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有没有数字计算型的验证?

Open alice1530 opened this issue 1 year ago • 7 comments

你好请问有没有计算型的验证,比如12+32=?,这种计算型的验证?

alice1530 avatar Jun 09 '23 02:06 alice1530

xbk?

xinjang avatar Jun 10 '23 04:06 xinjang

同问

evillazyer avatar Jul 05 '23 11:07 evillazyer

同问,有没有类似 2+3=?这样的验证呢?其他什么库能够做到?

zhonghai9967 avatar Jul 21 '23 00:07 zhonghai9967

#假设密码是x+y=? x -y=? #截图保存到d:/1.bmp with open('d:/1.bmp', 'rb') as f: .... img= f.read();res =ocr.classification(img) if len(res) <=3: .... pass #重新截图 else: #长度4|5 .... res=res.replace('o','0') #字母o替换为0 ....r0=res[0];r1=res[1];r2=res[2] ....if r1 == '+': ........ mima=int(r0)+int(r2) .... elif r1 == '-': ........mima=int(r0)-int(r2) ....else: ........mima='' # 重新截图

zhonghai9967 avatar Jul 21 '23 01:07 zhonghai9967

同问,很多这样的场景都是这样的验证码

9sis avatar Nov 24 '23 01:11 9sis

#假设密码是x+y=? x -y=? #截图保存到d:/1.bmp with open('d:/1.bmp', 'rb') as f: .... img= f.read();res =ocr.classification(img) if len(res) <=3: .... pass #重新截图 else: #长度4|5 .... res=res.replace('o','0') #字母o替换为0 ....r0=res[0];r1=res[1];r2=res[2] ....if r1 == '+': ........ mima=int(r0)+int(r2) .... elif r1 == '-': ........mima=int(r0)-int(r2) ....else: ........mima='' # 重新截图

十位数乘与十位数呢,你这个个位数相加太狭窄了

9sis avatar Nov 24 '23 01:11 9sis

顶~

Mason-mengze avatar Dec 05 '23 03:12 Mason-mengze