'test[0]test'

这是上面的 docs。

,

const txt = `/MCCI_IN200100UV01[@ITSVersion='XML_1.0'][@xsi:schemaLocation='urn:hl7-org:v3 MCCI_IN200100UV01.xsd']
/PORR_IN049016UV[r]/controlActProcess[@classCode='CACT']
[@moodCode='EVN']/subject[@typeCode='SUBJ'][1]/investigationEvent[@classCode='INVSTG']
[@moodCode='EVN']/outboundRelationship[@typeCode='SPRT'][relatedInvestigation/code[@code='2']
[@codeSystem='2.16.840.1.113883.3.989.2.1.1.22']][r]/relatedInvestigation[@classCode='INVSTG']
[@moodCode='EVN']/subjectOf2[@typeCode='SUBJ']/controlActEvent[@classCode='CACT']
    [@moodCode='EVN']/author[@typeCode='AUT']/assignedEntity[@classCode='ASSIGNED']/assignedPerson[@classCode='PSN']
        [@determinerCode='INSTANCE']/name/prefix[1]/@nullFlavor",`;
        
const count = (txt.match(/\[r\]/g) || []).length; // count occurrences using RegExp

let replacements; // set replacement values in-order
switch (count) {
  case 0:
    break
  case 1:
    replacements = ["a"];
    break;
  case 2:
    replacements = ["___REPLACEMENT_1___","___REPLACEMENT_2___"];
    break;
  case 3:
    replacements = ["d","e","f"];
    break;
}

let out = txt; // output variable
for (let i = 0; i < count; i++) {
  out = out.replace("[r]",replacements[i],1); // replace each occurrence one at a time
}

console.log(out);