René's URL Explorer Experiment

{"@context":"https://schema.org","@type":"DiscussionForumPosting","headline":"Getting list numbering","articleBody":"Hi all,\n\nI`m wondering how can I get list numbering of paragraphs with style \"Paragraph List\"\nSo far I got such thoughts: \n\n``` python\nnumId_val = paragraph._element.pPr.numPr.numId.val # getting value for \u003cw:num\u003e in numbering.xml\n```\n\nNext step should be finding \u003cw:num\u003e which corresponds to numId, that I got above\n\n``` python\nnum_obj = filter(lambda x: x.numId == numId_val, document.part.numbering_part.element.num_lst)[0]\n```\n\nThen look into \u003cw:AbstractNum\u003e to get ordering info (such as start value ...)\n\n``` python\nabsNum_obj = num_obj.abstractNumId\n```\n\nAnd that is fail, because it does not represent what I see in xml, \nI get smth like this\n\n``` python\nprint absNum_obj.xml\n```\n\n``` xml\n\u003cw:abstractNumId xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\" xmlns:wpc=\"http://schemas.microsoft.com/office/word/2010/wordprocessingCanvas\" xmlns:mc=\"http://schemas.openxmlformats.org/markup-compatibility/2006\" xmlns:o=\"urn:schemas-microsoft-com:office:office\" xmlns:r=\"http://schemas.openxmlformats.org/officeDocument/2006/relationships\" xmlns:m=\"http://schemas.openxmlformats.org/officeDocument/2006/math\" xmlns:v=\"urn:schemas-microsoft-com:vml\" xmlns:wp14=\"http://schemas.microsoft.com/office/word/2010/wordprocessingDrawing\" xmlns:wp=\"http://schemas.openxmlformats.org/drawingml/2006/wordprocessingDrawing\" xmlns:w10=\"urn:schemas-microsoft-com:office:word\" xmlns:w14=\"http://schemas.microsoft.com/office/word/2010/wordml\" xmlns:w15=\"http://schemas.microsoft.com/office/word/2012/wordml\" xmlns:wpg=\"http://schemas.microsoft.com/office/word/2010/wordprocessingGroup\" xmlns:wpi=\"http://schemas.microsoft.com/office/word/2010/wordprocessingInk\" xmlns:wne=\"http://schemas.microsoft.com/office/word/2006/wordml\" xmlns:wps=\"http://schemas.microsoft.com/office/word/2010/wordprocessingShape\" w:val=\"0\"/\u003e\n```\n\nInstead of \n\n``` xml\n\u003cw:abstractNum w:abstractNumId=\"0\"\u003e\n  \u003cw:nsid w:val=\"00000012\" /\u003e \n  \u003cw:multiLevelType w:val=\"hybridMultilevel\" /\u003e \n  \u003cw:tmpl w:val=\"31AE522E\" /\u003e \n- \u003cw:lvl w:ilvl=\"0\" w:tplc=\"0419000F\"\u003e\n  \u003cw:start w:val=\"1\" /\u003e \n  \u003cw:numFmt w:val=\"decimal\" /\u003e \n  \u003cw:lvlText w:val=\"%1.\" /\u003e \n  \u003cw:lvlJc w:val=\"left\" /\u003e \n- \u003cw:pPr\u003e\n- \u003cw:tabs\u003e\n  \u003cw:tab w:val=\"left\" w:pos=\"720\" /\u003e \n  \u003c/w:tabs\u003e\n  \u003cw:ind w:left=\"720\" w:hanging=\"360\" /\u003e \n  \u003c/w:pPr\u003e\n  \u003c/w:lvl\u003e\n....\n\u003c/w:abstractNum\u003e\n```\n\nAnd more, if I manage to get start value, I will have to manually count paragraphs with same level id, then increase start value and only after that I will get numbering of paragraph.\n\nMaybe there is more straightforward and easy way?\n\nThanks\n","author":{"url":"https://github.com/norecces","@type":"Person","name":"norecces"},"datePublished":"2015-04-29T16:35:57.000Z","interactionStatistic":{"@type":"InteractionCounter","interactionType":"https://schema.org/CommentAction","userInteractionCount":7},"url":"https://github.com/180/python-docx/issues/180"}