integer

问题 In my database I have a column "year" which is an integer. How can I search by using Criteria API (not by HQL) which records contain, for example "196..." in the year column? I think it should be Restrictions.like , but I got exception: SEVERE: java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Integer 回答1: a better and faster solution would be to use Criterion.sqlRestriction(String sql). You can form the sql as, "to_char(year) like '19%'", be aware that to_char is

问题 Say I have a number 0 that corresponds to the ASCII character a . How would I go about converting a number in the range 0 to 25 to letters in the alphabet? I have already tried adding 97 to the decimal value, but it just outputs the number+ 97 . typedef enum { a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z } set; void dispSet(set numbers[], int size_numbers) { int i; printf("[ "); for (i = 0; i < size_numbers-1; i++) { printf("%d, ", ((char) numbers[i])+97); }

问题 Say I have a number 0 that corresponds to the ASCII character a . How would I go about converting a number in the range 0 to 25 to letters in the alphabet? I have already tried adding 97 to the decimal value, but it just outputs the number+ 97 . typedef enum { a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z } set; void dispSet(set numbers[], int size_numbers) { int i; printf("[ "); for (i = 0; i < size_numbers-1; i++) { printf("%d, ", ((char) numbers[i])+97); }

问题 This question already has answers here : How to convert CryptoPP::Integer to char* (4 answers) Closed 2 years ago . I can't find the right code to convert a CryptoPP::Integer (from a RSA key generation) to a LPCTSTR (I want to store the key in the registry). Could you help me ? Thanks you ! 回答1: ... convert a CryptoPP::Integer (from a RSA key generation) to a LPCTSTR (I want to store the key in the registry). Could you help me ? Something like the following should do. The Integer class

问题 The best way to avoid SQL injection for defined value type such as numbers, is to validate the value; since it is easier to do so comparing with mysqli preparation. In PHP, we can do this by. 1. if(!is_numeric($value)) {$value=0;} 2. $value=floatval($value); 3. $value=intval($value); 4. $value=$value * 1; What is the most reliable one? or a better idea? UPDATE: Although I have stated in the original question, most of folks emphasized the usefulness of parameterized queries. Definitely, it is

问题 I'm trying to break down an integer with C on an 8-bit microcontroller (a PIC) into its ASCII equivalent characters. For example: convert 982 to '9','8','2' Everything I've come up with so far seems pretty brute force. This is the main idea of what I'm basically doing right now: if( (10 <= n) && (n < 100) ) { // isolate and update the first order of magnitude digit_0 = (n % 10); // isolate and update the second order of magnitude switch( n - (n % 10) ) { case 0: digit_1 = 0; break; case 10:

问题 the code below(in C++) is what I am trying the convert into C# DWORD Func_X_4(DWORD arg1, DWORD arg2, DWORD arg3) { LARGE_INTEGER result = {1, 0}; LARGE_INTEGER temp1 = {0}; LARGE_INTEGER temp2 = {0}; LARGE_INTEGER temp3 = {0}; LARGE_INTEGER temp4 = {0}; for(int x = 0; x < 32; ++x) { if(arg2 & 1) { temp1.LowPart = arg3; temp1.HighPart = 0; temp2.QuadPart = temp1.QuadPart * result.QuadPart; temp3.LowPart = arg1; temp3.HighPart = 0; temp4.QuadPart = temp2.QuadPart % temp3.QuadPart; result

问题 I have a problem with converting string to float. print gettype($value[$id]); //returns string var_dump($value[$id]);//returns string '34,7140' (length=7) $float = floatval($value[$id]); print gettype($float);//returns double var_dump($float);//returns float 34 echo $float;//returns 34 I don't understand why "34" ? Why $float is not '34,7140'? How can I get $float = 34,7140 ? 回答1: The problem is that floats are expected to be in the English format with a . separating the decimal part, not a

问题 I'm trying to divide an integer by an integer in Java and return a long result. I tried the following, but I keep getting "0". System.out.println(Long.valueOf(5) / Long.valueOf(18)); System.out.println((long)5 / (long)18); 回答1: It seems everyone has given the right answer, but you should note you can also do System.out.println(5 / 18f) Where f makes it a float System.out.println(5 / 18d) where d makes it a double 回答2: You don't need a long, you need a double. System.out.println(5 / 18.0); or

问题 I have a task that requires me to print out the following quantities: I need to write a java program that can calculate: 1. the number of integers read in 2. the average value—which need not be an integer! My final output should look like: $ javac SimpleCalc $ java SimpleCalc 3 4 6 8 1 5 8 4.4 7 I know i need to use a scanner class to read then as the user inputs then, i dont know how to make it display a count. I think that i can work this out, i can compute the average by adding up the