offset

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have a custom query that grabs data from the old system and maps it to models in the new system. The query looks like this: $companies = DB::connection('legacy')->select("..."); And since it's a lot of data, I'd like to use Eloquent's chunk feature (just sample code copied from their docs): User::chunk(200, function($users) { foreach ($users as $user) { // } }); How do I implement this? Edit: My code now looks like this, which results in no response: DB::connection('legacy')->select("SELECT * FROM companies")->chunk(200, function(

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I want to integrate sort, order, max and offset in a findAll query. The following works fine: def books = Book.findAll("from Book as b where b.approved=true order by b.dateCreated desc", [max: max, offset: offset]) But what I want is: def books = Book.findAll("from Book as b where b.approved=true", [sort: 'dateCreated', order: 'desc', max: max, offset: offset]) This does not work. How do I have to rewrite this? 回答1: HQL doesn't support sort and order as parameters, so you need to include the "order by" as part of the HQL expression def books

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I want to catch this error: $a[1] = 'jfksjfks'; try { $b = $a[0]; } catch (\Exception $e) { echo "jsdlkjflsjfkjl"; } Edit: in fact, I got this error on the following line: $parse = $xml->children[0]->children[0]->toArray(); 回答1: You need to define your custom error handler like: 回答2: You can't with a try/catch block, as this is an error, not an exception. Always tries offsets before using them: if( isset( $a[ 0 ] ) { $b = $a[ 0 ]; } 回答3: $a[1] = 'jfksjfks'; try { $offset = 0; if(isset($a[$offset])) $b = $a[$offset]; else throw new Exception(

I'm back again, frustrated and desperately searching for help :D. I am trying to create a cheat for a simple program, it's basically going to be a .dll file which will change an integer's value from the main program when it's injected to it using its base address. The thing is, I can't find it using cheat engine mainly because there are multiple level pointers with NEGATIVE? offsets. for example: //Starting pointer address 0x0033FCF0 -> 200 //Finding second level pointer using "Find out what's accessing this address" in Cheat Engine **(mov eax,[ebp+08])** // **EAX=0x00000007** , **EPB

https://www.cnblogs.com/fuckgiser/p/6027520.html 最近研究Cesium的实例化，尽管该技术需要在WebGL2.0，也就是OpenGL ES3.0才支持。调试源码的时候眼前一亮，发现VAO和glDrawBuffers都不是WebGL1.0的标准函数，都是扩展功能，看来WebGL2.0标准的推广势在必行啊。同时发现，通过ANGLE_instanced_arrays的扩展，也可以在WebGL1.0下实现实例化，创建实例化方法的代码如下： var glDrawElementsInstanced; var glDrawArraysInstanced; var glVertexAttribDivisor; var instancedArrays; // WebGL2.0标准直接提供了实例化接口 if (webgl2) { glDrawElementsInstanced = function(mode, count, type, offset, instanceCount) { gl.drawElementsInstanced(mode, count, type, offset, instanceCount); }; glDrawArraysInstanced = function(mode, first, count, instanceCount

写在前面 安全测试需要, 为防止后台响应数据返给前台过程中被篡改前台再拿被篡改后的数据进行接下来的操作影响正常业务, 决定采用RSA对响应数据进行签名和验签, 于是有了这篇<RSA后台签名前台验签的应用>. 我这里所谓的返给前台的数据只是想加密用户验证通过与否的字段success是true还是false, 前台拿这个success作为判断依据进行下一步的操作, 是进一步向后台发起请求还是直接弹出错误消息.照测试结果看这是个逻辑漏洞, 即使后台返回的是false, 在返回前台的过程中响应包被劫获, 将false改为true, 这样的操作也是能做到的(BurpSuit). 所以后台响应数据尽量不要再二次使用. 那既然能篡改, 如何防止流氓篡改呢? 说一下整体思路: 首先生成密钥对, 私钥存放在后台用于签名, 公钥存放在前台用于验签. 所谓签名就是指拿明文+私钥生成的 签名结果 , 返回数据给前台时将 明文+签名结果 一并返给前台, 前台用 公钥+接收到的明文+签名结果 进行验签, 这样即使响应包被劫获, 篡改明文后, 验证签名时也不会验证通过, 从而达到响应数据防篡改的目的. 接下来说一下具体步骤. 正文(具体步骤) 1.在线生成密钥对 采用在线工具生成密钥对, 私钥密码可填可不填, 网址: http://web.chacuo.net/netrsakeypair 生成密钥对备用. 2

8.Rotate String lintcode 题解: class Solution { public: /** * @param str: An array of char * @param offset: An integer * @return: nothing */ void rotateString(string &str, int offset) { string t_str; int len = str.length(); if (len == 0 || offset == 0)return; int new_offset = offset % len; int j = len - 1; for (int i = 0; i< new_offset; i++) { t_str.insert(0,1 ,str[j]); j--; } if (new_offset == 0)new_offset = len; for (int i=0;i!=j+1;i++) { t_str.push_back(str[i]); } for (int i = 0; i < len; i++) { str[i] = t_str[i]; } } }; 没能想出O(1)的题解 所以就直接暴力解了,复杂度勉强能接受 反思: 一开始没看清题目就直接编程,以为offset表示的是字符串的从0开始的下标

node-ip node.js用来获取id地址的工具 use var ip = require('ip'); ip.address() // my ip address ip.isEqual('::1', '::0:1'); // true ip.toBuffer('127.0.0.1') // Buffer([127, 0, 0, 1]) ip.toString(new Buffer([127, 0, 0, 1])) // 127.0.0.1 ip.fromPrefixLen(24) // 255.255.255.0 ip.mask('192.168.1.134', '255.255.255.0') // 192.168.1.0 ip.cidr('192.168.1.134/26') // 192.168.1.128 ip.not('255.255.255.0') // 0.0.0.255 ip.or('192.168.1.134', '0.0.0.255') // 192.168.1.255 ip.isPrivate('127.0.0.1') // true ip.isV4Format('127.0.0.1'); // true ip.isV6Format('::ffff:127.0.0.1'); // true // operate on buffers in-place

加密：在网络传输中，如果我们传输明文，会被人给截取，从而数据泄露。我们可以对他们进行加密 加密分为2大类，对称加密和非对称加密。 对称加密：对称加密就是加密和解密公用一个密钥。 DES：对称加密，速度较快，适用于加密大型数据的场合。 public class DesUtil { private final static String DES = "DES"; private final static String ENCODE = "UTF-8"; private final static byte[] ivByte= {1,2,3,4,5,6,7,8}; // 算法名称/加密模式/填充方式 public static final String CIPHER_ALGORITHM = "DES/CBC/PKCS5Padding"; //主要看填充方式 PKCS5Padding PKCS7Padding 还有其他的填充方式没用过 // public static final String CIPHER_ALGORITHM_CBC = "DES/CBC/ZerosPadding"; public static void main(String[] args) throws Exception { String data="123456"; String timeSend=