python正则表达式获取两段标记内的字符串
比如获取绿色字符串
ModelData.PayTableData =[{"SlotID":"","GroupID":"","GroupH":"","GroupType":"","CellID":"","CellH":"","Row":"","ResType":"","Res":"Slot1&slot1_wild_5.png","ResVar":null,"X":"","Y":"","Scale":"0.7","Zorder":""},
{"SlotID":"","GroupID":"","GroupH":"","GroupType":"","CellID":"","CellH":"","Row":"","ResType":"","Res":"Slot1&slot1_wild_4.png","ResVar":null,"X":"","Y":"","Scale":"0.7","Zorder":""},
{"SlotID":"","GroupID":"","GroupH":"","GroupType":"","CellID":"","CellH":"","Row":"","ResType":"","Res":"Slot1&slot1_wild_3.png","ResVar":null,"X":"","Y":"","Scale":"0.7","Zorder":""}];
ModelData.PayTableData1 =[{"SlotID":"","GroupID":"","GroupH":"","GroupType":"","CellID":"","CellH":"","Row":"","ResType":"","Res":"Slot1&slot1_wild_5.png","ResVar":null,"X":"","Y":"","Scale":"0.7","Zorder":""},
{"SlotID":"","GroupID":"","GroupH":"","GroupType":"","CellID":"","CellH":"","Row":"","ResType":"","Res":"Slot1&slot1_wild_4.png","ResVar":null,"X":"","Y":"","Scale":"0.7","Zorder":""},
{"SlotID":"","GroupID":"","GroupH":"","GroupType":"","CellID":"","CellH":"","Row":"","ResType":"","Res":"Slot1&slot1_wild_3.png","ResVar":null,"X":"","Y":"","Scale":"0.7","Zorder":""}];
ModelData.PayTableData2 =[{"SlotID":"","GroupID":"","GroupH":"","GroupType":"","CellID":"","CellH":"","Row":"","ResType":"","Res":"Slot1&slot1_wild_5.png","ResVar":null,"X":"","Y":"","Scale":"0.7","Zorder":""},
{"SlotID":"","GroupID":"","GroupH":"","GroupType":"","CellID":"","CellH":"","Row":"","ResType":"","Res":"Slot1&slot1_wild_4.png","ResVar":null,"X":"","Y":"","Scale":"0.7","Zorder":""},
{"SlotID":"","GroupID":"","GroupH":"","GroupType":"","CellID":"","CellH":"","Row":"","ResType":"","Res":"Slot1&slot1_wild_3.png","ResVar":null,"X":"","Y":"","Scale":"0.7","Zorder":""}];
只需要
#editor:pengyingh
#encoding:utf-
#!/usr/bin/python
import os
import sys
import json
import string
import re
jsPath = sys.argv[]
#appendPath = jsPath[:jsPath.rindex('/') + ]
fp = open(jsPath, 'r')
rawStr = fp.read()
#print(rawStr[:])
m = re.search(r'ModelData.PayTableData\s*=\s*(\[.+?\])', rawStr, re.S)
if m:
print m.group()
else:
print 'no match'
fp.close()
python正则表达式获取两段标记内的字符串的更多相关文章
- Java 正则表达式获取两个字符中间的内容
利用 正则表达式 获取两个字符串中间的值 直接上代码吧,不是很难. public static void main(String[] args) { // 内容 String value = &quo ...
- Python中使用正则表达式获取两个字符中间部分
问题背景:当我们爬取网页信息时,对于一些标签的提取是没有意义的,所以需要提取标签中间的信息. 解决办法:用到了re包下的函数 方法1:用到了research()方法和group()方法 方法2:用到了 ...
- python正则表达式获取代理IP网站上的IP地址
import urllib.request import re def open_url(url): req = urllib.request.Request(url) req.add_header( ...
- C# 获取一段日期内的工作日
/// <summary> /// 根据指定时间段计算工作日天数 /// </summary> /// <param name="firstDay"& ...
- python正则表达式应用 定义一个函数,求字符串中出现的所有整数之和
- Python正则表达式如何进行字符串替换实例
Python正则表达式如何进行字符串替换实例 Python正则表达式在使用中会经常应用到字符串替换的代码.有很多人都不知道如何解决这个问题,下面的代码就告诉你其实这个问题无比的简单,希望你有所收获. ...
- 【Python】【demo实验15】【练习实例】【两个数范围内素数的统计】
原题: 判断101-200之间有多少个素数,并输出所有素数. 关于素数的统计,之前已经做过相应的实验了,参考:[显示素数,显示两个数范围内的所有素数] 原题给出的解法,使用math的sqrt函数,这个 ...
- FZU-2105 Digits Count (两种标记成段更新)
题目大意:给n个0~15之间的数,有3种更新操作,1种询问操作.3种更新操作是:1.让某个闭区间的所有数字与一个0~15之间的数字进行逻辑与运算:2.让某个闭区间的所有数字与一个0~15之间的数字进行 ...
- python正则表达式re模块详细介绍--转载
本模块提供了和Perl里的正则表达式类似的功能,不关是正则表达式本身还是被搜索的字符串,都可以是Unicode字符,这点不用担心,python会处理地和Ascii字符一样漂亮. 正则表达式使用反斜杆( ...
随机推荐
- msf客户端渗透(一):payload利用简单示范
针对Windows 开启侦听 查看payload选项 将1.exe传到网页上 win7访问网页并下载1.exe 下载好之后双击运行,在服务器端就获得了一个shell 针对linux 先获取到一个软 ...
- Python全栈开发 列表, 元组 数据类型知识运用及操作 range知识
一.列表 1.什么是列表? 列表是一个可变类型,由 [ ] 表示,每一项元素用逗号隔开.列表能够装大量的数据,可以装对象的对象. 2.列表的索引和切片. 列表和字符串一样,也有索引和切片.只不过列表 ...
- day16 包的使用 json time 常用模块
复习 1.判断py文件的两种用途 提到判断__name__ == '__main__'时,会执行py文件, 直接输入main,在pycharm里按tab直接自动输入这条语句 2.解决模块相互导入的问题 ...
- Android学习路-Android Studio的工程目录
说明:下图为一个app的工程目录,如果在res下随便建立文件夹(比如test等名字)是不会显示在工程内的
- oracle 一致读原理
在Oracle数据库中,undo主要有三大作用:提供一致性读(Consistent Read).回滚事务(Rollback Transaction)以及实例恢复(Instance Recovery). ...
- 算法篇【递归2 -- N皇后问题】
问题:输入整数N,要求在N*N的棋盘上,互相不能攻击,不在同一行同一列上,切不在对角线上,输出全部方案. 输入: 4 输出: 2 4 1 3 3 1 4 2 思路: 假设在前k-1个摆好的 ...
- 14. Longest Common Prefix (截取字符串)
Write a function to find the longest common prefix string amongst an array of strings. char* longest ...
- 5. Longest Palindromic Substring (DP)
Given a string S, find the longest palindromic substring in S. You may assume that the maximum lengt ...
- C++中 top()与pop()
top()是取出栈顶元素,不会删掉栈里边的元素 pop()是删除栈顶元素.
- AngularJs中url参数的获取
前言: angular获取通过链接形式访问的页面,要获取url中的参数,就不能通过路由的方式传递获取了,使用原生js或者jquery,又显得比较麻烦,好在angular已经封装了获取url参数的方法, ...