concatenation

问题 I have some problem with concatenation mp3 files. For example, I have 5 mp3 files. And I want concatenate them to one file. I try this: try { InputStream in = new FileInputStream("C:/a.mp3"); byte[] buffer = new byte[1024]; OutputStream os = new FileOutputStream(new File("C:/output.mp3", true)); int count; while ((count = in.read(buffer)) != -1) { os.write(buffer, 0, count); os.flush(); } in.close(); in = new FileInputStream("C:/b.mp3");// second mp3 while ((count = in.read(buffer)) != -1) {

问题 I have two AFP files and I want to concatenate them together, how can I accomplish this. I have written java code to concatenate them, using BufferedInputStream and BufferedOutputStream and the result AFP is not correctly format. I even try to use linux cat but yield the same incorrect result. Please help. I dont think the problem is my java code, but I post the code below just in case. NOTE: One strange thing is that if I switch the order of the concatenation then it yield the right format

问题 I have two AFP files and I want to concatenate them together, how can I accomplish this. I have written java code to concatenate them, using BufferedInputStream and BufferedOutputStream and the result AFP is not correctly format. I even try to use linux cat but yield the same incorrect result. Please help. I dont think the problem is my java code, but I post the code below just in case. NOTE: One strange thing is that if I switch the order of the concatenation then it yield the right format

问题 Is there a memory-efficient way to concatenate gzipped files, using Python, on Windows, without decompressing them? According to a comment on this answer, it should be as simple as: cat file1.gz file2.gz file3.gz > allfiles.gz but how do I do this with Python, on Windows? 回答1: Just keep writing to the same file. with open(..., 'wb') as wfp: for fn in filenames: with open(fn, 'rb') as rfp: shutil.copyfileobj(rfp, wfp) 回答2: You don't need python to copy many files to one. You can use standard

问题 I am new to PHP and I want to concatenate a string with a variable without any space. I am using a variable $var and a string which is given below. $var // This is variable "Name\branch" //This is String I want to concatenate the string and the variable without any space. I am using code like this: $var2 = "Name\Branch\ $var" But a space is created between them. 回答1: Space is there, because you entered it ;) Use examples: $var2 = "Name\Branch\\$var"; $var2 = "Name\Branch\\" . $var; $var2 =

问题 I am new to PHP and I want to concatenate a string with a variable without any space. I am using a variable $var and a string which is given below. $var // This is variable "Name\branch" //This is String I want to concatenate the string and the variable without any space. I am using code like this: $var2 = "Name\Branch\ $var" But a space is created between them. 回答1: Space is there, because you entered it ;) Use examples: $var2 = "Name\Branch\\$var"; $var2 = "Name\Branch\\" . $var; $var2 =

问题 So I have two arrays Array ( [0] => test [1] => test 1 [2] => test 2 [3] => test 3 ) and Array ( [0] => test [1] => test 1 [2] => test 2 [3] => test 3 ) I want to combine them together so I get an array like this? Array ( [0] => test test [1] => test 1 test 1 [2] => test 2 test 2 [3] => test 3 test 3 ) I have found lots of functions like array_merge and array_combine but nothing that does what I want to do. Any ideas? Thanks in advance. Max 回答1: You could do it with array_map : $combined =

问题 How to use literal suffix for identifier transformed into literal string in MACRO by #identifier ? struct SomeType; SomeType operator "" _udl(const char* self); #define STRINGIFY_AS_UDL(id) /* #id _udl doesn't work */ /* How to have "id"_udl */ STRINGIFY_AS_UDL(foo) // -> "foo"_udl STRINGIFY_AS_UDL(bar) // -> "bar"_udl STRINGIFY_AS_UDL(42) // -> "42"_udl 回答1: UDL operators are also "regular" functions, so you can call them instead: #define STRINGIFY_AS_UDL(id) operator ""_udl(#id) but you can

问题 A carer can provide support to two different service at the same Start time and End time this is called 1:2 Service type. The unique ID for this type of shift type are generated as the same unique ID for this type of service. I want to add 2 at the end of the unique ID for every second service. How do I find this? Bare in mind I am using this within the UNION with other 3 select statement. The first query above the Union is the one My query below: SELECT DISTINCT CASE WHEN SHIFT_CODE = '1:2

问题 I would like to create a StringVar() that looks something like this: someText = "The Spanish Inquisition" # Here's a normal variable whose value I will change eventually TkEquivalent = StringVar() # and here's the StringVar() TkEquivalent.set(string(someText)) #and here I set it equal to the normal variable. When someText changes, this variable will too... HOWEVER: TkEquivalent.set("Nobody Expects " + string(someText)) If I do this, the StringVar() will no longer automatically update! How can