【实习记】2014-09-01从复杂到简单：一行命令区间查重+长整型在awk中的bug

 

 

9月1号，导出sql文件后，想到了awk，但很复杂。想到了用sed前期处理+python排序比较的区间查重法。编写加调试用了约3小时。

9月2号，编写C代码的sql语句过程中，发现排序可以交mysql，于是，又使用了约一个小时，学习并写出了一句命令行区间查重。

所以，方法不止一种，经验多了，才能快速找到最正确最快的方向。

重点记录第二种方法，此时是9月3号早上了。
一行命令查重：

$ mysql -upublic -ppublic ia_gcms_2014 -e "select Fgcms_product_id, Fissuer_acc_range_low, Fissuer_acc_range_high from t_IP0040T1 order by Fgcms_product_id, Fissuer_acc_range_low" | awk 'NR==1{print};NR>1{if(id==$1 && high>$2) {print ">>> two"; print pre; print;};pre=$0; id=$1; low=$2; high=$3;}'

输出：

Fgcms_product_id    Fissuer_acc_range_low    Fissuer_acc_range_high
>>> two
CIR    5215703100000000000    5215704099999999999
CIR    5215703600000000000    5215704099999999999
>>> two
MCW    5377950000000000000    5377959999999999999
MCW    5377950000500000000    5377950000599999999
>>> two
MPL    5220200000000000000    5220209999999999999
MPL    5220200000300000000    5220200000399999999
>>> two
MRG    5175900000000000000    5175909999999999999
MRG    5175900000200000000    5175900000599999999
>>> two
MRG    5215703100000000000    5215704099999999999
MRG    5215703600000000000    5215704099999999999

注意：上面只能是high>$2， 如果high>=$2不行，实践得知，这是是awk对长整支持不足所致。

$ echo "3560249999999999999 3560250000000000000" | awk '$1<$2{print "true"}' #
$ echo "3560249999999999999 3560250000000000010" | awk '$1<$2{print "true"}' #
$ echo "3560249999999999999 3560250000000000100" | awk '$1<$2{print "true"}' #
$ echo "3560249999999999999 3560250000000001000" | awk '$1<$2{print "true"}' # true

记录到find_cross.sh脚本以便使用

#!/usr/bin/env bash
#coding: utf-8

# filename:     find_cross.sh
# description:  数据库区间查重
# note:         Fgcms_product_id, Fissuer_acc_range_low复合唯一，排序由mysql做好，awk只要比较前后两行即可。

HOST=localhost
PORT=3306
USER=public
PASSWORD=public
DATABASE=ia_gcms_2014
# 来源表，表名要足够唯一，保证表名字不会被其他词命中，不然会错。
TABLE_SRC=t_IP0040T1
mysql -h${HOST} -P${PORT} -u${USER} -p${PASSWORD} ${DATABASE} \
    -e "select Fgcms_product_id, Fissuer_acc_range_low, Fissuer_acc_range_high \
        from ${TABLE_SRC} \
        order by Fgcms_product_id, Fissuer_acc_range_low" \
    | awk 'NR==1{print}; \
        NR>1{\
            if(id==$1 && high>$2) \
                {print ">>> two";print pre; print;} \
            pre=$0; id=$1; low=$2; high=$3; \
        }'

第一种方法现在看来很笨重，导出数据表到sql文件，再使用sed+python做文本处理。

#!/usr/bin/env python
#-*- coding: utf-8 -*-

# 使用：
# mysqldump -upublic -ppublic ia_gcms_2014 t_IP0040T1 > ia_gcms_2014_t_IP0040T1.sql
# sed '/INSERT INTO.*VALUES.*;$/{s/(/\n(/g}' ia_gcms_2014_t_IP0040T1.sql ia_gcms_2014_t_IP0040T1_sed.sql
# python find_cross.py ia_gcms_2014_t_IP0040T1_sed.sql

'''
filename:       find_cross.py
description:    找出区间交叉的记录
'''
import os, sys
from operator import itemgetter

def find_across(filename):
    '''
    寻交叉区间
    '''
    sql_file = open(filename, 'r')
    # 列表解析，一句抵10句      # with '\n'
    tuple_rows = [ eval(line[:-2]) for line in sql_file.readlines()
            if line[0]=='(' and line[-3]==')' and line[-2] in (',',';') ]   # line[-1]=='\n'

    tuple_rows.sort(key=lambda x:x[4])      # key=itemgetter(4)
    # 不用变成int了，等长就直接比字符串
    cmper = lambda a,b:\
                -1 if a[4]<b[4] else \
                1 if a[4]>b[4] else \
                -1 if a[5]<b[2] else \
                1 if a[2]>b[5] else \
                0
    tuple_rows.sort(cmper)

    # print tuple_rows

    begin_end = 0
    for i in range(len(tuple_rows)-1):
        if cmper(tuple_rows[i], tuple_rows[i+1])==0:
            if begin_end == 0:
                begin_end = 1
                print "============== begin =============="
            # if tuple_rows[i][4]==tuple_rows[i][4]:   # 多余
            print tuple_rows[i]
        else:
            if begin_end == 1:
                print tuple_rows[i]
                # print "==============  end  =============="
                begin_end = 0
    if begin_end == 1:
        print tuple_rows[-1]
        # print "==============  end  =============="

if __name__ == "__main__":
    if len(sys.argv) < 2:
        print "<py> <filename>"
        exit(0)
    find_across(sys.argv[1])