Umwandlung von XML zu Json

Question

Hallo Ich habe xml so -

Dies ist Knoten dieses xml-

<item col_sort='20160102' Month='Jan-2016' Week='Week 1' Date='01/02/2016' Name='A' Id='1' 
        Column1='0' Column1Agg1='1410.00' Column1Agg2='405.00' Column1Agg3='0' 
        Column2='38' Column2Agg1='2976' Column2Agg2='79' Column2Agg3='79' 
        Column3='0.32' Column3Agg1='0.56' Column3Agg2='0.41' Column3Agg3='0.41' 
        Column4='11.42' Column4Agg1='8.48' Column4Agg2='9.55' Column4Agg3='9.55' 
        Column5='8.76' Column5Agg1='2.97' Column5Agg2='9.81' Column5Agg3='9.81' 
        Column6='-1' Column6Agg1='-2' Column6Agg2='-1' Column6Agg3='-1' 
        Column7='-99' Column7Agg1='99.9' Column7Agg2='-99' Column7Agg3='-99' 
        Column8='-99.00' Column8Agg1='8.00' Column8Agg2='-99.00' Column8Agg3='-99.00' 
        Column9='17.82' Column9Agg1='9.96' Column9Agg2='17.84' Column9Agg3='17.84' 
        Column10='18.18' Column10Agg1='10.51' Column10Agg2='18.28' Column10Agg3='18.28' 
        Column11='-2' Column11Agg1='-2' Column11Agg2='-2' Column11Agg3='-2' 
        Column12='7.9' Column12Agg1='44.6' Column12Agg2='7.6' Column12Agg3='7.6' 
        Column13='12.86' Column13Agg1='10.48' Column13Agg2='12.68' Column13Agg3='12.68' /> 

Nun, hier die Agg dem Kind Knoten vorangestellt wird. Also, hier in einem Knoten haben wir alle Kinder mit dem Elternwert bewertet. Hier repräsentiert Agg1 ein Elternteil. Agg2 ist das Kind von Agg1 und Agg3 ist das Kind von Agg2. Und die Knoten, die kein Präfix haben, sind der Blattknoten.

json = [ { 
     "Month": "Dec-2016", 
     "Column1Agg1": "1410.00", 
     "Column2Agg1": "2976", 
     "Column3Agg1": "0.56", 
     "Column4Agg1": "8.48",, 
     "Column5Agg1": "2.97", 
     "Column6Agg1": "-2", 
     "Column7Agg1": "-99.9", 
     "Column8Agg1": "8.00", 
     "Column9Agg1": "9.96", 
     "Column10Agg1": "10.51", 
     "Column11Agg1": "-2", 
     "Column12Agg1": "44.6", 
     "Column13Agg1": "10.48", 
     "children" : [ 
     {"Week": "Week 1", 
     "Column1Agg2": "405.00", 
     "Column2Agg2": "79", 
     "Column3Agg2": "0.41", 
     "Column4Agg2": "9.55", 
     "Column5Agg2": "9.81", 
     "Column6Agg2": "-1", 
     "Column7Agg2": "-99", 
     "Column8Agg2": "-99.00", 
     "Column9Agg2": "17.84", 
     "Column10Agg2": "18.28", 
     "Column11Agg2": "-2", 
     "Column12Agg2": "7.6", 
     "Column13Agg2": "12.68", 
     "children" : [{ 
      "Date": "01/02/2016", 
     "Column1Agg3": "405.00", 
     "Column2Agg3": "79", 
     "Column3Agg3": "0.41", 
     "Column4Agg3": "9.55", 
     "Column5Agg3": "9.81", 
     "Column6Agg3": "-1", 
     "Column7Agg3": "-99", 
     "Column8Agg3": "-99.00", 
     "Column9Agg3": "17.84", 
     "Column10Agg3": "18.28", 
     "Column11Agg3": "-2", 
     "Column12Agg3": "7.6", 
     "Column13Agg3": "12.68", 
     "children" : [{ 
     "Name" : "A", 
     "Id" : "1", 
     "Column1": "0", 
     "Column2": "38", 
     "Column3": "0.32", 
     "Column4": "11.42", 
     "Column5": "8.76", 
     "Column6": "-1", 
     "Column7": "-99", 
     "Column8": "-99.00", 
     "Column9": "17.82", 
     "Column10": "18.18", 
     "Column11": "-2", 
     "Column12": "7.9", 
     "Column13": "12.86" 

      }] 
      }] 
      }] 
      }] 

Wie Sie diese Art von XML in JSON mit Kindern konvertieren.

