如何在OpenRefine ReST-API的“创建项目”发布请求中传递选项JSON?

问题描述 投票:1回答:1

我目前正在尝试将Excel表格(如.xls)上传到Ontotexts GraphDB的OpenRefine(或OntoRefine)模块。由于我在上传xls时遇到问题,我决定先将xls文件转换为csv文件然后上传。不幸的是,每次将文件作为CSV时,OpenRefine都不会自动识别。因此,每行中的所有数据都存储在一列中。例如。:

--------------------------------------------------
|      Col1,     Col2,     Col3,     Col4        |
--------------------------------------------------
|      Row11,     Row12,     Row13,     Row14    |
--------------------------------------------------
|      Row21,     Row22,     Row23,     Row24    |
--------------------------------------------------

代替:

--------------------------------------------------
|      Col1    |  Col2    |  Col3    |  Col4     |
--------------------------------------------------
|      Row11   |  Row12   |  Row13   |  Row14    |
--------------------------------------------------
|      Row21   |  Row22   |  Row23   |  Row24    |
--------------------------------------------------

随着发布请求

POST /command/core/create-project-from-upload

'format'参数中的文件格式和'options'参数中带分隔符的json可以添加到POST请求中。但是,这也不起作用,官方OpenRefine文档(https://github.com/OpenRefine/OpenRefine/wiki/OpenRefine-API)不包含任何关于'options'JSON语法的提示。

我当前的代码如下所示:

import os
import xlrd
import csv
import requests
import re

xls_file_name_ext = os.path.basename('excel_file.xls')

# create the filename with path to the new csv file (path + name stays the same)
csv_file_path = os.path.dirname(xls_file_name_ext) + '/' + os.path.splitext(xls_file_name_ext)[0] + '.csv'

# remove all comma in xls file
xls_wb = xlrd.open_workbook(xls_file_name_ext)
xls_sheet = xls_wb.sheet_by_index(0)
for col in range(xls_sheet.ncols):
    for row in range(xls_sheet.nrows):
        _new_cell_val = str(xls_sheet.cell(row, col).value).replace(",", " ")
        xls_sheet._cell_values[row][col] = _new_cell_val

# write to csv
with open(csv_file_path, 'w', newline='', encoding='utf-8') as csv_file:
    c_w = csv.writer(csv_file, delimiter=',')
    for row in range(xls_sheet.nrows):
        c_w.writerow(xls_sheet.row_values(row))

ontorefine_server = 'http://localhost:7200/orefine'

# filename of csv as project name in OntoRefine
onterefine_project_name = os.path.splitext(os.path.basename(csv_file_path))[0]

# the required paraneters for the post request
ontorefine_data = {"project-name": onterefine_project_name,
                   "format": "text/line-based/*sv",
                   "options": {
                       "separator": ","
                                }
                   }
ontorefine_file = {'project-file': open(csv_file_path, "rb")}

# execute the post request
ontorefine_response = requests.post(
    ontorefine_server + '/command/core/create-project-from-upload', data=ontorefine_data, files=ontorefine_file
)

我假设我错误地传递了POST请求参数。

openrefine graphdb
1个回答
0
投票

当然,这完全取决于您的输入数据,但格式看起来还不错。如果您尝试从UI导入,这就是OntoRefine在“幕后”所做的事情。您可以通过拦截网络流量为自己查看相同的有效负载:

{
"format": "text/line-based/*sv",
"options": {
    "project-name":"Your-project-here",
    "separator":","
}

从这一点来看,看起来项目名称位置是唯一的区别。这是一个curl命令,它执行相同的操作:

curl 'http://localhost:7200/orefine/command/core/importing-controller?controller=core%2Fdefault-importing-controller&jobID=1&subCommand=create-project' -H 'Content-Type: application/x-www-form-urlencoded; charset=UTF-8' --data 'format%3Dtext%2Fline-based%2F*sv%26options%3D%7B%22separator%22%3A%22%2C%22%22projectName%22%3A%22Your-project-name%22%7D'
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