Please help, I can't find the solution
Situation. I have busybox httpd server. In cgi-bin folder is an cgi-executable, which sends to client formatted text by printf command.
Problem is that the text format should look like a column, but client receives only a string. Despite the fact that in "printf" I use "\n" and "(char) 13".
Another words executable doesn't return "return" symbol
I wrote following
for (i=0; i<4;i++)
printf ("%9.8g%c\n", lTemp[i]*dTemp[i], (char) 13 );
The text that is sent from your CGI program to the web client is treated as HTML text, not plain text.
When HTML is processed for display in the browser, newline and carriage return (what you simply call "return") characters are ignored.
To cause the displayed text to perform a line break, the HTML break tag, "< br />" should be inserted into the output string:
printf("%9.8g <br />\r\n", lTemp[i] * dTemp[i]);
The use of newlines and whitespace in the text that your CGI programs generates will have little bearing on the actual HTML page that gets displayed. Use newlines and whitespace to format the HTML so that the source is readable, and use HTML tags to control the displayed text in the client's browser.
BTW
Using a numeric constant and a character conversion in a printf is not the preferred method of outputting a carriage-return character.
Use the defined escape sequence \r in the format.
Related
I am writing markdown files in Obsidian.md and trying to convert them via Pandoc and LaTeX to PDF. Text itself works fine doing this, howerver, in Obsidian I use ==equal signs== to highlight something, however this doesn't work in LaTeX.
So I'd like to create a filter that either removes the equal signs entirely, or replaces it with something LaTeX can render, e.g. \hl{something}. I think this would be the same process.
I have a filter that looks like this:
return {
{
Str = function (elem)
if elem.text == "hello" then
return pandoc.Emph {pandoc.Str "hello"}
else
return elem
end
end,
}
}
this works, it replaces any instance of "hello" with an italicized version of the word. HOWEVER, it only works with whole words. e.g. if "hello" were part of a word, it wouldn't touch it. Since the equal signs are read as part of one word, it won't touch those.
How do I modify this (or, please, suggest another filter) so that it CAN replace and change parts of a word?
Thank you!
this works, it replaces any instance of "hello" with an italicized version of the word. HOWEVER, it only works with whole words. e.g. if "hello" were part of a word, it wouldn't touch it. Since the equal signs are read as part of one word, it won't touch those.
How do I modify this (or, please, suggest another filter) so that it CAN replace and change parts of a word?
Thank you!
A string like Hello, World! becomes a list of inlines in pandoc: [ Str "Hello,", Space, Str "World!" ]. Lua filters don't make matching on that particularly convenient: the best method is currently to write a filter for Inlines and then iterate over the list to find matching items.
For a complete example, see https://gist.github.com/tarleb/a0646da1834318d4f71a780edaf9f870.
Assuming we already found the highlighted text and converted it to a Span with with class mark. Then we can convert that to LaTeX with
function Span (span)
if span.classes:includes 'mark' then
return {pandoc.RawInline('latex', '\\hl{')} ..
span.content ..
{pandoc.RawInline('latex', '}')}
end
end
Note that the current development version of pandoc, which will become pandoc 3 at some point, supports highlighted text out of the box when called with
pandoc --from=markdown+mark ...
E.g.,
echo '==Hi Mom!==' | pandoc -f markdown+mark -t latex
⇒ \hl{Hi Mom!}
So I query that pulls text from a column called "description."
Each description contains a list like so:
1) Some text here
2) Some text here
3) Some text here
The problem is when I run the query in my zsh script the new lines return a "+" symbol instead of a carriage return.
1) Some text here + 2) Some text here + 3) Some text here...
The code is
tempdescriptionday4=$(/Applications/Postgres.app/Contents/Versions/12/bin/psql -h 1.1.1.1 -p5555 -U myuser mytable -t -c "SELECT description FROM cycle_10 WHERE air_date = 2020-11-10"
I'm trying to inject this list into an XML file for an RSS feed but I'm stuck on how to format this properly. I tried replacing the + symbol with
but that didn't work--I still am not getting a new line. Any ideas?
