I have a char buffer which will be of the following format.
Somecontent
.
.
Content-Length: 1570
MoreContent
.
.
EndofContent
I want to replace The more content part with something else say Xyz . But now I need to change content-length value to the new length of xyz. Also more content has to be replaced by xyz.
I have been able to extract the content below content-length and modify it, but placing it back into the original buffer has been an issue for me.
The output should be:
Somecontent
.
.
Content-Length: 100
xyz
I tried using strstr to find location of content-length, and use memcpy to copy contents before that, but it hasn;t worked. Can anyone suggest a method. The buffer is in char *buf;
I assume that, when you replace "More content" with "xyz", you're replacing, the end of "xyz" is the new "End of content". Correct? I'm also assuming the new "xyz" cannot overflow your buffer, correct?
In that case:
1) Search for "Content length". Save the offset (for example, variable "ofs1").
2) Search for the start of "More content". Save the offset (for example, variable "ofs2").
3) If you don't already know the new Content-Length, you can easily compute it as length(More Content) - Length(xyz)
4) Copy xyz over More content (e.g. memcpy)
5) Update "Content-Length". Blank-pad, to keep the same #/digits.
6) Voila! Done.
Related
I wonder how to convert a dm3 file into .jpg/jpeg images? there is test annotation and scale bar on the image. I setup a script but it always show that "the format cannot contain the data to be saved". This can be done via file/batch convert function. So how to realize the same function in script？ Thanks
image test:=IntegerImage("test",2,1,100,100)
test.ShowImage()
image frontimage:=GetFrontImage()
string filename=getname(frontimage)
imagedisplay disp = frontImage.ImageGetImageDisplay(0)
disp.applydatabar()
ImageDocument frontDoc = GetFrontImageDocument()
string directoryname, pathname
number length
if(!SaveAsDialog("","Do Not Change Me",directoryname)) exit(0)
length=len(directoryname)-16
directoryname=mid(directoryname,0,length)
pathname=directoryname+filename
frontDoc.ImageDocumentSaveToFile( "JPG Format", pathname )
To convert to jpg you have to use "JPEG/JFIF Format" as the handler (=format).
It has to be exactly this string in the ImageDocument.ImageDocumentSaveToFile() function. Other formats are mentioned in the help (F1 > Scripting > Objects > Document Object Model > ImageDocument Object > ImageDocumentSaveToFile() function). Those are (for example):
'Gatan Format'
'Gatan 3 Format'
'GIF Format'
'BMP Format'
'JPEG/JFIF Format'
'Enhanced Metafile Format'
In your code you are using the SaveAsDialog() to get a directory. This is not necessary. You can use GetDirectoryDialog() to get a directory. This saves you the name operation for the directoryname and avoids problems when users do change your filename.
Also for concatinating paths I prefer using PathConcatenate(). On the first hand this makes your code a lot more readable since its name tells what you are doing. On the other hand this also takes care of the directory ending with \ or not and other path related things.
The following code is what I think you need:
Image test := IntegerImage("test", 2, 1, 100, 100);
test.ShowImage();
Image frontimage := GetFrontImage();
ImageDisplay disp = frontImage.ImageGetImageDisplay(0);
disp.applydatabar();
ImageDocument frontDoc = GetFrontImageDocument();
string directoryname;
if(!GetDirectoryDialog("Select directory", "C:\\\\", directoryname)){
// ↑
// You can of course use something else as the start point for selection here
exit(0);
}
string filename = GetName(frontimage);
string pathname = directoryname.PathConcatenate(filename);
frontDoc.ImageDocumentSaveToFile("JPEG/JFIF Format", pathname);
This answer is correct and should be accepted. Your problem is the wrong file-type string. You want to use "JPEG/JFIF Format"
A bit more general information on image file saving in DigitalMicrograph.
One doesn't save images but always imageDocuments that can contain one, more, or even zero image objects in them. Script-commands that save an image like SaveAsGatan() really just call things like: ImageGetOrCreateImageDocument().ImageDocumentSaveToFile()
The difference doesn't really matter for simple one-image-in-document type images, but it can make a difference when there are multiple images in a document, or when a single image is displayed multiple times simultaneously (which can be done.) So it is always good to know what "really" goes on.
