作为我的发布过程的一部分，我必须比较我的应用程序使用的一些JSON配置数据。作为第一次尝试，我只是简单地打印了JSON并对它们进行了差异化（使用kdiff3或只是diff）。
然而，随着数据的增长，kdiff3会混淆输出中的不同部分，使得添加看起来像巨型修改，奇数删除等。这使得很难弄清楚有什么不同。我也尝试过其他差异工具（meld，kompare，diff，还有其他一些），但它们都有同样的问题。
尽管我付出了最大的努力，但我似乎无法以diff工具可以理解的方式格式化JSON。
示例数据：
[
{
"name": "date",
"type": "date",
"nullable": true,
"state": "enabled"
},
{
"name": "owner",
"type": "string",
"nullable": false,
"state": "enabled",
}
...lots more...
]
上述情况可能不会导致问题（当问题开始出现数百行时会出现问题），但这就是比较内容的要点。
这只是一个样本;完整对象是4-5个属性，有些属性中有4-5个属性。属性名称非常统一，但它们的值非常不同。
一般来说，似乎所有差异工具都会将结束“}”与关闭“}”的下一个对象混淆。我似乎无法打破这种习惯。
我尝试添加空格，更改缩进，并在各个对象之前和之后添加一些“BEGIN”和“END”字符串，但该工具仍然感到困惑。
答案 0 :(得分：21)
如果您的任何工具都有此选项，Patience Diff可以为您提供更好的帮助。我会尝试用它找到一个工具（其他是Git和Bazaar）并报告回来。
编辑：似乎the implementation in Bazaar可用作独立工具，只需极少的更改。
Edit2：WTH，为什么不粘贴你让我破解的新酷炫差异脚本的来源？在这里，我没有版权声明，只是重新安排了Bram / Canonical的代码。
#!/usr/bin/env python
# Copyright (C) 2005, 2006, 2007 Canonical Ltd
# Copyright (C) 2005 Bram Cohen, Copyright (C) 2005, 2006 Canonical Ltd
#
# This program is free software; you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation; either version 2 of the License, or
# (at your option) any later version.
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program; if not, write to the Free Software
# Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
import os
import sys
import time
import difflib
from bisect import bisect
__all__ = ['PatienceSequenceMatcher', 'unified_diff', 'unified_diff_files']
py3k = False
try:
xrange
except NameError:
py3k = True
xrange = range
# This is a version of unified_diff which only adds a factory parameter
# so that you can override the default SequenceMatcher
# this has been submitted as a patch to python
def unified_diff(a, b, fromfile='', tofile='', fromfiledate='',
tofiledate='', n=3, lineterm='\n',
sequencematcher=None):
r"""
Compare two sequences of lines; generate the delta as a unified diff.
Unified diffs are a compact way of showing line changes and a few
lines of context. The number of context lines is set by 'n' which
defaults to three.
By default, the diff control lines (those with ---, +++, or @@) are
created with a trailing newline. This is helpful so that inputs
created from file.readlines() result in diffs that are suitable for
file.writelines() since both the inputs and outputs have trailing
newlines.
For inputs that do not have trailing newlines, set the lineterm
argument to "" so that the output will be uniformly newline free.
The unidiff format normally has a header for filenames and modification
times. Any or all of these may be specified using strings for
'fromfile', 'tofile', 'fromfiledate', and 'tofiledate'. The modification
times are normally expressed in the format returned by time.ctime().