Found this in the postgresql12 manual — it's called:linestyle
"Sets the border line drawing style to one of ascii, old-ascii, or unicode. Unique abbreviates are allowed. (that would mean one letter is enough.) The default setting is ascii. This option only affects the aligned and wrapped output formats"
So for me since I have multiple lines in a cell it "wraps" the text at the end making it a wrapped output format.
"ascii style uses plain ASCII characters. Newlines in data are shown using a + symbol in the right-hand margin." When the wrapped format wraps data from one line to the next without a new line character, a dot"." is shown in in the right-hand margin of the first line, and again in the left-hand margin of the following line."
There's also "old-ascii" style that uses a ":" for wrapping and a "unicode" style which uses an ellipsis symbol in the right and margin of first line.
So the problem I have is that the output format is using ASCII by default when there's no available option for XML output style for new lines. Bummer.
Turning linestyle off fixes removing the + symbol but I want the outputted linestyle to be XML so if it strips any new line indication there's no way for me to format it. "-A --no-align Switches to unaligned output mode. (The default output mode is aligned.) This is equivalent to \pset format unaligned."
So I used awk to substitute the + symbol for
and also tried
but those are not creating new lines for me? It validates as proper XML.
I have a text file containing unwanted null characters (ASCII NUL, \0). When I try to view it in vi I see ^# symbols, interleaved in normal text. How can I:
Identify which lines in the file contain null characters? I have tried grepping for \0 and \x0, but this did not work.
Remove the null characters? Running strings on the file cleaned it up, but I'm just wondering if this is the best way?
I’d use tr:
tr < file-with-nulls -d '\000' > file-without-nulls
If you are wondering if input redirection in the middle of the command arguments works, it does. Most shells will recognize and deal with I/O redirection (<, >, …) anywhere in the command line, actually.
Use the following sed command for removing the null characters in a file.
sed -i 's/\x0//g' null.txt
this solution edits the file in place, important if the file is still being used. passing -i'ext' creates a backup of the original file with 'ext' suffix added.
A large number of unwanted NUL characters, say one every other byte, indicates that the file is encoded in UTF-16 and that you should use iconv to convert it to UTF-8.
I discovered the following, which prints out which lines, if any, have null characters:
perl -ne '/\000/ and print;' file-with-nulls
Also, an octal dump can tell you if there are nulls:
od file-with-nulls | grep ' 000'
If the lines in the file end with \r\n\000 then what works is to delete the \n\000 then replace the \r with \n.
tr -d '\n\000' <infile | tr '\r' '\n' >outfile
Here is example how to remove NULL characters using ex (in-place):
ex -s +"%s/\%x00//g" -cwq nulls.txt
and for multiple files:
ex -s +'bufdo!%s/\%x00//g' -cxa *.txt
For recursivity, you may use globbing option **/*.txt (if it is supported by your shell).
Useful for scripting since sed and its -i parameter is a non-standard BSD extension.
See also: How to check if the file is a binary file and read all the files which are not?
I used:
recode UTF-16..UTF-8 <filename>
to get rid of zeroes in file.
I faced the same error with:
import codecs as cd
f=cd.open(filePath,'r','ISO-8859-1')
I solved the problem by changing the encoding to utf-16
f=cd.open(filePath,'r','utf-16')
Remove trailing null character at the end of a PDF file using PHP, . This is independent of OS
This script uses PHP to remove a trailing NULL value at the end of a binary file, solving a crashing issue that was triggered by the NULL value. You can edit this script to remove all NULL characters, but seeing it done once will help you understand how this works.
Backstory
We were receiving PDF's from a 3rd party that we needed to upload to our system using a PDF library. In the files being sent to us, there was a null value that was sometimes being appended to the PDF file. When our system processed these files, files that had the trailing NULL value caused the system to crash.
Originally we were using sed but sed behaves differently on Macs and Linux machines. We needed a platform independent method to extract the trailing null value. Php was the best option. Also, it was a PHP application so it made sense :)
This script performs the following operation:
Take the binary file, convert it to HEX (binary files don't like exploding by new lines or carriage returns), explode the string using carriage return as the delimiter, pop the last member of the array if the value is null, implode the array using carriage return, process the file.
//In this case we are getting the file as a string from another application.
// We use this line to get a sample bad file.
