代码： spgz='{:.2f}'.format(env ['hr.payroll.function'] .get_emp_sumvalue_by_meta(employee.id, 1, ['spgz'] )) cdict = {1: u'', 2: u'拾', 3: u'佰', 4: u'仟'} result = spgz xdict = {1: u'元', 2: u'万', 3: u'亿', 4: u'兆'} # 数字标识符 gdict = {0: u'零', 1: u'壹', 2: u'贰', 3: u'叁', 4: u'肆', 5: u'伍', 6: u'陆', 7: u'柒', 8: u'捌', 9: u'玖'} cdata = str(spgz).split('.') cki = cdata [0] ckj = cdata [1] chk = u'' g = len(cki) % 4 cski = [] lx = len(cki) - 1 if g > 0: cski.append(cki [0:g] ) k = g while k <= lx: cski.append(cki [k:k + 4] ) k += 4 ikl = len(cski) # 获取拆分后的List长度 大写合并 for z in range(ikl): lenki = len(cski [z] ) lk = lenki cschange_cski = u'' for i in range(lenki): if int(cski [z] [i] ) == 0: if i < lenki - 1: if int(cski [z] [i + 1] ) != 0: cschange_cski = cschange_cski + gdict[int(cski [z] [i] )] else: cschange_cski = cschange_cski + gdict[int(cski [z] [i] )] + cdict [lk] lk -= 1 if cschange_cski == '': # 有可能一个字符串全是0的情况 chk = chk + cschange_cski # 此时不需要将数字标识符引入 else: chk = chk + cschange_cski + xdict [ikl - z] # 合并：前字符串大写+当前字符串大写+标识符 处理小数部分 lenkj = len(ckj) if lenkj == 1: # 若小数只有1位 if int(ckj [0] ) == 0: chk = chk + u'整' else: chk = chk + gdict[int(ckj [0] )] + u'角整' else: # 若小数有两位的四种情况 if int(ckj [0] ) == 0 and int(ckj [1] ) != 0: chk = chk + u'零' + gdict[int(ckj [1] )] + u'分' elif int(ckj [0] ) == 0 and int(ckj [1] ) == 0: chk = chk + u'整' elif int(ckj [0] ) != 0 and int(ckj [1] ) != 0: chk = chk + gdict[int(ckj [0] )] + u'角' + gdict[int(ckj [1] )] + u'分' else: chk = chk + gdict[int(ckj [0] )] + u'角整' result = chk