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I am using Laravel + Backpack for admin panel and I am trying to display a chart with 12 labels (for each month) and integer data corresponding for each month. My problem is that all datasets go under the first label and I cannot figure out how to put each dataset under the correct label.
Code (keep it short):
$array; // array of eloquent models
$array2; // array of eloquent models
$this->chart->labels([
'January',
'February',
]);
$this->chart->dataset('January Exp', 'bar', count($array))
->color('rgba(205, 32, 31, 1)')
->backgroundColor('rgba(205, 32, 31, 0.4)');
$this->chart->dataset('February Exp', 'bar', count($array2))
->color('rgba(205, 32, 31, 1)')
->backgroundColor('rgba(205, 32, 31, 0.4)');
I have tried looking up on Google of a way I can bind each dataset to a particular label, but I was not able to find a solution. Perhaps I am going at it from the wrong direction.
Following Backpack's docs and Laravel Charts' docs I should be able to declare the labels and the datasets and those should show up as expected.
I have also tried declaring the datasets in the setup() function as well as the data() function. Both ways lead to the same result.
Found a solution that seems to work for my use case.
Declare the labels then one dataset with an array for the values.
Code:
$array; // array of eloquent models
$array2; // array of eloquent models
$this->chart->labels([
'January',
'February',
]);
$this->chart->dataset('Exp', 'bar', [count($array), count($array2)])
->color('rgba(205, 32, 31, 1)')
->backgroundColor('rgba(205, 32, 31, 0.4)');
I hope someone finds this useful in the future.
I have the following example of PDF417 barcode:
which can be decoded with online tool like zxing
as the following result: 5wwwwwxwww0app5p3pewi0edpeapifxe0ixiwwdfxxi0xf5e�¼ô���������¬‚C`Ìe%�æ‹�ÀsõbÿG)=‡x‚�qÀ1ß–[FzùŽûVû�É�üæ±RNI�Y[.H»Eàó¼åñüì²�tØ¿ªWp…Ã�{�Õ*
or online-qrcode-generator
as 5wwwwwxwww0app5p3pewi0edpeapifxe0ixiwwdfxxi0xf5e~|~~~~~~~~~~d~C`~e%~~~~;To~B~{~dj9v~~Z[Xm~~"HP3~~LH~~~O~"S~~,~~~~~~~k1~~~u~Iw}SQ~fqX4~mbc_ (I don't know which encoding is used to encode this)
The first part of the encoded key that contains barcode is always known and it is 5wwwwwxwww0app5p3pewi0edpeapifxe0ixiwwdfxxi0xf5e
The second part of it can be decoded from the base64string and it always contains 88 bytes. In my case it is:
Frz0DAAAAAAAAAAArIJDYMxlJQDmiwHAc/Vi/0cpPYd4ghlxwDHflltGevmO+1b7GckT/OZ/sVJOSRpZWy5Iu0Xg87zl8fzssg502L+qV3CFwxZ/ewjVKg==
I'm using Swift on iOS device to generate this PDF417 barcode by decoding the provided base64 string like this:
let base64Str = "Frz0DAAAAAAAAAAArIJDYMxlJQDmiwHAc/Vi/0cpPYd4ghlxwDHflltGevmO+1b7GckT/OZ/sVJOSRpZWy5Iu0Xg87zl8fzssg502L+qV3CFwxZ/ewjVKg=="
let knownKey = "5wwwwwxwww0app5p3pewi0edpeapifxe0ixiwwdfxxi0xf5e"
let decodedData = Data(base64Encoded: base64Str.replacingOccurrences(of: "-", with: "+")
.replacingOccurrences(of: "_", with: "/"))
var codeData=knownKey.data(using: String.Encoding.ascii)
codeData?.append(decodedData)
let image = generatePDF417Barcode(from: codeData!)
let imageView = UIImageView(image: image!)