$fd = file_get_contents($filename);
//We trim leading and tailing whitespace and convert the string into hex
$bin2hex = trim(bin2hex($fd));
//We create an array using carriage return as the delminiter
$bin2hex_ex = explode('0d0a', $bin2hex);
//look at the last element. if the last element is equal to 00 we pop it off
$end = end($bin2hex_ex);
if($end === '00') {
array_pop($bin2hex_ex);
}
//we implode the array using carriage return as the glue
$bin2hex = implode('0d0a', $bin2hex_ex);
//the new string no longer has the null character at the EOF
$fd = hex2bin($bin2hex);
I have a text file which shows a Line Break in UltraEdit if we replace a special character in text file manually it works fine. Unknown Line Break. I have to change it manually and then process the files.
Please let me know some way how to remove all occurrences of this character with VB.Net code.
If I replace ♀ in UltraEdit, it replaces line break with my desired string. But in my VB string I cannot use this character or line break.
The character you have in your file is the form-feed character usually used as control character for a page break.
In UltraEdit in Page Setup configuration dialog (a printing related dialog) there is the option Page break code which has by default the decimal value 12 (hexadecimal 0C) which is the form-feed character.
A page break can be displayed in UltraEdit with a horizontal line across the document window on enabling Show Page Breaks as Lines in menu/ribbon View.
The form-feed character can be removed in UltraEdit with searching for ^b on using a normal, non regular expression or an UltraEdit regular expression replace, or with searching for \f on using a Unix or Perl regular expression replace.
In VB.Net code ChrW(12) can be used to reference the form-feed control character as suggested already by Hans Passant.
I have a column in my database that contains a string like this:
"Warning set for 7 days.\nCritical Notice - Last Time Machine backup was 118 days ago at 2012-11-16 20:40:52\nLast Time Machine Destination was FreeAgent GoFlex Drive\n\nDefined Destinations:\nDestination Name: FreeAgent GoFlex Drive\nBackup Path: Not Specified\nLatest Backup: 2012-11-17"
I am displaying this data in an e-mail to users. I have be able to easily format the field in my html e-mails perfectly by doing the following:
simple_format(#servicedata.service_exit_details.gsub('\n', '<br>'))
The above code replaces the "\n" with "<br>" tags and simple_format handles the rest.
My issues arises with how to format it properly in the plain text template. Initially I thought I could just call the column, seeing as it has "\n" I assumed the plain text would interpret and all would be well. However this simply spits out the string with "\n" intact just as displayed above rather than created line breaks as desired.
In an attempt to find a way to parse the string so the line breaks are acknowledged. I have tried:
#servicedata.service_exit_details.gsub('\n', '"\r\n"')
#servicedata.service_exit_details.gsub('\n', '\r\n')
raw #servicedata.service_exit_details
markdown(#servicedata.service_exit_details, autolinks: false) # with all the necessary markdown setup
simple_format(#servicedata.service_exit_details.html_safe)
none of which worked.
Can anyone tell me what I'm doing wrong or how I can make this work?
What I want is for the plain text to acknowledge the line breaks and format the string as follows:
Warning set for 7 days.
Critical Notice - Last Time Machine backup was 118 days ago at 2012-11-16 20:40:52
Last Time Machine Destination was FreeAgent GoFlex Drive
Defined Destinations:
Destination Name: FreeAgent GoFlex Drive
Backup Path: Not Specified\nLatest Backup: 2012-11-17"
I see.
You need to differentiate a literal backslash followed by a letter n as a sequence of two characters, and a LF character (a.k.a. newline) that is usually represented as \n.
You also need to distinguish two different kinds of quoting you're using in Ruby: singles and doubles. Single quotes are literal: the only thing that is interpreted in single quotes specially is the sequence \', to escape a single quote, and the sequence \\, which produces a single backslash. Thus, '\n' is a two-character string of a backslash and a letter n.
Double quotes allow for all kinds of weird things in it: you can use interpolation with #{}, and you can insert special characters by escape sequences: so "\n" is a string containing the LF control character.
Now, in your database you seem to have the former (backslash and n), as hinted by two pieces of evidence: the fact that you're seeing literal backslash and n when you print it, and the fact that gsub finds a '\n'. What you need to do is replace the useless backslash-and-n with the actual line separator characters.
#servicedata.service_exit_details.gsub('\n', "\r\n")