Example:
>>> for line in unified_diff('one two three four'.split(),
... 'zero one tree four'.split(), 'Original', 'Current',
... 'Sat Jan 26 23:30:50 1991', 'Fri Jun 06 10:20:52 2003',
... lineterm=''):
... print line
--- Original Sat Jan 26 23:30:50 1991
+++ Current Fri Jun 06 10:20:52 2003
@@ -1,4 +1,4 @@
+zero
one
-two
-three
+tree
four
"""
if sequencematcher is None:
import difflib
sequencematcher = difflib.SequenceMatcher
if fromfiledate:
fromfiledate = '\t' + str(fromfiledate)
if tofiledate:
tofiledate = '\t' + str(tofiledate)
started = False
for group in sequencematcher(None,a,b).get_grouped_opcodes(n):
if not started:
yield '--- %s%s%s' % (fromfile, fromfiledate, lineterm)
yield '+++ %s%s%s' % (tofile, tofiledate, lineterm)
started = True
i1, i2, j1, j2 = group[0][3], group[-1][4], group[0][5], group[-1][6]
yield "@@ -%d,%d +%d,%d @@%s" % (i1+1, i2-i1, j1+1, j2-j1, lineterm)
for tag, i1, i2, j1, j2 in group:
if tag == 'equal':
for line in a[i1:i2]:
yield ' ' + line
continue
if tag == 'replace' or tag == 'delete':
for line in a[i1:i2]:
yield '-' + line
if tag == 'replace' or tag == 'insert':
for line in b[j1:j2]:
yield '+' + line
def unified_diff_files(a, b, sequencematcher=None):
"""Generate the diff for two files.
"""
mode = 'rb'
if py3k: mode = 'r'
# Should this actually be an error?
if a == b:
return []
if a == '-':
file_a = sys.stdin
time_a = time.time()
else:
file_a = open(a, mode)
time_a = os.stat(a).st_mtime
if b == '-':
file_b = sys.stdin
time_b = time.time()
else:
file_b = open(b, mode)
time_b = os.stat(b).st_mtime
# TODO: Include fromfiledate and tofiledate
return unified_diff(file_a.readlines(), file_b.readlines(),
fromfile=a, tofile=b,
sequencematcher=sequencematcher)
def unique_lcs_py(a, b):
"""Find the longest common subset for unique lines.
:param a: An indexable object (such as string or list of strings)
:param b: Another indexable object (such as string or list of strings)
:return: A list of tuples, one for each line which is matched.
[(line_in_a, line_in_b), ...]
This only matches lines which are unique on both sides.
This helps prevent common lines from over influencing match
results.
The longest common subset uses the Patience Sorting algorithm:
http://en.wikipedia.org/wiki/Patience_sorting
"""
# set index[line in a] = position of line in a unless
# a is a duplicate, in which case it's set to None
index = {}
for i in xrange(len(a)):
line = a[i]
if line in index:
index[line] = None
else:
index[line]= i
# make btoa[i] = position of line i in a, unless
# that line doesn't occur exactly once in both,
# in which case it's set to None
btoa = [None] * len(b)
index2 = {}
for pos, line in enumerate(b):
next = index.get(line)
if next is not None:
if line in index2:
# unset the previous mapping, which we now know to
# be invalid because the line isn't unique
btoa[index2[line]] = None
del index[line]
else:
index2[line] = pos
btoa[pos] = next
# this is the Patience sorting algorithm
# see http://en.wikipedia.org/wiki/Patience_sorting
backpointers = [None] * len(b)
stacks = []
lasts = []
k = 0
for bpos, apos in enumerate(btoa):
if apos is None:
continue
# as an optimization, check if the next line comes at the end,
# because it usually does
if stacks and stacks[-1] < apos:
k = len(stacks)
# as an optimization, check if the next line comes right after
# the previous line, because usually it does
elif stacks and stacks[k] < apos and (k == len(stacks) - 1 or
stacks[k+1] > apos):
k += 1
else:
k = bisect(stacks, apos)
if k > 0:
backpointers[bpos] = lasts[k-1]
if k < len(stacks):
stacks[k] = apos
lasts[k] = bpos
else:
stacks.append(apos)
lasts.append(bpos)
if len(lasts) == 0:
return []
result = []
k = lasts[-1]
while k is not None:
result.append((btoa[k], k))
k = backpointers[k]
result.reverse()
return result
def recurse_matches_py(a, b, alo, blo, ahi, bhi, answer, maxrecursion):
"""Find all of the matching text in the lines of a and b.
