substr

How do I get a substring from this example value: john.abc_1234 I want it to return john.abc .So basically we need to remove all the information after _ . More examples: 1234_abc You can use SUBSTR and INSTR : select substr('john.abc_1234', 1, instr('john.abc_1234', '_') -1) from dual Warning: This is only guaranteed to work if your string actually has an underscore in it Update Additionally, if you are running from Oracle 10g on, you could take the Regex path, which would more powerfully handle exceptions. Here are some links on how to do it in Oracle: http://psoug.org/reference/regexp.html

背景代码： std::string topic = topic_name.substr(topic_name.find_last_of('/') + 1); // topic_name中没有'/',这个长度 涉及： 1、find_last_of（）； 解析：从string末端开始寻找子串或者字符，如果找到返回字串首字符的索引（其实就是字符中的第几位），如果没有匹配则返回 std:: string ::npos（实际上为 -1） size_t find_last_of (const string& str, size_t pos = npos) const noexcept; size_t find_last_of (const char* s, size_t pos = npos) const; size_t find_last_of (const char* s, size_t pos, size_t n) const; size_t find_last_of (char c, size_t pos = npos) const noexcept; http://www.cplusplus.com/reference/string/string/find_last_of/ 版权声明：本文为CSDN博主「NearXDU」的原创文章，遵循 CC 4.0 BY-SA 版权协议

GIF二次渲染： 　　　　关于绕过gif的二次渲染,我们只需要找到渲染前后没有变化的位置,然后通过16进制编辑器把php代码写进去,就可以成功上传带有php代码的图片了. PNG图片二次渲染 国外大牛写的脚本,直接拿来运行即可. <?php $p = array(0xa3, 0x9f, 0x67, 0xf7, 0x0e, 0x93, 0x1b, 0x23, 0xbe, 0x2c, 0x8a, 0xd0, 0x80, 0xf9, 0xe1, 0xae, 0x22, 0xf6, 0xd9, 0x43, 0x5d, 0xfb, 0xae, 0xcc, 0x5a, 0x01, 0xdc, 0x5a, 0x01, 0xdc, 0xa3, 0x9f, 0x67, 0xa5, 0xbe, 0x5f, 0x76, 0x74, 0x5a, 0x4c, 0xa1, 0x3f, 0x7a, 0xbf, 0x30, 0x6b, 0x88, 0x2d, 0x60, 0x65, 0x7d, 0x52, 0x9d, 0xad, 0x88, 0xa1, 0x66, 0x44, 0x50, 0x33); $img = imagecreatetruecolor(32, 32); for ($y = 0; $y < sizeof($p); $y += 3) { $r = $p[$y]; $g = $p[$y+1];

前序中序遍历建立二叉树 1 Node * createTree1(string pres, string cens) { 2 if(pres.size() == 0 || cens.size() == 0) { 3 return nullptr; 4 } 5 Node * node = new Node(); 6 node->data = pres.at(0); 7 int key = cens.find(pres.at(0)); 8 9 string mpres = pres.substr(1, key); 10 string mcens = cens.substr(0, key); 11 node->left = createTree1(mpres,mcens); 12 13 mpres = pres.substr(key+1, pres.size()-key-1); 14 mcens = cens.substr(key+1, cens.size()-key-1); 15 node->right = createTree1(mpres,mcens); 16 return node; 17 }; View Code 中序后序遍历建立二叉树 后序中序遍历就是将前序中序遍历中的对前序的操作转换为对后序的操作，前序从第一个开始取，后序从后往前取。对中序操作不变 来源： https:/

1、mysql查询区分大小写 方法1、select * from `user` where binary `email`='zhang_hao@163.com'; 方法2、修改表字段 或建表的时候加上 binary 2、mysql统计一个数据库中每张表的行数、表所占空间大小(数据长度+索引长度) use information_schema; select table_name,table_rows,data_length+index_length as data_size from tables where TABLE_SCHEMA = '数据库名' order by data_size desc,table_rows desc; 3、mysql多表联合查询加分页 SELECT `un`.crid,cr.crname,sum(total) successtotalfee,sum(count) taketimes from ((SELECT crid,sum(total) total,count(1) count from ebh_takes where state=1 AND del=0 group by crid) UNION ALL (SELECT crid,sum(moneyaftertax) total,count(1) count from ebh_jsapplys

Medimu Compare two version numbers version1 and version2 . If version1 > version2 return 1; if version1 < version2 return -1; otherwise return 0 . You may assume that the version strings are non-empty and contain only digits and the . character. The . character does not represent a decimal point and is used to separate number sequences. For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision. You may assume the default revision number for each level of a version number to be 0 . For example, version

其中高精度乘法通过了POJ2389，其他没有测过，不过应该是没有问题的。 其中高精度除法返回一对string，分别表示商和余数。 代码： #include <bits/stdc++.h> using namespace std; const int maxn = 100010; int a[maxn], b[maxn], res[maxn]; string add(string s1, string s2) { // under condition: s1,s2>=0 // 初始化部分 int n = s1.length(), m = s2.length(); for (int i = 0; i < n; i ++) a[i] = s1[n-1-i] - '0'; for (int i = 0; i < m; i ++) b[i] = s2[m-1-i] - '0'; int len = max(n, m) + 1; for (int i = n; i < len; i ++) a[i] = 0; for (int i = m; i < len; i ++) b[i] = 0; for (int i = 0; i < len; i ++) res[i] = 0; // 处理部分 for (int i = 0; i < len; i ++) { res[i] += a[i] + b

题目描述: Given a string s1, we may represent it as a //分割 binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 ="great": great / \ gr eat / \ / \ g r e at / \ a t //爬行字符串 To scramble the string, we may choose any non-leaf node and swap its two children. For example, if we choose the node"gr"and swap its two children, it produces a scrambled string"rgeat". rgeat / \ rg eat / \ / \ r g e at / \ a t We say that"rgeat"is a scrambled string of"great". Similarly, if we continue to swap the children of nodes"eat"and"at", it produces a