I'm new to GTK4, an open source toolkit for C++ binding. I want to send text from current TextView buffer (from a saved file) to a PDF file using GTK libraries (gtkmm4), but couldn't get anything printed out.
This is the code I have started from reading the documentation:
void MainWindow:export_note() {
auto op = Gtk::PrintOperation::create();
// setup op
cout << save_file_path << endl;
string content = editor.get_buffer()->get_text();
ofstream out(work_dir + save_file_path);
out << content;
out.close();
curr_state = edit_file;
op->set_export_filename("test.pdf");
auto res = op->run(Gtk::PrintOperation::Action::EXPORT);
return;
}
This only exports to a blank PDF, but I'm expecting text to show up on PDF.
It looks like you are not using any binary application to attempt conversion from text to text/pdf or more commonly binary application/pdf. You cannot simply stuff text data into a container called Test.pdf
There are simple means to convert Text to PDF, traditionally by using PostScript Printer files, but more commonly recently using a PDF printer driver direct
so start at the most basic level Hello World needs a file something like this, where the body is built up from stacked vectors or font labelled strings at X&Y co-ordinates.
Test.pdf
%PDF-1.1
%âãÏÓ
1 0 obj<</Type/Catalog/Pages 2 0 R>>endobj
2 0 obj<</Type/Pages/Kids [3 0 R]/Count 1/MediaBox [0 0 594 792]>>endobj
3 0 obj<</Type/Page/Parent 2 0 R/Resources<</Font<</F1<</Type/Font/Subtype/Type1/BaseFont/Helvetica>>>>>>/Contents 4 0 R>>endobj
4 0 obj<</Length 81
>>
stream
BT /F1 18 Tf 036 740 Td (Hello) Tj ET
BT /F1 18 Tf 036 720 Td (World!) Tj ET
endstream
endobj xref
0 5
0000000000 65535 f
0000000019 00000 n
0000000063 00000 n
0000000137 00000 n
0000000267 00000 n
trailer<</Root 1 0 R /Size 5>>startxref
399 %%EOF
I sign a PDF and I add update version in which I write the DSS with its CRLs, Certs, VRI.
19 0 obj
[15 0 R 16 0 R]
endobj
20 0 obj
[13 0 R 14 0 R]
endobj
11 0 obj
[15 0 R 16 0 R]
endobj
12 0 obj
[13 0 R 14 0 R]
endobj
17 0 obj
<<
/CRL 11 0 R
/Cert 12 0 R
>>
endobj
18 0 obj
<<
/5F44CF6F351DFD45FB62F3D0ED046408BC892797 17 0 R
>>
endobj
21 0 obj
<<
/VRI 18 0 R
/CRLs 19 0 R
/Certs 20 0 R
>>
I am confused about how should I write the Certificate and CRL streams.
15 0 obj
<<
/Length 1454
/Filter /FlateDecode
>>
stream
xÚ3hb0hb{ÅÄÈhÀÉƪÍÇÌ$ÅÊ`àcÈä2‡²° 3…Šˆ€8\¼®y%E¥Å%:žyÉz†ªÊ
ZbXd{0%KW÷ýY¯’ó‚-ØÂÛ„OÏó½z•î ‰`®•® K-›2}tÖ§^_8;xÉì¥Ó®~›.g9A'Õüê½—
ZbXd{0%KW÷ýY¯’ó‚-ØÂÛ„OÏó½z•î ‰`®•® K-›2}tÖ§^_8;xÉì¥Ó®~›.g9A'Õüê½—
endstream
endobj
16 0 obj
<<
/Length 1477
/Filter /FlateDecode
>>
stream
„kâR7Å41*!‡#8Íñ3 Ź˜#‰o=«‡çƒ#yë:X]r\~}¼)/Ñmç×£¦³äsËê]ÓÕ_+µ¥$Ô¿}¾ÜÏiÁÝT!¹ôi–Í9üÀ}Š¸|
ìŒH¿GÓø^ú¿ÔVÜK–qõ†µ®“¸»Ý*Žh¾JzåU7c~÷•ÔêýK*îú®¹¸DcÁ³·NtV~Vóåíé5\‚&½|¶NäïŽ[K
î›NRZbXd{0%KW÷ýY¯’ó‚-ØÂÛ„OÏó½z•î ‰`®•® K-›2}tÖ§^_8;xÉì¥Ó®~›.g9A'Õüê½—›oÇ:ç-¶?