:param a: A sequence
:param b: Another sequence
:param alo: The start location of a to check, typically 0
:param ahi: The start location of b to check, typically 0
:param ahi: The maximum length of a to check, typically len(a)
:param bhi: The maximum length of b to check, typically len(b)
:param answer: The return array. Will be filled with tuples
indicating [(line_in_a, line_in_b)]
:param maxrecursion: The maximum depth to recurse.
Must be a positive integer.
:return: None, the return value is in the parameter answer, which
should be a list
"""
if maxrecursion < 0:
print('max recursion depth reached')
# this will never happen normally, this check is to prevent DOS attacks
return
oldlength = len(answer)
if alo == ahi or blo == bhi:
return
last_a_pos = alo-1
last_b_pos = blo-1
for apos, bpos in unique_lcs_py(a[alo:ahi], b[blo:bhi]):
# recurse between lines which are unique in each file and match
apos += alo
bpos += blo
# Most of the time, you will have a sequence of similar entries
if last_a_pos+1 != apos or last_b_pos+1 != bpos:
recurse_matches_py(a, b, last_a_pos+1, last_b_pos+1,
apos, bpos, answer, maxrecursion - 1)
last_a_pos = apos
last_b_pos = bpos
answer.append((apos, bpos))
if len(answer) > oldlength:
# find matches between the last match and the end
recurse_matches_py(a, b, last_a_pos+1, last_b_pos+1,
ahi, bhi, answer, maxrecursion - 1)
elif a[alo] == b[blo]:
# find matching lines at the very beginning
while alo < ahi and blo < bhi and a[alo] == b[blo]:
answer.append((alo, blo))
alo += 1
blo += 1
recurse_matches_py(a, b, alo, blo,
ahi, bhi, answer, maxrecursion - 1)
elif a[ahi - 1] == b[bhi - 1]:
# find matching lines at the very end
nahi = ahi - 1
nbhi = bhi - 1
while nahi > alo and nbhi > blo and a[nahi - 1] == b[nbhi - 1]:
nahi -= 1
nbhi -= 1
recurse_matches_py(a, b, last_a_pos+1, last_b_pos+1,
nahi, nbhi, answer, maxrecursion - 1)
for i in xrange(ahi - nahi):
answer.append((nahi + i, nbhi + i))
def _collapse_sequences(matches):
"""Find sequences of lines.
Given a sequence of [(line_in_a, line_in_b),]
find regions where they both increment at the same time
"""
answer = []
start_a = start_b = None
length = 0
for i_a, i_b in matches:
if (start_a is not None
and (i_a == start_a + length)
and (i_b == start_b + length)):
length += 1
else:
if start_a is not None:
answer.append((start_a, start_b, length))
start_a = i_a
start_b = i_b
length = 1
if length != 0:
answer.append((start_a, start_b, length))
return answer
def _check_consistency(answer):
# For consistency sake, make sure all matches are only increasing
next_a = -1
next_b = -1
for (a, b, match_len) in answer:
if a < next_a:
raise ValueError('Non increasing matches for a')
if b < next_b:
raise ValueError('Non increasing matches for b')
next_a = a + match_len
next_b = b + match_len
class PatienceSequenceMatcher_py(difflib.SequenceMatcher):
"""Compare a pair of sequences using longest common subset."""
_do_check_consistency = True
def __init__(self, isjunk=None, a='', b=''):
if isjunk is not None:
raise NotImplementedError('Currently we do not support'
' isjunk for sequence matching')
difflib.SequenceMatcher.__init__(self, isjunk, a, b)
def get_matching_blocks(self):
"""Return list of triples describing matching subsequences.
Each triple is of the form (i, j, n), and means that
a[i:i+n] == b[j:j+n]. The triples are monotonically increasing in
i and in j.
