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where 1=1; This condition is always true, in the case of an indefinite number of query conditions, 1 = 1 The specification statement can be very convenient. One, without the trouble of where 1=1 in the multi-condition query　　For

where 1=1; This condition is always true, and1=1 can be a very convenient specification statement in the case of an indefinite number of query conditions. One, without the trouble of where 1=1 in the multi-condition queryFor example, if you make a query page, and, there are

Excerpt from: http://blog.sina.com.cn/s/blog_afe616ab0101camd.htmlThe role of SQL where 1=1 and 0=1where 1=1; This condition is always true, and in the case of an indefinite number of query conditions, the 1=

Http://www.cnblogs.com/brolanda/p/4548612.html See conditions in the WHERE statement: where 1=1 and 1=2 or 1 String sql = "SELECT * from emp where 1=1"; if (username!=null) sql

Input 1: Output 0; input 0: Output 1 I had an interview that day. I encountered such a problem: input 0, output 1, and input 1, and output 0

# Include Int main (void){Int A [100] [200]; Printf ("% d \ n", A [0] [1], A [0] [0], A [0] [1]- A [0] [0]);Printf ("% x % d \ n", A

"Code"""" all elements greater than 0 are converted to 1 """ = Np.array ([[1, 2, 3, 4]])print(" pre-conversion:")Print (np_arr)print(" after conversion:")print"Result"before conversion:[[1 2 3 4]] after conversion: [[1 1

Oop = $. Transform$. Rotation = (quat 0 0 0 1) Subobjectlevel = 2$. Modifiers [1]. Center = pos_posSubobjectlevel = 0 $. Modifiers [1]. Gizmo = OOP -- I used the following method to

SQL where 1 = 1 and 0 = 1Where 1 = 1; this condition is always True. In the case of an indefinite number of query conditions, 1 = 1 can be a convenient standard statement.I. You do not

PHP greedy algorithm solves the 0-1 knapsack problem instance analysis and 0-1 instance analysis. An example of PHP greedy algorithm for solving the 0-1 knapsack problem. An example of this article describes the PHP greedy algorit

01 the problem of obtaining and not obtaining a backpack is actually an opportunity cost problem. If you take something, although the current value has increased for the moment, you have paid the opportunity cost. If not, the space left may be more valuable in the future. Therefore, space and value are always in conflict. Our goal is to use limited space to load the best value. Given weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in th

PHP dynamic planning solves the 0-1 knapsack problem instance analysis and 0-1 instance analysis. PHP dynamic planning solves the problem of 0-1 backpacking instance analysis, 0-

Before encountering a very interesting topic, is about the problem of sorting algorithm:Known: An array: array element: 0 or 1 or 2Solve: Sort the array by 0-1-2?1Template classT>2 voidSwap (t t1, tT2)3 {4T tmp =T1;5T1 =T2;6t2 =tmp;7 }8 9 //sorting a array which the elements is 0