variables

问题 I have a number of Bash and Perl scripts which are unrelated in functionality, but are related in that they work within the same project. The fact that they work in the same project means that I commonly specify the same directories, the same project specific commands, the same keywords at the top of every script. Currently, this has not bitten me, but I understand that it would be easier to have all of these values in one place, then if something changes I can change a value once and have

问题 readfile() says it outputs the content but I want the content to be saved in a variable. How do I do that? I tried $content=readfile("file.txt") but that doesn't come out right. I want to do like this: $data=readfile("text.txt"); echo htmlentities($data); That should give you an idea of how I want that to work. 回答1: That is what file_get_contents is for: $data = file_get_contents("text.txt"); echo htmlentities($data); If you must use readfile you can use output buffering to get the contents

问题 readfile() says it outputs the content but I want the content to be saved in a variable. How do I do that? I tried $content=readfile("file.txt") but that doesn't come out right. I want to do like this: $data=readfile("text.txt"); echo htmlentities($data); That should give you an idea of how I want that to work. 回答1: That is what file_get_contents is for: $data = file_get_contents("text.txt"); echo htmlentities($data); If you must use readfile you can use output buffering to get the contents

问题 This question already has answers here : What is the difference between Const and Static in C#? (5 answers) Closed 4 years ago . what is the difference between "static" and "const" when it comes to declare global variables; namespace General { public static class Globals { public const double GMinimum = 1e-1; public const double GMaximum = 1e+1; } } which one is better (considering that these variables wont be changing ever) namespace General { public static class Globals { public static

问题 How can I achieve below in Scheme REPL? Create a variable name from a string. =>(define (string->variable-name "foo") 12) =>foo 12 =>(+ foo 8) 20 In Common Lisp, this should be => (set (intern "ANY-TEXT") 5) => ANY-TEXT 5 How do I build a #procedure like "string->variable-name" (and "variable-name->string") ? Thanks a lot. 回答1: If what you're passing to string->variable-name is always a string literal (i.e., not a variable that contains a string), you can do that using a syntax-case macro

问题 I need help with something that might be fairly simple in R. I want to refer to a range of columns in a data frame (e.g., extracting a few select variables). However, I don't know their column numbers. Normally, if I wanted to extract columns 4-10 i would say mydata[,4:10]. However, given that I don't know the column numbers, I would want to refer to them by name. Is there an easy way to do this? in sas or spss it is fairly easy to refer to a range of variables by name. Alternatively, is

问题 Eventually I understand this and it works. bash script: #!/bin/bash #$ -V #$ -cwd #$ -o $HOME/sge_jobs_output/$JOB_ID.out -j y #$ -S /bin/bash #$ -l mem_free=4G c=$SGE_TASK_ID cd /home/xxx/scratch/test/ FILENAME=`head -$c testlist|tail -1` python testpython.py $FILENAME python script: #!/bin/python import sys,os path='/home/xxx/scratch/test/' name1=sys.argv[1] job_id=os.path.join(path+name1) f=open(job_id,'r').readlines() print f[1] thx 回答1: Exported bash variables are actually environment

问题 so the string is like this: "bla bla bla {VARIABLE} bla bla" when I use this string somewhere in a function I want to replace {VARIABLE} with $variable (or any other uppercase strings with wrapped within {} charcters). $variable (and any other variables) will be defined inside that function Can i do this? 回答1: $TEST = 'one'; $THING = 'two'; $str = "this is {TEST} a {THING} to test"; $result = preg_replace('/\{([A-Z]+)\}/e', "$$1", $str); 回答2: Use a regular expression to find all substitutions

问题 So my question is about variable accessing speed in Java. Today in my "CS" (if you can call it that) the teacher presented a similar example to the following of a List: public class ListExample<T> { private Node<T> head; private Node<T> tail; private class Node<T> { /* ... */ } public void append(T content) { if (!isEmpty()) { Node<T> dummy = new Node<T>(content); head = dummy; tail = dummy; head.setNext(head); // or this dummy.setNext(dummy); } else { /* ... */ } } // more methods // ... }

问题 So my question is about variable accessing speed in Java. Today in my "CS" (if you can call it that) the teacher presented a similar example to the following of a List: public class ListExample<T> { private Node<T> head; private Node<T> tail; private class Node<T> { /* ... */ } public void append(T content) { if (!isEmpty()) { Node<T> dummy = new Node<T>(content); head = dummy; tail = dummy; head.setNext(head); // or this dummy.setNext(dummy); } else { /* ... */ } } // more methods // ... }