Querydict一个字符串

nel*_*ela 2 python django

我有以下字符串:

string = 'amount=2000|captureDay=0|captureMode=AUTHOR_CAPTURE|currencyCode=978|merchantId=002020000000001|orderId=|transactionDateTime=2012-04-12T12:09:56+02:00|transactionReference=1212943|keyVersion=1|authorisationId=0020000006791167|complementaryCode=|maskedPan=|paymentMeanBrand=IDEAL|paymentMeanType=CREDIT_TRANSFER|responseCode=00'
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我该如何制作一个querydict呢?

我希望能够使用item['amount']哪种回报2000

编辑:

到目前为止我尝试过的:

dict = string.split('|')
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输出是:

['amount=2000', 'captureDay=0', 'captureMode=AUTHOR_CAPTURE', 'currencyCode=978', 'merchantId=002020000000001', 'orderId=', 'transactionDateTime=2012-04-12T12:09:56+02:00', 'transactionReference=1212943', 'keyVersion=1', 'authorisationId=0020000006791167', 'complementaryCode=', 'maskedPan=', 'paymentMeanBrand=IDEAL', 'paymentMeanType=CREDIT_TRANSFER', 'responseCode=00']
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Mat*_*ttH 7

你可以替换|使用&,并给整个字符串Django的的QueryDict.

请参阅QueryDict文档

例如

In [1]: from django.http import QueryDict

In [2]: sample = 'amount=2000|captureDay=0|captureMode=AUTHOR_CAPTURE|currencyCode=978|merchantId=002020000000001|orderId=|transactionDateTime=2012-04-12T12:09:56+02:00|transactionReference=1212943|keyVersion=1|authorisationId=0020000006791167|complementaryCode=|maskedPan=|paymentMeanBrand=IDEAL|paymentMeanType=CREDIT_TRANSFER|responseCode=00'

In [3]: qdict = QueryDict(sample.replace('|','&'))

In [4]: qdict
Out[4]: <QueryDict: {u'orderId': [u''], u'keyVersion': [u'1'], u'transactionReference': [u'1212943'], u'paymentMeanType': [u'CREDIT_TRANSFER'], u'maskedPan': [u''], u'currencyCode': [u'978'], u'paymentMeanBrand': [u'IDEAL'], u'complementaryCode': [u''], u'amount': [u'2000'], u'authorisationId': [u'0020000006791167'], u'responseCode': [u'00'], u'captureMode': [u'AUTHOR_CAPTURE'], u'captureDay': [u'0'], u'transactionDateTime': [u'2012-04-12T12:09:56 02:00'], u'merchantId': [u'002020000000001']}>
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  • 注意:如果您的某些字符串值首先包含"&"符号,则此方法无效. (3认同)