Re: [xwiki-users] Implement Jsontool.serialize in groovy

[Eduardo Abritta]

Hi Thomas, I instantiate the jsonbuilder as follows in the example yet returned 
to previous error that was not allow adding elements in my json method. Add 
(). 
I have also realized by placing the method within an array as criuado test 
(data.projetos [i]. Id.add ()).

{{groovy}}def hql0 = , BaseObject as obj where doc.fullName = obj.name and 
obj.className = 'E-nova.E-novaClass';def listProjetos = 
services.query.hql(hql0).execute();def i = 0;def data = new 
groovy.json.JsonBuilder();data = [totalprojetos: 
listProjetos.size(),projetos: [ id: , tag: , label: , size: , color: 
, x: , y :  ],edges:[ id: , source: , target:  ]];
for(item in listProjetos){ def hql = select distinct list from BaseObject as 
obj, DBStringListProperty as prop join prop.list list where 
obj.className='XWiki.TagClass' and obj.id=prop.id.id and prop.id.name='tags' 
and obj.name ='$item' ;def tag = services.query.hql(hql).execute(); i = i + 1; 
data.projetos.id.add()=1; data.projetos.tag.add()=tags; 
data.projetos.label.add()=Label do projeto; data.projetos.size.add()=5; 
data.projetos.color.add()=#696969; data.projetos.x.add()=1; 
data.projetos.y.add()= 1;}
def json = new groovy.json.JsonBuilder(data)println 
json.toPrettyString(){{/groovy}}
Atenciosamente,
Eduardo Abritta
e-mail: eduardo.abri...@outlook.com | cel: (32)8472-9631

 From: eduardo.abri...@outlook.com
 To: users@xwiki.org
 Date: Mon, 30 Jun 2014 15:23:37 -0300
 Subject: Re: [xwiki-users] Implement Jsontool.serialize in groovy
 
 I'm trying to implement a function that told me but I can not add the data 
 online, I can only add one. How do I do that?
{{groovy}}def hql0 = , BaseObject as obj where doc.fullName = obj.name and 
obj.className = 'E-nova.E-novaClass';def listProjetos = 
services.query.hql(hql0).execute();def i = 0;def data = new 
groovy.json.JsonBuilder();data = [totalprojetos: 
listProjetos.size(),projetos: [  id: ,  tag: ,
  label: ,  size: ,  color: , 
 x: ,  y : ],edges:[id: ,   
 source: ,target: ]];
for(item in listProjetos){ def hql = select distinct list from BaseObject as 
obj, DBStringListProperty as prop join prop.list list where 
obj.className='XWiki.TagClass' and obj.id=prop.id.id and prop.id.name='tags' 
and obj.name ='$item' ;def tag = services.query.hql(hql).execute(); i = i + 1; 
data.projetos.id=1; data.projetos.tag=tags; data.projetos.label=Label do 
projeto; data.projetos.size=5; data.projetos.color=#696969; 
data.projetos.x=1; data.projetos.y= 1;}
def json = new groovy.json.JsonBuilder(data)println 
json.toPrettyString(){{/groovy}}
 Atenciosamente,
 Eduardo Abritta
 e-mail: eduardo.abri...@outlook.com | cel: (32)8472-9631
 
 
 
  Date: Fri, 27 Jun 2014 11:51:47 +0200
  From: thomas.morta...@xwiki.com
  To: users@xwiki.org
  Subject: Re: [xwiki-users] Implement Jsontool.serialize in groovy
  
  You should probably use JsonBuilder which is the standard tool to
  manipulate JSON in Groovy.
  
  On Fri, Jun 27, 2014 at 10:33 AM, Clemens Klein-Robbenhaar
  c.robbenh...@espresto.com wrote:
  
   The velocity tools are not automatically created for groovy scripts
   The closest you can get is to create them manually in your script, like:
  
def jsontool = new org.xwiki.velocity.tools.JSONTool()
  
   I feel there should be a better way to do that, but it works this way.
   (it is likely that using groovy.json.JsonBuilder() works, too, but I have 
   not tried)
  
   Clemens
  
   On 06/24/2014 09:21 PM, Eduardo Abritta wrote:
   Hello Sirs,
   I need to implement the function jsontool.serialize in the code below, 
   groovy, and I do not understand the error that is accusing, can help me 
   solve this?
   {{groovy}}def hql0 = , BaseObject as obj where doc.fullName = obj.name 
   and obj.className = 'E-nova.E-novaClass';def listProjetos = 
   services.query.hql(hql0).execute();def json = ['totalprojetos': 
   listProjetos.size(), 'projetos':[1000] , 'edges':[1000] ];def i = 0;
   for(item in listProjetos){def hql = select distinct list from 
   BaseObject as obj, DBStringListProperty as prop join prop.list list 
   where obj.className='XWiki.TagClass' and obj.id=prop.id.id and 
   prop.id.name='tags' and obj.name ='$item' ;def tag = 
   services.query.hql(hql).execute();i = i + 1;
   json.projetos.add({id: xwiki.getDocument(item).getName()
   })}
   jsontool.serialize(json);
   {{/groovy}}
   Ty.
   Atenciosamente,
   Eduardo Abritta
   e-mail: eduardo.abri...@outlook.com | cel: (32)8472-9631
  