ImageDocuments contain some properties relating to saving:
A save format (“Gatan Format”, “TIFF Format”, …)
Default value: What it was opened with, or last used save-format in case of creation
Script commands: ImageDocumentGetCurrentFileSaveFormat() ImageDocumentSetCurrentFileSaveFormat()
A current file path:
Default value: What it was opened from, or empty
Script commands: ImageDocumentGetCurrentFile() ImageDocumentSetCurrentFile()
A dirty-state:
Default value: clean when opened, dirty when created
Script commands: ImageDocumentIsDirty() ImageDocumentClean()
A linked-to-file state:
Default value: true when opened, false when created
Script commands: ImageDocumentIsLinkedToFile()
There are two ways of saving an imageDocument:
Saving the current document itself to disc:
void ImageDocumentSave( ImageDocument imgDoc, Number save_style ) This utilizes the current properties of the imageDocument to save it to current path in current format, marking it clean in the process. The save_style parameter determines how the program deals with missing info:
0 = never ask for path
1 = ask if not linked (or empty path)
2 = always ask
Saving a copy of the current document to disc:
void ImageDocumentSaveToFile( ImageDocument imgDoc, String handler, String fileName ) This makes a copy and save the file under provided path in the provided format. The imageDocument in memory does not change its properties. Most noticeable: It does not become clean, and it is not linked to the provided file on disc. The filename parameter specifies the saving location including the filename. If a file extension is provided, it has to match the file-format, but it can be left out. The handler parameter specified the file-format and can be anything GMS currently supports, such as:
Gatan Format
Gatan 3 Format
GIF Format
BMP Format
JPEG/JFIF Format
Enhanced Metafile Format
In short:
To save the currently opened imageDocument with a different format, you would want to do:
imageDocument doc = GetFrontImageDocument()
doc.ImageDocumentSetCurrentFileSaveFormat("TIFF Format")
doc.ImageDocumentSave(0)
While to just save a copy of the current state you would use:
imageDocument doc = GetFrontImageDocument()
string path = doc.ImageDocumentGetCurrentFile() // full path including extension!
path = PathExtractDirectory(path,0) + PathExtractBaseName(path,0) // path without file extension
doc.ImageDocumentSaveToFile("TIFF Format", path )
Got quite a head-scratcher....
I'm using the VBScript function REPLACE to replace spaces in a decrypted field from a MSSQL DB with "/".
But the REPLACE function isn't "seeing" the spaces.
For example, if I run any one of the following, where the decrypted value of the field "ITF_U_ClientName_Denc" is "Johnny Carson":
REPLACE(ITF_U_Ledger.Fields("ITF_U_ClientName_Denc")," ","/")
REPLACE(ITF_U_Ledger.Fields("ITF_U_ClientName_Denc")," ","/")
REPLACE(ITF_U_Ledger.Fields("ITF_U_ClientName_Denc"),"Chr(160)","/")
REPLACE(CSTR(ITF_U_Ledger.Fields("ITF_U_ClientName_Denc"))," ","/")
REPLACE(ITF_U_Ledger.Fields("ITF_U_ClientName_Denc")," ","/",1,-1,1)
REPLACE(ITF_U_Ledger.Fields("ITF_U_ClientName_Denc")," ","/",1,-1,0)
The returned value is "Johnny Carson" (space not replaced with /)
The issue seems to be exclusively with spaces, because when I run this:
REPLACE(ITF_U_Ledger.Fields("ITF_U_ClientName_Denc"),"a","/")
I get "Johnny C/rson".
Also, the issue seems to be exclusively with spaces in the decrypted value, because when I run this:
REPLACE("Johnny Carson"," ","/")
Of course, the returned value is "Johnny/Carson".
I have checked what is being written to the source of the page and it is simply "Johnny Carson" with no encoding or special characters.
I have also tried the SPLIT function to see if it would "see" the space, but it doesn't.
Finally, thanks to a helpful comment, I tried VBS REGEX searching for \s.
Set regExp = New RegExp
regExp.IgnoreCase = True
regExp.Global = True
regExp.Pattern = "\s" 'Add here every character you don't consider as special character
strProcessed = regExp.Replace(ITF_U_Ledger.Fields("ITF_U_ClientName_Denc"), "?")
Unfortunately, strProcessed retruns "Johnny Carson" (ie. spaces not detected/removed).
If I replace regExp.Pattern = "a", strProcessed returns "Johnny C?rson".
Many thanks for your help!!
As we found, the right character code is 160, and that did the trick:
replace(..., ChrW(160), "...")
This seems to be data specific and, additionally, as an alternative you can try to get same encoding of the source script (i.e. save with Save As with Encoding), or convert received database value into a different target encoding.
Problem:
Hello, I have been struggling recently in my programming endeavours. I have managed to receive the output below from Google Speech to Text, but I cannot figure out how draw data from this block.
Excerpt 1:
[VoiceMain]: Successfully initialized
{"result":[]}
{"result":[{"alternative":[{"transcript":"hello","confidence":0.46152416},{"transcript":"how low"},{"transcript":"how lo"},{"transcript":"how long"},{"transcript":"Polo"}],"final":true}],"result_index":0}
[VoiceMain]: Successfully initialized
{"result":[]}
{"result":[{"alternative":[{"transcript":"hello"},{"transcript":"how long"},{"transcript":"how low"},{"transcript":"howlong"}],"final":true}],"result_index":0}
Objective:
My goal is to extract the string "hello" (without the quotation marks) from the first transcript of each block and set it equal to a variable. The problem arises when I do not know what the phrase will be. Instead of "hello", the phrase may be a string of any length. Even if it is a different string, I would still like to set it to the same variable to which the phrase "hello" would have been set to.