endstream
endobj
13 0 obj
<<
/Length 1240
/Filter /FlateDecode
>>
stream
%ŸwC[í2×¾Iej©úkŽ-:ݳÔ<¼a£ƒô/5›‡~zÒ•7ü9uãcfk?ËÅ`ßÃ:Èb—’‚Ÿõ{ÏÅ—¢{]HçQ”9w(ÂB#í×g¥ìþè
^–F«š/r§š¿ì=#,^pëO€{äú=}RÎêð¦ÉŠ7or¼±Ëtë–x·˜§LÌŒŒ‹› Cd0€eùÿ³°03±>0P ñUY$
endstream
endobj
14 0 obj
<<
/Length 1159
/Filter /FlateDecode
>>
stream
4!>T‚êPpÎI,.V0Ò™#ûœºƒ=LÍš•ãˆ‘•¹‰‘Ÿ(ÎÅÔÄÈÈplŽ÷A¯¹7k/[‡O\}
öe™¨îö£œ¶ä'¶ÌpžªweÞª[¡$¼ØÍþþtó[½xÉO4ÞZ¥ØŸ^g ø,mu„_Rz™_PÏê.||º¶*þîÝxv½"»êôó»ø%Ü%ý
endstream
endobj
Please ignore the lengths and content of the streams above. I truncated them so the lengths don't correspond anymore. The streams are bigger than that.
The issue is that my PDF is not LTV enabled and I tested some scenarios from which I concluded that
my stream are not being written the right way.
I use the following structure from WinCrypt.h:
typedef struct _CERT_CONTEXT {
DWORD dwCertEncodingType;
BYTE *pbCertEncoded;
DWORD cbCertEncoded;
PCERT_INFO pCertInfo;
HCERTSTORE hCertStore;
} CERT_CONTEXT, *PCERT_CONTEXT;
typedef const CERT_CONTEXT *PCCERT_CONTEXT;
I go through them and get the bytes this way:
PCCERT_CONTEXT cngContext = (PCCERT_CONTEXT)(*itChain);
ByteArray certBytes(cngContext->pbCertEncoded, (size_t)cngContext->cbCertEncoded);
Then I just apply FlateDecode on the obtained bytes and write them into the PDF like a stream as you can see in the second block of code.
Am I missing any step? Like a conversion or something? I saw that the stream should be BER-Encoded. So should I transform the bytes into BER-Encoded and then apply FlateDecode?
Edit:
You can find My File here
SOLUTION
The problem was the stream of CRLs that I was writing in the PDF file.
Having the CRL_CONTEXT structures from each Certificate, I just took the pbCrlEncoded variable and write it directly in the stream of the CRL.
It seemed correct but I noticed I didn't have any CRL_ENTRY in the CRL_INFO of this structure so the encoded BYTEs didn't contain any list of revoked certificates.
Therefore, found that the certificates have a URL from where you can download the updated CRL. You can do that by opening Manage Computer Certificates in Windows -> find your Certificate and Open the Certificate -> Details -> CRL Distribution Points -> URL = "..". By accessing this url, the browser automatically downloads the CRL Info. You can access it and see some informations like Next Update which is the last day that this list is valid. After that, I'm assuming that you need to download it again for getting an updated version. Also you can see the list itself of revoked certificates.
This is the list I needed to put into the CRLs streams in PDF.