//the function to generate PDF417 UIMAGE from parsed Data
func generatePDF417Barcode(from codeData: Data) -> UIImage? {
if let filter = CIFilter(name: "CIPDF417BarcodeGenerator") {
filter.setValue(codeData, forKey: "inputMessage")
let transform = CGAffineTransform(scaleX: 3, y: 3)
if let output = filter.outputImage?.transformed(by: transform) {
return UIImage(ciImage: output)
}
}
return nil
}
But I always get the wrong barcodes generated. It can be seen visually.
Please help me correct the code to get the same result as the first barcode image.
I also have the another example of barcode:
The first part of key is the same but it's second part is known as int8 byte array and I also don't have an idea how to generate the PDF417 barcode from it (with prepended key) correctly.
Here's how I try:
let knownKey = "5wwwwwxwww0app5p3pewi0edpeapifxe0ixiwwdfxxi0xf5e"
let secretArray: [Int8] = [22, 124, 24, 12, 0, 0, 0, 0, 0, 0, 0, 0, 100, 127, 67, 96, -52, 101, 37, 0, -85, -123, 1, -64, 111, -28, 66, -27, 123, -25, 100, 106, 57, 118, -4, 16, 90, 91, 88, 109, -105, 126, 34, 72, 80, 51, -116, 28, 76, 72, -37, -24, -93, 79, -115, 34, 83, 18, -61, 44, -12, -13, -8, -59, -107, -9, -128, 107, 49, -50, 126, 13, -59, 50, -24, -43, 127, 81, -85, 102, 113, 88, 52, -60, 109, 98, 99, 95]
let secretUInt8 = secretArray.map { UInt8(bitPattern: $0) }
let secretData = Data(secretUInt8)
let keyArray: [UInt8] = Array(knownKey.utf8)
var keyData = Data(keyArray)
keyData.append(secretData)
let image = generatePDF417Barcode(from: keyData!)
let imageView = UIImageView(image: image!)
There are a lot of things going on here. Gereon is correct that there are a lot of parameters. Choosing different parameters can lead to very different bar codes that decode identically. Your current barcode is "correct" (though a bit messy due to an Apple bug). It's just different.
I'll start with the short answer of how to make your data match the barcode you have. Then I'll walk through what you should probably actually do, and finally I'll get to the details of why.
First, here's the code you're looking for (but probably not the code you want, unless you have to match this barcode):
filter.setValue(codeData, forKey: "inputMessage")
filter.setValue(3, forKey: "inputCompactionMode") // This is good (and the big difference)
filter.setValue(5, forKey: "inputDataColumns") // This is fine, but probably unneeded
filter.setValue(0, forKey: "inputCorrectionLevel") // This is bad
PDF 417 defines several "compaction modes" to let it pack a truly impressive amount of information into a very small space while still offering excellent error detection and correction, and handling a lot of real-world scanning concerns. The default compaction mode only supports Latin text and basic punctuation. (It compacts even more if you only use uppercase Latin letters and space.) The first part of your string can be stored with text compaction, but the rest can't, so it has to switch to byte compaction.
Core Image actually does this switch shockingly badly by default (I opened FB9032718 to track). Rather than encoding in text and then switching to bytes, or just doing it all in bytes, it switches to bytes over and over again unnecessarily.
There's no way for you to configure multiple compaction methods, but you can just set it to byte, which is what value 3 is. And that's also how your source is doing it.
The second difference is the number of data columns, which drive how wide the output is. Your source is using 5, but Core Image is choosing 6 based on its default rules (which aren't fully documented).
Finally, your source has set the error correction level to 0, which is not recommended. For a message of this size, the minimum recommended error correction level is 3, which is what Core Image chooses by default.
If you just want a good barcode, and don't have to match this input, my recommendation would be to set inputCompactionMode to 3, and leave the rest as defaults. If you want a different aspect ratio, I'd use inputPreferredAspectRatio rather than modifying the number of data columns directly.