Furthermore, I would like to extract the number after the word "confidence". In this case, it is 0.46152416. Data type does not matter for the confidence variable. The confidence variable appears to be more difficult to extract from the blocks because it may or may not be present. If it is not present, it must be ignored. If it is present however, it must be detected and stored as a variable.
Also please note that this text block is stored within a file named "CurlOutput.txt".
All help or advice related to solving this problem is greatly appreciated.
You could do this with regex, but then I am assuming you will want to use this as a dict later in your code. So here is a python approach to building this result as a dictionary.
import json
with open('CurlOutput.txt') as f:
lines = f.read().splitlines()
flag = '{"result":[]} '
for line in lines: # Loop through each lin in file
if flag in line: # check if this is a line with data on it
results = json.loads(line.replace(flag, ''))['result'] # Load data as a dict
# If you just want to change first index of alternative
# results[0]['alternative'][0]['transcript'] = 'myNewString'
# If you want to check all alternative for confidence and transcript
for result in results[0]['alternative']: # Loop over each alternative
transcript = result['transcript']
confidence = None
if 'confidence' in result:
confidence = result['confidence']
# now do whatever you want with confidence and transcript.
how to extract a specific string from an array of characters?
in my case buff contain what's shown in the image and i want to copy the file name 'hi.jpg' to another array of char, keep in mind that the file name will change depending on the http request sent by the browser.
You can use strstr(buff,"GET ") and strstr(buff," HTTP/1.1") to find the start end end pos, then use strncpy to copy whats in between.
I'm currently trying to attach image files to a model directly from a zip file (i.e. without first saving them on a disk). It seems like there should be a clearer way of converting a ZipEntry to a Tempfile or File that can be stored in memory to be passed to another method or object that knows what to do with it.
Here's my code:
def extract (file = nil)
Zip::ZipFile.open(file) { |zip_file|
zip_file.each { |image|
photo = self.photos.build
# photo.image = image # this doesn't work
# photo.image = File.open image # also doesn't work
# photo.image = File.new image.filename
photo.save
}
}
end
But the problem is that photo.image is an attachment (via paperclip) to the model, and assigning something as an attachment requires that something to be a File object. However, I cannot for the life of me figure out how to convert a ZipEntry to a File. The only way I've seen of opening or creating a File is to use a string to its path - meaning I have to extract the file to a location. Really, that just seems silly. Why can't I just extract the ZipEntry file to the output stream and convert it to a File there?
So the ultimate question: Can I extract a ZipEntry from a Zip file and turn it directly into a File object (or attach it directly as a Paperclip object)? Or am I stuck actually storing it on the hard drive before I can attach it, even though that version will be deleted in the end?
UPDATE
Thanks to blueberry fields, I think I'm a little closer to my solution. Here's the line of code that I added, and it gives me the Tempfile/File that I need:
photo.image = zip_file.get_output_stream image
However, my Photo object won't accept the file that's getting passed, since it's not an image/jpeg. In fact, checking the content_type of the file shows application/x-empty. I think this may be because getting the output stream seems to append a timestamp to the end of the file, so that it ends up looking like imagename.jpg20110203-20203-hukq0n. Edit: Also, the tempfile that it creates doesn't contain any data and is of size 0. So it's looking like this might not be the answer.
So, next question: does anyone know how to get this to give me an image/jpeg file?
UPDATE:
I've been playing around with this some more. It seems output stream is not the way to go, but rather an input stream (which is which has always kind of confused me). Using get_input_stream on the ZipEntry, I get the binary data in the file. I think now I just need to figure out how to get this into a Paperclip attachment (as a File object). I've tried pushing the ZipInputStream directly to the attachment, but of course, that doesn't work. I really find it hard to believe that no one has tried to cast an extracted ZipEntry as a File. Is there some reason that this would be considered bad programming practice? It seems to me like skipping the disk write for a temp file would be perfectly acceptable and supported in something like Zip archive management.
Anyway, the question still stands:
Is there a way of converting an Input Stream to a File object (or Tempfile)? Preferably without having to write to a disk.
Try this
Zip::ZipFile.open(params[:avatar].path) do |zipfile|
zipfile.each do |entry|
filename = entry.name
basename = File.basename(filename)
tempfile = Tempfile.new(basename)
tempfile.binmode
tempfile.write entry.get_input_stream.read
user = User.new
user.avatar = {
:tempfile => tempfile,
:filename => filename
}
user.save
end
end
Check out the get_input_stream and get_output_stream messages on ZipFile.