So I found a method to do that download process by code. This is a snippet of code used:
PCERT_CHAIN_ELEMENT chainElement; // this is the certification in the chain
pExtension = CertFindExtension(szOID_CRL_DIST_POINTS, chainElement->pCertContext->pCertInfo->cExtension, chainElement->pCertContext->pCertInfo->rgExtension);
if (!pExtension)
return ByteArray();
if (!CryptDecodeObject(X509_ASN_ENCODING, szOID_CRL_DIST_POINTS, pExtension->Value.pbData, pExtension->Value.cbData, 0, 0, &cbStructInfo))
return ByteArray();
if (!(pvStructInfo = LocalAlloc(LMEM_FIXED, cbStructInfo)))
return ByteArray();
CryptDecodeObject(X509_ASN_ENCODING, szOID_CRL_DIST_POINTS, pExtension->Value.pbData, pExtension->Value.cbData, 0, pvStructInfo, &cbStructInfo);
pInfo = (CRL_DIST_POINTS_INFO*)pvStructInfo;
Net::HttpRequest req;
Net::HttpRequestOptions ops;
ops.verb = Net::GET;
crllist = req.send(pInfo->rgDistPoint->DistPointName.FullName.rgAltEntry->pwszURL);
This way I obtained the Bytes that I could paste in PDF after applying FlateDecode on them.
Now the PDF is LTV Enabled.
what is the appropriate regex to match any number greater than "750 active" in the below text at the last line ?
=============================================================================
c0000000a7470b30 Y--P--- 4207362 weblogic - c000000098078f98 0 1 1832876 0
c0000000a74853f0 Y--P--- 4376431 krizzsa LL2CE414 c0000000a19479b8 0 8 70173 170
c0000000b3a1f2c8 Y--P--- 3996541 weblogic - c0000000acd54f90 0 1 64112 0
c0000000b3a22418 Y--P--- 4371951 tinpatel tK c000000098385b70 0 1 62 0
c0000000b3a286b8 B--PR-- 4385816 ayaw SL5CG752 c00000001b0bdde0 0 5 14452 701
c0000000b3a2afd0 Y--P--- 4383560 sognenov t3 c000000099afe900 0 1 100 0
c0000000b3a2b808 Y--P--- 4368082 wenpli 66 c00000009e6f8260 0 1 461 0
c0000000b3a2c878 Y--P--- 4228342 sarbrar tc c0000000a62da668 0 1 0 0
c0000000b3a2f9c8 Y--P--- 4384060 weblogic - c0000000a2deb910 0 1 0 0
c0000000b3a35430 Y--P--- 4383243 nakahmed t1 c00000009e0b9ce8 0 1 17 0
c0000000b3a38580 Y--P--- 3937012 mvvinay 76 c0000000a162a888 0 1 0 0
c0000000b3a3d7b0 Y--P--- 616042 neminhas LD2UA442 c0000000aea25ca8 0 2 0 0
c0000000b3a43218 Y--P--- 4383236 nakahmed t1 c0000000981b4570 0 1 37 22
c0000000b3a473d8 Y--P--- 4382647 viwang 2UA40922 c0000000a268a2d0 0 7 75275 4856
412 active, 2176 total, 441 maximum concurrent
================================================================
in the last line if the active connections to a DB goes more than 750 we need to match it and do further processing.So can someone please help with regex ?
Please note that there is a space in the beginning of the last line.
If it's Perl, there's no need to use matching for everything.
if (/ ([0-9]+) active/ && $1 > 750) {
print "Matching!\n";
If you need a single regex, it's
/^\ (?: 7 (?: 5 [1-9]
| [6-9][0-9])
| [89] [0-9]{2}
| [1-9][0-9]{3,})\ active/x
or shortly
/^ (?:7(?:5[1-9]|[6-9][0-9])|[89][0-9]{2}|[1-9][0-9]{3,}) active/
It may not be the best idea, yet if we might have to, probably our expression would be:
^\h*(7[5-9]\d|[89]\d\d|[1-9]\d{3,})\h+active\b
((7[5-9]\d|[89]\d\d|[1-9]\d{3,})\sactive)
((7[5-9]\d|[89]\d\d|[1-9]\d{3,})\s+active)
((7[5-9]\d|[89]\d\d|[1-9]\d{3,}) active)
based on bobble-bubble's advice,
DEMO
or something similar to:
([7][5-9][0-9]|[8-9][0-9][0-9]|[1-9][0-9]{3,}) active
(([7][5-9][0-9]|[8-9][0-9][0-9]|[1-9][0-9]{3,}) active)
if we might want to capture things, separately.