You may want to stop reading now. This was a very enjoyable puzzle to spend the morning on, so I'm going to dump a lot of details here.
If you want a deep dive into how this format works, I don't know anything currently available other than the ISO 15438 Spec, which will cost you around US$200. But there used to be some pages at GeoCities that explained a lot of this, and they're still available through the Wayback Machine.
There also aren't a lot of tools for decoding this stuff on the command line, but pdf417decode does a reasonable job. I'll use output from it to explain how I knew all the values.
The last tool you need is a way to turn jpeg output into black-and-white pbm files so that pdf417decode can read them. For that, I use the following (after installing netpbm):
cat /tmp/barcode.jpeg | jpegtopnm | ppmtopgm | pamthreshold | pamtopnm > new.pbm && ./pdf417decode -c -e new.pbm
With that, let's decode the first three rows of your existing barcode (with my commentary to the side). Everywhere you see "function output," that means this value is the output of some function that takes the other thing as the input:
0 7f54 0x02030000 (0) // Left marker
0 6a38 0x00000007 (7) // Number of rows function output
0 218c 0x00000076 (118) // Total number of non-error correcting codewords
0 0211 0x00000385 (901) // Latch to Byte Compaction mode
0 68cf 0x00000059 (89) // Data
0 18ec 0x0000021c (540)
0 02e7 0x00000330 (816)
0 753c 0x00000004 (4) // Number of columns function output
0 7e8a 0x00030001 (1) // Right marker
1 7f54 0x02030000 (0) // Left marker
1 7520 0x00010002 (2) // Security Level function output
1 704a 0x00010334 (820) // Data
1 31f2 0x000101a7 (423)
1 507b 0x000100c9 (201)
1 5e5f 0x00010319 (793)
1 6cf3 0x00010176 (374)
1 7d47 0x00010007 (7) // Number of rows function output
1 7e8a 0x00030001 (1) // Right marker
2 7f54 0x02030000 (0) // Left marker
2 6a7e 0x00020004 (4) // Number of columns function output
2 0fb2 0x0002037a (890) // Data
2 6dfa 0x000200d9 (217)
2 5b3e 0x000200bc (188)
2 3bbc 0x00020180 (384)
2 5e0b 0x00020268 (616)
2 29e0 0x00020002 (2) // Security Level function output
2 7e8a 0x00030001 (1) // Right marker
The next 3 lines will continue this pattern of function outputs. Note that the same information is encoded on the left and right, but in a different order. The system has a lot of redundancy, and can detect that it's seeing a mirror image of the barcode.
We don't care about the number of rows for this purpose, but given a current row of n and a total number of rows of N, the function is:
30 * (n/3) + ((N-1)/3)
Where / always means "integer, truncating division." Given there are 24 rows, on row 0, this is 0 + (24-1)/3 = 7.
The security level function's output is 2. Given a security level of e, the function is:
30 * (n/3) + 3*e + (N-1) % 3
=> 0 + 3*e + (23%3) = 2
=> 3*e + 2 = 2
=> 3*e = 0
=> e = 0
Finally, the number of columns can just be counted off in the output. For completeness, given a number of columns c, the function is:
30 * (n/3) + (c - 1)
=> 0 + c - 1 = 4
=> c = 5
If you look at the Data lines, you'll notice that they don't match your input data at all. That's because they have a complex encoding that I won't detail here. But for Byte compaction, you can think of it as similar to Base64 encoding, but instead of 64, it's Base900. Where Base64 encodes 3 bytes of data into 4 characters, Base900 encodes 6 bytes of data into 5 codewords.
In the end, all these codewords get converted to symbols (actual lines and spaces). Which symbol is used depends on the line. Lines divisible by 3 use one symbol set, the lines after use a second, and the lines after that use a third. So the same codewords will look completely different on line 7 than on line 8.
Taken together, all these things make it very difficult to look at a barcode and decide how "different" it is from another barcode in terms of content. You just have to decode them and see what's going on.
