Message211581
The RE compiler will not error out, with a back reference in there...
It treats the {\1} as a literal {\1} in the string.
In [180]: re.search("(\d) fo.{\1}", '3 foo{\1}').group(0)
Out[180]: '3 foo{\x01}' |
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Date |
User |
Action |
Args |
2014-02-19 01:04:02 | hardkrash | set | recipients:
+ hardkrash, ezio.melotti, mrabarnett |
2014-02-19 01:04:02 | hardkrash | set | messageid: <1392771842.8.0.300525728381.issue20678@psf.upfronthosting.co.za> |
2014-02-19 01:04:02 | hardkrash | link | issue20678 messages |
2014-02-19 01:04:02 | hardkrash | create | |
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