Here is my minimal working example:
list1 = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20] #len = 21
list2 = [1,1,1,0,1,0,0,1,0,1,1,0,1,0,1,0,0,0,1,1,0] #len = 21
list3 = [0,0,1,0,1,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1] #len = 21
list4 = [1,0,0,1,1,0,0,0,0,1,0,1,1,1,1,0,1,0,1,0,1] #len = 21
I have four lists and I want to "clean" my list 1 using the following rule: "if any of list2[i] or list3[i] or list4[i] are equal to zero, then I want to eliminate the item I from list1. SO basically I only keep those elements of list1 such that the other lists all have ones there.
here is the function I wrote to solve this
def clean(list1, list2,list3,list4):
for i in range(len(list2)):
if (list2[i]==0 or list3[i]==0 or list4[i]==0):
list1.pop(i)
return list1
however it doesn't work. If you apply it, it give the error
Traceback (most recent call last):line 68, in clean list1.pop(I)
IndexError: pop index out of range
What am I doing wrong? Also, I was told Pandas is really good in dealing with data. Is there a way I can do it with Pandas? Each of these lists are actually columns (after removing the heading) of a csv file.
EDIT
For example at the end I would like to get: list1 = [4,9,11,15]
I think the main problem is that at each iteration, when I pop out the elements, the index of all the successor of that element change! And also, the overall length of the list changes, and so the index in pop() is too large. So hopefully there is another strategy or function that I can use
This is definitely a job for pandas:
import pandas as pd
df = pd.DataFrame({
'l1':list1,
'l2':list2,
'l3':list3,
'l4':list4
})
no_zeroes = df.loc[(df['l2'] != 0) & (df['l3'] != 0) & (df['l4'] != 0)]
Where df.loc[...] takes the full dataframe, then filters it by the criteria provided. In this example, your criteria are that you only keep the items where l2, l3, and l3 are not zero (!= 0).
Gives you a pandas dataframe:
l1 l2 l3 l4
4 4 1 1 1
9 9 1 1 1
12 12 1 1 1
18 18 1 1 1
or if you need just list1:
list1 = df['l1'].tolist()
if you want the criteria to be where all other columns are 1, then use:
all_ones = df.loc[(df['l2'] == 1) & (df['l3'] == 1) & (df['l4'] == 1)]
Note that I'm creating new dataframes for no_zeroes and all_ones and that the original dataframe stays intact if you want to further manipulate the data.
Update:
Per Divakar's answer (far more elegant than my original answer), much the same can be done in pandas:
df = pd.DataFrame([list1, list2, list3, list4])
list1 = df.loc[0, (df[1:] != 0).all()].astype(int).tolist()
Here's one approach with NumPy -
import numpy as np
mask = (np.asarray(list2)==1) & (np.asarray(list3)==1) & (np.asarray(list4)==1)
out = np.asarray(list1)[mask].tolist()
Here's another way with NumPy that stacks those lists into rows to form a 2D array and thus simplifies things quite a bit -
arr = np.vstack((list1, list2, list3, list4))
out = arr[0,(arr[1:] == 1).all(0)].tolist()
Sample run -
In [165]: arr = np.vstack((list1, list2, list3, list4))
In [166]: print arr
[[ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20]
[ 1 1 1 0 1 0 0 1 0 1 1 0 1 0 1 0 0 0 1 1 0]
[ 0 0 1 0 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 1]
[ 1 0 0 1 1 0 0 0 0 1 0 1 1 1 1 0 1 0 1 0 1]]
In [167]: arr[0,(arr[1:] == 1).all(0)].tolist()
Out[167]: [4, 9, 12, 18]
I have a problem with saving a binary representation of a file to a file...
