IE7与对象
IE7与对象
提问于 2010-10-07 13:11:05
对于IE7不希望通过使用HTMLTMPL的ajax调用传递一个已被腌制的对象，我有一个问题。它可以在IE8 (兼容模式)和火狐中工作。
我使用以下命令对一个对象进行了腌制：
newhash['pickled'] = pickle.dumps(hash)由于JS不喜欢换行符，所以我使用以下方法对它们进行了正则化：
newhash['pickled'] = re.sub('\n', 'LINEBREAK', newhash['pickled'])当我在Python中捕获我的腌制对象时，它取出换行并放回\n。
$.getJSON('/folder/MyPython.py', {'mode':'MyFunction', 'pickled':"<TMPL_VAR pickled ESCAPE="NONE">"}, function(data){我就在它之前和之后发出警告，一切都像预期的一样。如果我在函数(数据)部分发出警报，则不会发出任何警报。但是，如果我取出ajax调用中的“泡菜”键，一切都会正常工作。泡沫化对象是我的python进行计算所需的所有变量，因此，简单地说，不传递它不是一个选项。
我的腌制对象看起来如下：
(dp0LINEBREAKS'rlbool'LINEBREAKp1LINEBREAKL1LLINEBREAKsS'class7'LINEBREAKp2LINEBREAKS'50'LINEBREAKp3LINEBREAKsS'fedxbool'LINEBREAKp4LINEBREAKL1LLINEBREAKsS'weight1'LINEBREAKp5LINEBREAKS'1500'LINEBREAKp6LINEBREAKsS'conwaybool'LINEBREAKp7LINEBREAKL1LLINEBREAKsS'originzip'LINEBREAKp8LINEBREAKS'37130'LINEBREAKp9LINEBREAKsS'company'LINEBREAKp10LINEBREAKS''LINEBREAKp11LINEBREAKsS'destinationzip'LINEBREAKp12LINEBREAKS'37130'LINEBREAKp13LINEBREAKsS'class6'LINEBREAKp14LINEBREAKS'50'LINEBREAKp15LINEBREAKsS'mode'LINEBREAKp16LINEBREAKS'Crawl'LINEBREAKp17LINEBREAKsS'averitlogin'LINEBREAKp18LINEBREAKS'QVSINC'LINEBREAKp19LINEBREAKsS'accessories'LINEBREAKp20LINEBREAK(lp21LINEBREAKsS'address'LINEBREAKp22LINEBREAKS'330%20Robert%20Rose%20Blvd.'LINEBREAKp23LINEBREAKsS'active'LINEBREAKp24LINEBREAKL1LLINEBREAKsS'averittbool'LINEBREAKp25LINEBREAKL1LLINEBREAKsS'id'LINEBREAKp26LINEBREAKL19LLINEBREAKsS'averitpass'LINEBREAKp27LINEBREAKS'MERIDIAN'LINEBREAKp28LINEBREAKsS'shipmentdate'LINEBREAKp29LINEBREAKS'2010-10-08'LINEBREAKp30LINEBREAKsS'city'LINEBREAKp31LINEBREAKS'Murfreesboro'LINEBREAKp32LINEBREAKsS'class5'LINEBREAKp33LINEBREAKS'50'LINEBREAKp34LINEBREAKsS'last'LINEBREAKp35LINEBREAKS'Morgan'LINEBREAKp36LINEBREAKsS'originstate'LINEBREAKp37LINEBREAKS'TN'LINEBREAKp38LINEBREAKsS'zip'LINEBREAKp39LINEBREAKS'37129'LINEBREAKp40LINEBREAKsS'phone'LINEBREAKp41LINEBREAKS'615%20713-5432'LINEBREAKp42LINEBREAKsS'destinationstate'LINEBREAKp43LINEBREAKS'TN'LINEBREAKp44LINEBREAKsS'Accessories'LINEBREAKp45LINEBREAK(lp46LINEBREAKsS'comments'LINEBREAKp47LINEBREAKg11LINEBREAKsS'estesbool'LINEBREAKp48LINEBREAKL1LLINEBREAKsS'origincity'LINEBREAKp49LINEBREAKS'Murfreesboro'LINEBREAKp50LINEBREAKsS'class8'LINEBREAKp51LINEBREAKS'50'LINEBREAKp52LINEBREAKsS'state'LINEBREAKp53LINEBREAKS'TN'LINEBREAKp54LINEBREAKsS'email'LINEBREAKp55LINEBREAKS'chris2'LINEBREAKp56LINEBREAKsS'vitranbool'LINEBREAKp57LINEBREAKL1LLINEBREAKsS'saiabool'LINEBREAKp58LINEBREAKL1LLINEBREAKsS'destinationcity'LINEBREAKp59LINEBREAKS'Murfreesboro'LINEBREAKp60LINEBREAKsS'class3'LINEBREAKp61LINEBREAKS'50'LINEBREAKp62LINEBREAKsS'class4'LINEBREAKp63LINEBREAKS'50'LINEBREAKp64LINEBREAKsS'class1'LINEBREAKp65LINEBREAKS'50'LINEBREAKp66LINEBREAKsS'class2'LINEBREAKp67LINEBREAKS'50'LINEBREAKp68LINEBREAKsS'address2'LINEBREAKp69LINEBREAKg11LINEBREAKsS'first'LINEBREAKp70LINEBREAKS'Chris'LINEBREAKp71LINEBREAKs.我只能假设这里有一些IE7有问题的字符。
提前谢谢你的帮助。
回答 3
Stack Overflow用户
回答已采纳
发布于 2010-10-07 13:31:53
首先，阅读Python Pickle为何不安全。不要使用用户可以修改的已被腌制的对象。
相反，为什么不简单地使用JSON，这显然是为JavaScript开发的。它包含在Python2.6中，也可用于较早的版本。因为您的数据只是一个字典，所以JSON应该工作得很好。
另一个选项是base-64编码，这不会给您带来特殊字符的麻烦。
Stack Overflow用户
发布于 2010-10-07 13:13:47
这意味着可以在用户端更改已被腌制的数据？会有很高的安全风险。
Stack Overflow用户
发布于 2010-10-07 14:16:09
答案是URL长度。最大请求url长度为2083，我正在检查它。修好了！
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/3881958
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