本帖最后由 bobo 于 2013-9-14 09:16 编辑
如何在新表单建成后,直接进行流程启动和提交的脚本操作?- //生成故障分析报告
- var process = new Packages.cn.myapps.core.dynaform.document.ejb.DocumentProcessBean(getApplication());
- var params = new Packages.cn.myapps.base.action.ParamsTable();
- var user = getWebUser();
- var doc = new Packages.cn.myapps.core.dynaform.document.ejb.Document();
- doc.setFormid('11e3-1135-56c0edcf-9b85-a1c246d4e64f');//表单ID
- doc.setFormname('艾诺demo/FRACAS/故障分析报告');
- doc.setAuthor(user.getId());
- doc.setIstmp(false);
- doc.setFlowid("11e3-1135-56c0edcf-9b85-a1c246d4e64f");//flowid为对应的流程Id.
- doc.setApplicationid(getApplication());
- doc.setDomainid(user.getDomainid());
- doc.addStringItem("故障分析报告编号", gzfxbgbh );
- doc.addStringItem("故障报告表编号", gzbgbh );
- doc.addStringItem("域", Yu );
- //var t=process.doStartFlow(doc,params,user);//启动流程的同时会新建document对象
- //process.doCreate(doc);
- //process.doStartFlowOrUpdate(doc,params,user);//启动流程
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