  
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  -- 
  Thomas Mortagne
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Re: [xwiki-users] Implement Jsontool.serialize in groovy

[Eduardo Abritta]

I'm trying to implement a function that told me but I can not add the data 
online, I can only add one. How do I do that?
{{groovy}}def hql0 = , BaseObject as obj where doc.fullName = obj.name and 
obj.className = 'E-nova.E-novaClass';def listProjetos = 
services.query.hql(hql0).execute();def i = 0;def data = new 
groovy.json.JsonBuilder();data = [totalprojetos: 
listProjetos.size(),projetos: [  id: ,  tag: ,
  label: ,  size: ,  color: , 
 x: ,  y : ],edges:[id: ,   
 source: ,target: ]];
for(item in listProjetos){  def hql = select distinct list from BaseObject as 
obj, DBStringListProperty as prop join prop.list list where 
obj.className='XWiki.TagClass' and obj.id=prop.id.id and prop.id.name='tags' 
and obj.name ='$item' ;def tag = services.query.hql(hql).execute();i = 
i + 1;data.projetos.id=1;data.projetos.tag=tags;
data.projetos.label=Label do projeto;data.projetos.size=5;
data.projetos.color=#696969;data.projetos.x=1;data.projetos.y= 1;}
def json = new groovy.json.JsonBuilder(data)println 
json.toPrettyString(){{/groovy}}
Atenciosamente,
Eduardo Abritta
e-mail: eduardo.abri...@outlook.com | cel: (32)8472-9631



 Date: Fri, 27 Jun 2014 11:51:47 +0200
 From: thomas.morta...@xwiki.com
 To: users@xwiki.org
 Subject: Re: [xwiki-users] Implement Jsontool.serialize in groovy
 
 You should probably use JsonBuilder which is the standard tool to
 manipulate JSON in Groovy.
 
 On Fri, Jun 27, 2014 at 10:33 AM, Clemens Klein-Robbenhaar
 c.robbenh...@espresto.com wrote:
 
  The velocity tools are not automatically created for groovy scripts
  The closest you can get is to create them manually in your script, like:
 
   def jsontool = new org.xwiki.velocity.tools.JSONTool()
 
  I feel there should be a better way to do that, but it works this way.
  (it is likely that using groovy.json.JsonBuilder() works, too, but I have 
  not tried)
 
  Clemens
 
  On 06/24/2014 09:21 PM, Eduardo Abritta wrote:
  Hello Sirs,
  I need to implement the function jsontool.serialize in the code below, 
  groovy, and I do not understand the error that is accusing, can help me 
  solve this?
  {{groovy}}def hql0 = , BaseObject as obj where doc.fullName = obj.name 
  and obj.className = 'E-nova.E-novaClass';def listProjetos = 
  services.query.hql(hql0).execute();def json = ['totalprojetos': 
  listProjetos.size(), 'projetos':[1000] , 'edges':[1000] ];def i = 0;
  for(item in listProjetos){def hql = select distinct list from 
  BaseObject as obj, DBStringListProperty as prop join prop.list list where 
  obj.className='XWiki.TagClass' and obj.id=prop.id.id and 
  prop.id.name='tags' and obj.name ='$item' ;def tag = 
  services.query.hql(hql).execute();i = i + 1;
  json.projetos.add({id: xwiki.getDocument(item).getName()})}
  jsontool.serialize(json);
  {{/groovy}}
  Ty.
  Atenciosamente,
  Eduardo Abritta
  e-mail: eduardo.abri...@outlook.com | cel: (32)8472-9631
 
 
  ___
  users mailing list
  users@xwiki.org
  http://lists.xwiki.org/mailman/listinfo/users
 
 
 
  ___
  users mailing list
  users@xwiki.org
  http://lists.xwiki.org/mailman/listinfo/users
 
 
 
 -- 
 Thomas Mortagne
 ___
 users mailing list
 users@xwiki.org
 http://lists.xwiki.org/mailman/listinfo/users
  
___
users mailing list
users@xwiki.org
http://lists.xwiki.org/mailman/listinfo/users

Re: [xwiki-users] Implement Jsontool.serialize in groovy

[Clemens Klein-Robbenhaar]

The velocity tools are not automatically created for groovy scripts
The closest you can get is to create them manually in your script, like:

 def jsontool = new org.xwiki.velocity.tools.JSONTool()

I feel there should be a better way to do that, but it works this way.
(it is likely that using groovy.json.JsonBuilder() works, too, but I have not 
tried)