Irgendwelche Ideen, wie man dies tut

Answer 1

Ich habe dies mit Regex codiert. Sie können bei Bedarf zusätzliche Prüfungen hinzufügen.

var data = "<item col_sort='20160102' Month='Jan-2016' Week='Week 1' Date='01/02/2016' Name='A' Id='1' \ 
 
        Column1='0' Column1Agg1='1410.00' Column1Agg2='405.00' Column1Agg3='0' \ 
 
        Column2='38' Column2Agg1='2976' Column2Agg2='79' Column2Agg3='79' \ 
 
        Column3='0.32' Column3Agg1='0.56' Column3Agg2='0.41' Column3Agg3='0.41' \ 
 
        Column4='11.42' Column4Agg1='8.48' Column4Agg2='9.55' Column4Agg3='9.55'\ 
 
        Column5='8.76' Column5Agg1='2.97' Column5Agg2='9.81' Column5Agg3='9.81'\ 
 
        Column6='-1' Column6Agg1='-2' Column6Agg2='-1' Column6Agg3='-1'\ 
 
        Column7='-99' Column7Agg1='99.9' Column7Agg2='-99' Column7Agg3='-99'\ 
 
        Column8='-99.00' Column8Agg1='8.00' Column8Agg2='-99.00' Column8Agg3='-99.00'\ 
 
        Column9='17.82' Column9Agg1='9.96' Column9Agg2='17.84' Column9Agg3='17.84' \ 
 
        Column10='18.18' Column10Agg1='10.51' Column10Agg2='18.28' Column10Agg3='18.28'\ 
 
        Column11='-2' Column11Agg1='-2' Column11Agg2='-2' Column11Agg3='-2'\ 
 
        Column12='7.9' Column12Agg1='44.6' Column12Agg2='7.6' Column12Agg3='7.6' \ 
 
        Column13='12.86' Column13Agg1='10.48' Column13Agg2='12.68' Column13Agg3='12.68' />"; 
 

 
    var getChildren = function(i, search){ 
 
\t \t var result = {}; 
 
\t  var re = new RegExp("Column([0-9]+)Agg"+i+"='([0-9.-]+)'", "g"), m, p = 0; \t \t \t  
 
\t \t do { 
 
\t \t  m = re.exec(search); \t \t  \t \t  
 
\t \t  if (m) { \t \t  \t 
 
\t \t   result["Column" + m[1] + "Agg" + i] = m[2]; 
 
\t \t   p++; 
 
\t \t  } 
 
\t \t } while (m); \t \t \t 
 
    \t \t if(p>0){ 
 
\t \t \t result.children = getChildren(i+1, search); 
 
\t \t } 
 
\t \t return result; 
 
\t } 
 

 
    var m; 
 
\t var results = {}; 
 
\t 
 
\t results = getChildren(1, data); \t 
 
\t results.Month = /Month='([^']+)'/g.exec(data)[1]; 
 
\t results.children.Week = /Week='([^']+)'/g.exec(data)[1]; 
 
\t results.children.children.Date = /Date='([^']+)'/g.exec(data)[1]; 
 
\t 
 
\t console.log(results);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 2

Wenn ich dies tat ich durch die Umwandlung Ihrer verrückt XML-Eingabe in etwas mehr Sanitär-, entlang der Linien beginnen würde:

<item col_sort='20160102' Month='Jan-2016' Week='Week 1' Date='01/02/2016' Name='A' Id='1'> 
    <column nr='1' value='0'> 
    <agg nr='1'>1410.00</agg> 
    <agg nr='2'>405.00</agg> 
    <agg nr='3'>0</agg> 
    </column> 
    <column nr='2' value='38'> 
    <agg nr='1'>2976</agg> 
    <agg nr='2'>79</agg> 
    <agg nr='3'>79</agg> 
    </column> 
etc 

Das ist leicht genug getan mit XSLT . Hier ist eine Skizze:

<xsl:for-each-group select="@*" group-by="f:colNr(name())"> 
    <xsl:sort select="number(current-grouping-key())"/> 
    <column nr="{current-grouping-key()}"> 
    <xsl:for-each select="current-group()"> 
     <xsl:sort select="f:aggNr(name())"> 
     <agg nr="{f:aggNr(name())}"> 
     <xsl:value-of select="."/>