floor

I have a SQLite database with a table containing the scores of some league players in a bowling center. I'm trying to get the average of the Score column for each player ID. The problem with this is I only need the whole part of the average, and it should not be rounded (example: an average of 168.99 should return 168, not 169). When trying to find something like this on Google, I only found solutions for SQL Server and some others, but not SQLite, so if anyone can help me with this, I'd really appreciate it! So far, I'm using ROUND(AVG(s1.Score),2) and I'm truncating the extra part in my Java

I need to get the left hand side integer value from a decimal or double. For Ex: I need to get the value 4 from 4.6. I tried using Math.Floor function but it's returning a double value, for ex: It's returning 4.0 from 4.6. The MSDN documentation says that it returns an integer value. Am I missing something here? Or is there a different way to achieve what I'm looking for? The range of double is much wider than the range of int or long . Consider this code: double d = 100000000000000000000d; long x = Math.Floor(d); // Invalid in reality The integer is outside the range of long - so what would

In C, is there a difference between integer division a/b and floor(a/b) where both a and b are integers? More specifically what happens during both processes? Pete Becker a/b does integer division. If either a or b is negative, the result depends on the compiler (rounding can go toward zero or toward negative infinity in pre-C99; in C99+, the rounding goes toward 0). The result has type int . floor(a/b) does the same division, converts the result to double, discards the (nonexistent) fractional part, and returns the result as a double. ouah floor returns a double while a / b where both a and b

盲注查询 盲注其实就是没有回显，不能直观地得到结果来调整注入数据，只能通过其他方式来得到是否注入成功，主要是利用了一些数据库内置函数来达到的 布尔盲注 布尔很明显Ture跟Fales，也就是说它只会根据你的注入信息返回Ture跟Fales 其实登录处的注入就是布尔型的，万能密码就是构造一个永真的查询，比如下面的 select user from test where passwd=‘{injuct}’； #构造永真，即令where的条件用于为真 select user from test where passwd=‘aa‘or’1’=‘1’； #注入的数据是aa‘or’1’=‘1 密码输入无论是否正确，查询都成立。 布尔盲注其实就是利用了这种，我们什么时候需要采用这种呢 1）当没有数据输出点时，我们没有办法直观的判断注入的sql执行情况， 2）有正确或者错误的两种返回，比如查询正确返回一个页面，失败返回另一个页面，但是没有数据 时间盲注 界面返回值只有一种,true 无论输入任何值 返回情况都会按正常的来处理。加入特定的时间函数， 通过查看web页面返回的时间差来判断注入的语句是否正确 。 利用的内置函数 sleep(n)：将程序挂起一段时间 n为n秒 if(expr1,expr2,expr3):判断语句 如果第一个语句正确就执行第二个语句如果错误执行第三个语句 注入的语句

刚开始学习sql注入，遇见了 select count(*) from table group by floor(rand(0)*2); 这么条语句。本着刚开始学习一定要弄明白的态度，在此做个总结。 (更好的阅读体验可访问 这里 ) 首先，只要该语句明白了，那么类似 select count(*),(floor(rand(0)*2))x from table group by x; 这样的变形语句基本上都可以变通（这里只是起了个别名）。 基本的查询 select 不必多说，剩下的几个关键字有 count 、group by 、floor、rand。 几个关键函数的说明 rand(0)*2 rand() 可以产生一个在0和1之间的随机数。 可见，每次产生的都不一样。当我们提供一个种子参数 0 后，再次查看： 可以发现，每次产生的值都是一样的。也可以称之为伪随机（产生的数据都是可预知的）。 查看多个数据看一下。（ test 是我之前创建的一个拥有9条数据的表） 发现第一条数据与刚才查看的单个数据相符合，其它的数据也完全一样。 为什么要乘以 2 呢？这就要配合 floor 函数来说了。 floor(rand(0)*2) floor() 返回小于等于该值的最大整数。 之前我们了解到，rand() 是返回 0 到 1 之间的随机数，那么乘 2 后自然是返回 0 到 2 之间的随机数，再配合

练习1 ： 开发一个自定义指令v-birthday ，接收一个出生日期的时间戳，将它转换为己经出生 了xxx 天。 练习2 ： 扩展练习1 的自定义指令v-birthday ，将出生了xxx 天转换为具体年龄，比如25 岁 8 个月10 天。 Demo在线效果浏览 解答： 为简便起见，在time.js中，重新改造了下Time对象，由2个函数组成，分别接受一个时间戳为参数，返回字符串tip。 同时，写了了个指令，一个是 birthday 指令，输出显示出生了多少天，另一个是updatebirthday，显示多少岁多少月多少天。 export var Time={ getFormatDates(timestamp){ var dates = Math.floor(timestamp / 86400); var tip = '您已经出生了 ' + dates +' 天'; return tip; }, getFormatAges(timestamp){ var dates= Math.floor(timestamp / 86400); var year = Math.floor(dates / 365); var remains = Math.floor(dates % 365); var month = Math.floor(remains / 31); var day = Math

In the accepted answer on my earlier question ( What is the fastest way to generate a random integer in javascript? ), I was wondering how a number loses its decimals via the symbol | . For example: var x = 5.12042; x = x|0; How does that floor the number to 5 ? Some more examples: console.log( 104.249834 | 0 ); //104 console.log( 9.999999 | 0 ); // 9 Because, according to the ECMAScript specifications, bitwise operators operators call ToInt32 on each expression to be evaluated. See 11.10 Binary Bitwise Operators : The production A : A @B , where @ is one of the bitwise operators in the

I'd like a floor function with the syntax int floor(double x); but std::floor returns a double . Is static_cast <int> (std::floor(x)); guaranteed to give me the correct integer, or could I have an off-by-one problem? It seems to work, but I'd like to know for sure. For bonus points, why the heck does std::floor return a double in the first place? The range of double is way greater than the range of 32 or 64 bit integers, which is why std::floor returns a double . Casting to int should be fine so long as it's within the appropriate range - but be aware that a double can't represent all 64 bit