Let me show you my pain:
Everything starts with a file, file.pdf
Then the file is send via POST to a website with some additional data:
curl --data "sector=4&name=John&surname=Smith&email=john#smith.com&isocode=PL&theFile=$(cat file.pdf | base64)" http://localhost/awesomeUpload
then the data is received and decoded:
var decoded = BinaryDecode(data.theFile, "Base64");
then I attempt to save it by:
var theFilePath = ExpandPath("/localserver/temp/theFile.pdf");
fileWrite(theFilePath , data.theFile);
or:
var file_output_steam = CreateObject("java","java.io.FileOutputStream").init(theFilePath);
file_output_steam.write(data.theFile);
file_output_steam.close();
My files does not match ;(
the original one looks like
%PDF-1.5
%µµµµ
1 0 obj
<</Type/Catalog/Pages 2 0 R/Lang(pl-PL) /StructTreeRoot 13 0 R/MarkInfo<</Marked true>>>>
endobj
2 0 obj
<</Type/Pages/Count 1/Kids[ 3 0 R] >>
endobj
3 0 obj
<</Type/Page/Parent 2 0 R/Resources<</Font<</F1 5 0 R/F2 10 0 R>>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 595.32 841.92] /Contents 4 0 R/Group<</Type/Group/S/Transparency/CS/DeviceRGB>>/Tabs/S/StructParents 0>>
where as the copy that went through ColdFusion looks like:
%PDF-1.5
%µµµµ
1 0 obj
<</Type/Catalog/Pages 2 0 R/Lang(pl-PL) /StructTreeRoot 13 0 R/MarkInfo<</Marked true>> B™[™ŘšBŚŘšBŹŐ\KÔYŮ\ËĐŰÝ[ťKŇÚYÖČČ—H€Đ¦VćFö& ĐŁ2ö& ĐŁĂÂőG—RővRő&VçB""ő&W6÷W&6W3ĂÂôföçCĂÂôcR"ôc"#ŕ˝AÉ˝ŤM•Ńl˝A˝Q•áĐ˝%µ…ť•˝%µ…ť•˝%µ…ť•%t€>/MediaBox[ 0 0 595.32 841.92] /Contents 4 0 R/Group<</Type/Group/S/Transparency/CS/DeviceRGB>>/Tabs/S/StructParents 0>B™[™ŘšBŤŘšBŹŃš[\‹Ń›]QXŰŮKÓ[™ÝMŚOŹBśÝ™X[CBž'cłB°Ś!8Ě1Ď]CsôŘQ&‰2 PäV˝ëËöĽ¨QŰge•ź
ďÂŃ,đť#"aKR•˘<1™[ä¸
(ÄňĄyoâ9S\Śĺ <ę8I±D¬‰#…Ć”ťLé‘ا÷ÍnU|WŸ‰t`ýuşąĽ\hlu&âĂ7ß
ů"Ĺ\Ŕ>pÇč÷÷.°ß’Ř——•‹ĚB™[™Ý™X[CB™[™ŘšBŤHŘšBŹŐ\Kћ۝ÔÝXť\KŐ\LĐ\ŮQ›ŰťĐPŃQJĐŘ[XśšKŃ[ŰŮ[™ËŇY[ť]KRŃ\ŘŮ[™[ť›ŰťČ
‹ŐŐ[šXŰŮHŚŹŹB™[™ŘšBŤŘšB–Č
Č—HB™[™ŘšBŤČŘšBŹĐ\ŮQ›ŰťĐPŃQJĐŘ[XśšKÔÝXť\KĐŇQ›Űť\L‹Ő\Kћ۝ĐŇQŃŇQX\ŇY[ť]KŃČLĐŇQŢ\Ý[R[™›Č‹Ń›Űť\ŘÜš\ÜH‹ŐČŚŕЦVćFö& ĐŁ‚ö& ĐŁĂÂô÷&FW&–ćr„–FVçF—G’’ő&Vv—7G'’„Fö&R’ő7WĆVÖVçBăŕЦVćFö& ĐŁ’ö& ĐŁĂÂőG—RôföçDFW67&—F÷"ôföçDć
please help