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python - 我有一个无效的语法错误，该错误出现在以前未编辑的工作函数中

问题描述

我有一系列旨在从库存网站提取数据的功能。这些功能按预期工作，以遍历清单中的一系列文件夹以提取我打算使用的数据项。在处理了我在这里询问的函数不依赖的代码中其他地方的错误后出现了错误，尽管我在此错误之前更改的函数确实使用了出现错误的函数。

我尝试删除第 73 行中出现语法错误的“返回项”，但随后错误被推到第 75 行的下一行代码“def sortly_pull():”。我使用 spyder 编译和写进去，它今天早些时候崩溃了，所以我想知道这是否更像是一个硬件问题......关于在哪里寻找或使用工具来解决这个问题的任何提示或线索将不胜感激。谢谢大家的时间！

编辑：错误是'文件“C：stuff\sortly_pull_final.py”，第 73 行返回项目 ^ SyntaxError：无效语法'

'''

        def folder_scraper(url):
    r = requests.get(url, headers=headers)
    t = r.json()
    folders = []
    for j in range(len(t['data'])):
        if t['data'][j]["type"] == 'folder':
            folders.append(t['data'][j]["id"])
    for i in range(len(folders)):
        if folders[i] is not str:
            folders[i] = str(folders[i])
    return folders
'''



'''
        def item_scraper(url):
    url = url + '&include=custom_attributes%2Cphotos'
    r = requests.get(url, headers=headers)
    t = r.json()
    items = []
    for j in range(len(t['data'])):
        if t[['data'][j]["type"] == 'item':
            items.append(t['data'][j])
    return items
'''

'''
       def sortly_pull():
    folders_list = folder_scraper(url5)
    subfolders_list = []
    subfolders_list2 = []
    subfolders_list3 = []
    subfolders_list4 = []
    subfolders_list5 = []
    subfolders_list6 = []
    subfolders_list7 = []
    subfolders_list8 = []
    subfolders_list9 = []
    subfolders_list10 = []
    sortly_data = []
    for i in range(len(folders_list)):
        url = url_combiner_folder(folders_list[i])
        subfolders_list.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for j in range(len(subfolders_list)):
        url = url_combiner_folder(subfolders_list[j])
        subfolders_list2.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for a in range(len(subfolders_list2)):
        url = url_combiner_folder(subfolders_list2[a])
        subfolders_list3.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for b in range(len(subfolders_list3)):
        url = url_combiner_folder(subfolders_list3[b])
        subfolders_list4.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for c in range(len(subfolders_list4)):
        url = url_combiner_folder(subfolders_list4[c])
        subfolders_list5.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for d in range(len(subfolders_list5)):
        url = url_combiner_folder(subfolders_list5[d])
        subfolders_list6.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for e in range(len(subfolders_list6)):
        url = url_combiner_folder(subfolders_list6[e])
        subfolders_list7.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for f in range(len(subfolders_list7)):
        url = url_combiner_folder(subfolders_list7[f])
        subfolders_list8.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for g in range(len(subfolders_list8)):
        url = url_combiner_folder(subfolders_list8[g])
        subfolders_list9.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for h in range(len(subfolders_list9)):
        url = url_combiner_folder(subfolders_list9[h])
        subfolders_list10.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    for k in range(len(subfolders_list10)):
        if len(subfolders_list10) == 0:
            pass
        url = url_combiner_folder(subfolders_list10[k])
        #subfolders_list11.append(folder_scraper(url))
        sortly_data.append(item_scraper(url))
    #print(len(subfolders_list))
    print(len(subfolders_list10))
    print(len(sortly_data))
    return sortly_data

'''

标签： python

解决方案

我不是 python 专家，但以下看起来它在t从未关闭之后有一个额外的括号：

if t[['data'][j]["type"] == 'item':

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