在文本上区分JSON
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，我只是打印了JSON并对它们进行了区分（使用kdiff3或diff）。
然而，随着数据的增长，kdiff3会混淆输出中的不同部分，使得增加看起来像是巨大的修改，奇怪的删除等等。这使得很难弄清楚什么是不同的。 我也尝试了其他差异工具（meld，kompare，diff和其他几个），但它们都有同样的问题。
尽管我尽了最大的努力，但似乎无法通过diff工具可以理解的方式来格式化JSON。
示例数据：
[
{
"name": "date",
"type": "date",
"nullable": true,
"state": "enabled"
},
{
"name": "owner",
"type": "string",
"nullable": false,
"state": "enabled",
}
...lots more...
]
上述可能不会导致问题（问题发生在有数百条线路时），但那是正在比较的内容的要点。
这只是一个样本; 完整对象是4-5个属性，并且一些属性具有4-5个属性。 属性名称非常统一，但它们的值很不一样。
一般来说，似乎所有diff工具都会将结束“}”与关闭“}”的下一个对象混淆。 我似乎无法打破他们的这种习惯。
我尝试添加空白，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，但该工具仍然令人困惑。
如果您的工具有任何选项，Patience Diff可以为您提供更好的工作。 我会尝试找到一个工具（其他tha Git和Bazaar）并回报。
编辑：看来，集市中的实施可以作为一个独立的工具使用，只需很少的更改。
编辑2：WTH，为什么不粘贴新的酷差异脚本的来源，你让我破解？ 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：我也做了Neil Fraser的Diff Match和Patch的最低限度独立版本，我对你的用例的结果比较非常感兴趣。 再次，我声称没有版权。
编辑4：我刚刚找到DataDiff，这可能是另一个尝试的工具。
DataDiff是一个库，用于提供Python数据结构的可读的差异。 它可以处理序列类型（列表，元组等），集合和字典。
在适用的时候，字典和序列将被递归地区分。
所以，我写了一个工具来完成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出现在github页面上。
您应该从substack签出difflet。 它既是一个node.js模块，也是一个命令行实用程序，完全可以做到这一点：
https://github.com/substack/difflet
链接地址: http://www.djcxy.com/p/47113.html