The last triple is a dummy, (len(a), len(b), 0), and is the only
triple with n==0.
>>> s = PatienceSequenceMatcher(None, "abxcd", "abcd")
>>> s.get_matching_blocks()
[(0, 0, 2), (3, 2, 2), (5, 4, 0)]
"""
# jam 20060525 This is the python 2.4.1 difflib get_matching_blocks
# implementation which uses __helper. 2.4.3 got rid of helper for
# doing it inline with a queue.
# We should consider doing the same for recurse_matches
if self.matching_blocks is not None:
return self.matching_blocks
matches = []
recurse_matches_py(self.a, self.b, 0, 0,
len(self.a), len(self.b), matches, 10)
# Matches now has individual line pairs of
# line A matches line B, at the given offsets
self.matching_blocks = _collapse_sequences(matches)
self.matching_blocks.append( (len(self.a), len(self.b), 0) )
if PatienceSequenceMatcher_py._do_check_consistency:
if __debug__:
_check_consistency(self.matching_blocks)
return self.matching_blocks
unique_lcs = unique_lcs_py
recurse_matches = recurse_matches_py
PatienceSequenceMatcher = PatienceSequenceMatcher_py
def main(args):
import optparse
p = optparse.OptionParser(usage='%prog [options] file_a file_b'
'\nFiles can be "-" to read from stdin')
p.add_option('--patience', dest='matcher', action='store_const', const='patience',
default='patience', help='Use the patience difference algorithm')
p.add_option('--difflib', dest='matcher', action='store_const', const='difflib',
default='patience', help='Use python\'s difflib algorithm')
algorithms = {'patience':PatienceSequenceMatcher, 'difflib':difflib.SequenceMatcher}
(opts, args) = p.parse_args(args)
matcher = algorithms[opts.matcher]
if len(args) != 2:
print('You must supply 2 filenames to diff')
return -1
for line in unified_diff_files(args[0], args[1], sequencematcher=matcher):
sys.stdout.write(line)
if __name__ == '__main__':
sys.exit(main(sys.argv[1:]))
编辑3：我还制作了minimally standalone version Neil Fraser Diff Match and Patch的DataDiff，我对您的用例结果进行比较非常感兴趣。我再次声称没有版权。
编辑4：我刚刚找到{{3}}，这可能是另一种尝试的工具。
DataDiff是一个提供的库 人类可读的python数据差异 结构。它可以处理序列 类型（列表，元组等），集合和 字典。
字典和序列将是 在适用时以递归方式进行差异化。
答案 1 :(得分：4)
所以，我之前写了一个工具来做JSON文件的统一差异，可能会引起一些兴趣。
https://github.com/jclulow/jsondiff
该工具的输入和输出的一些示例显示在github页面上。
答案 2 :(得分：2)
您应该从子包中检出difflet。它既是node.js模块又是命令行实用程序，它完全按照以下方式执行：
答案 3 :(得分：2)
我知道这是一个非常古老的问题，但是python模块＆＃34; JSON工具＆＃34;为扩散json文件提供了另一种解决方案：
https://pypi.python.org/pypi/json_tools https://bitbucket.org/vadim_semenov/json_tools/src/75cc15381188c760badbd5b66aef9941a42c93fa?at=default
答案 4 :(得分：0)
Eclipse可能做得更好。在eclipse项目中打开这两个文件，选择它们，然后右键单击 - ＆gt;比较 - ＆gt;彼此。
答案 5 :(得分：0)
除了格式化更改之外，diffing工具还应该以稳定的方式（例如按字母顺序）对JSON对象属性进行排序，因为属性的顺序在语义上是无意义的。也就是说，重新排序属性不应该改变内容的含义。
除此之外，解析和漂亮打印的方式最多只能将一个条目放在一行上，这可能允许使用文本差异。 如果没有，任何适用于树的diff算法（用于xml diffing）都应该更好。