Clemens

On 06/24/2014 09:21 PM, Eduardo Abritta wrote:
 Hello Sirs,
 I need to implement the function jsontool.serialize in the code below, 
 groovy, and I do not understand the error that is accusing, can help me solve 
 this?
 {{groovy}}def hql0 = , BaseObject as obj where doc.fullName = obj.name and 
 obj.className = 'E-nova.E-novaClass';def listProjetos = 
 services.query.hql(hql0).execute();def json = ['totalprojetos': 
 listProjetos.size(), 'projetos':[1000] , 'edges':[1000] ];def i = 0;
 for(item in listProjetos){def hql = select distinct list from BaseObject 
 as obj, DBStringListProperty as prop join prop.list list where 
 obj.className='XWiki.TagClass' and obj.id=prop.id.id and prop.id.name='tags' 
 and obj.name ='$item' ;def tag = services.query.hql(hql).execute();i 
 = i + 1;
 json.projetos.add({id: xwiki.getDocument(item).getName()})}
 jsontool.serialize(json);
 {{/groovy}}
 Ty.
 Atenciosamente,
 Eduardo Abritta
 e-mail: eduardo.abri...@outlook.com | cel: (32)8472-9631
 
 
 ___
 users mailing list
 users@xwiki.org
 http://lists.xwiki.org/mailman/listinfo/users
 


___
users mailing list
users@xwiki.org
http://lists.xwiki.org/mailman/listinfo/users

Re: [xwiki-users] Implement Jsontool.serialize in groovy

[Thomas Mortagne]

You should probably use JsonBuilder which is the standard tool to
manipulate JSON in Groovy.

On Fri, Jun 27, 2014 at 10:33 AM, Clemens Klein-Robbenhaar
c.robbenh...@espresto.com wrote:

 The velocity tools are not automatically created for groovy scripts
 The closest you can get is to create them manually in your script, like:

  def jsontool = new org.xwiki.velocity.tools.JSONTool()

 I feel there should be a better way to do that, but it works this way.
 (it is likely that using groovy.json.JsonBuilder() works, too, but I have not 
 tried)

 Clemens

 On 06/24/2014 09:21 PM, Eduardo Abritta wrote:
 Hello Sirs,
 I need to implement the function jsontool.serialize in the code below, 
 groovy, and I do not understand the error that is accusing, can help me 
 solve this?
 {{groovy}}def hql0 = , BaseObject as obj where doc.fullName = obj.name and 
 obj.className = 'E-nova.E-novaClass';def listProjetos = 
 services.query.hql(hql0).execute();def json = ['totalprojetos': 
 listProjetos.size(), 'projetos':[1000] , 'edges':[1000] ];def i = 0;
 for(item in listProjetos){def hql = select distinct list from 
 BaseObject as obj, DBStringListProperty as prop join prop.list list where 
 obj.className='XWiki.TagClass' and obj.id=prop.id.id and prop.id.name='tags' 
 and obj.name ='$item' ;def tag = services.query.hql(hql).execute();
 i = i + 1;
 json.projetos.add({id: xwiki.getDocument(item).getName()})}
 jsontool.serialize(json);
 {{/groovy}}
 Ty.
 Atenciosamente,
 Eduardo Abritta
 e-mail: eduardo.abri...@outlook.com | cel: (32)8472-9631


 ___
 users mailing list
 users@xwiki.org
 http://lists.xwiki.org/mailman/listinfo/users



 ___
 users mailing list
 users@xwiki.org
 http://lists.xwiki.org/mailman/listinfo/users



-- 
Thomas Mortagne
___
users mailing list
users@xwiki.org
http://lists.xwiki.org/mailman/listinfo/users

[xwiki-users] Implement Jsontool.serialize in groovy

[Eduardo Abritta]

Hello Sirs,
I need to implement the function jsontool.serialize in the code below, 
groovy, and I do not understand the error that is accusing, can help me solve 
this?
{{groovy}}def hql0 = , BaseObject as obj where doc.fullName = obj.name and 
obj.className = 'E-nova.E-novaClass';def listProjetos = 
services.query.hql(hql0).execute();def json = ['totalprojetos': 
listProjetos.size(), 'projetos':[1000] , 'edges':[1000] ];def i = 0;
for(item in listProjetos){  def hql = select distinct list from BaseObject 
as obj, DBStringListProperty as prop join prop.list list where 
obj.className='XWiki.TagClass' and obj.id=prop.id.id and prop.id.name='tags' 
and obj.name ='$item' ;def tag = services.query.hql(hql).execute();i = 
i + 1;
json.projetos.add({id: xwiki.getDocument(item).getName()})}
jsontool.serialize(json);
{{/groovy}}
Ty.
Atenciosamente,
Eduardo Abritta
e-mail: eduardo.abri...@outlook.com | cel: (32)8472-9631

  
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users mailing list
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