CIPDF417BarcodeGenerator has a few more input parameters besides inputMessage that can have an influence on how the generated barcode looks - see the documentation. Visual inspection/comparison of two codes only makes sense when you know that all these parameters, most importantly inputCorrectionLevel were equal for both generators.
So, instead of a visual comparison, simply try decoding the barcodes using one of the many scanner apps out there, and compare the decoded bytes.
For your second example, try this:
// ...
var keyData = knownKey.data(using: .isoLatin1)!
keyData.append(secretData)
let image = generatePDF417Barcode(from: keyData)
I have given string 1100011110000011111000000. I have to encode it by Basic Lempel-Ziv algorithm. I know how to do it. The problem is two first numbers are 1. I parse it into anordered dictionary of substrings. So result is:
1, 10, 0, 01, 11, 100, 00, 011, 111, 000
I want to number new substring, but then 1=No.1, 10=No.2 and 0=No.3., that means, I cant encode substring 10, because here is not value of 0.
Another problem is - at the end here are three 0 more, but I have one substring 000. What I have to do with last 000?
So, is there any different way to parse the sting in substrings?
I am using the following rest end point while trying to get permission levels for a particular file/document.
https://<web url>/_api/Web/GetFileByServerRelativeUrl('<path to the file>')/getlimitedwebpartmanager(scope=1 or 0)
I am able to get hold of the file successfully. But how should I get hold of the permission levels now?
What I do to get the permission levels is by using the "$expand" query parameter equal to "ListItemAllFields/RoleAssignments/XXXX" where XXXX is Member, RoleDefinitionBindings, and so forth expanding out the chain as far as I need to get the user and permission level info that I need. For example,
.../GetFileByServerRelativeUrl('')?$expand=ListItemAllFields/RoleAssignments/Member,ListItemAllFields/RoleAssignments/RoleDefinitionBindings,ListItemAllFields/RoleAssignments/Member/Users
It took a lot of web searching to figure out that the "$expand" query parameter even existed but it's very useful to get all the info needed in one GET. I subsequently add a "$select" parameter to the query to filter out just the pieces that my application uses.
Edit:
Also look at Radu's question and my follow-up answer below. The answer to the question depends on what you're trying to accomplish. To quote part of my answer:
If you're dredging for all permissions had by all users, use my approach. However, you'll need sufficient rights to get that information. If you want to know the base permissions of the user you're using to make the request, use his approach. I have both use cases in my work so I, in fact, do both.
The accepted solution didn't work for me. It seems that a user doesn't by default have permission to see RoleAssignments. But here's what I ended up doing.
I grabbed the effectiveBasePermissions by getting a response from
/_api/web/getFileByServerRelativeUrl('my/relative/path')/ListItemAllFields/effectiveBasePermissions
I think this works for any item/folder in sharepoint by appending the /ListItemAllFields/effectiveBasePermissions to the url)
So, this returns the permission that the current user has for that file/item. The Permissions "object" is a set of 35 flags which are encoded in 64 bits (not all 64 bits are used - only 35). And these bits are provided by the endpoint as two integers (32 bits each) named:
Low - representing the first 32 bits
High - the next 32 bits in sequence, up to 64
Now, to see if the user has, for instance, editing permissions on that file, we need to look at the corresponding bit in this Low-High sequence. You can find what each bit in the sequence means, below:
ViewListItems = 1,
AddListItems = 2,
EditListItems = 3,
DeleteListItems = 4,
ApproveItems = 5,
OpenItems = 6,
ViewVersions = 7,
DeleteVersions = 8,
CancelCheckout = 9,
ManagePersonalViews = 10,
ManageLists = 12,
ViewFormPages = 13,
AnonymousSearchAccessList = 14,
Open = 17,
ViewPages = 18,
AddAndCustomizePages = 19,
ApplyThemeAndBorder = 20,
ApplyStyleSheets = 21,
ViewUsageData = 22,
CreateSSCSite = 23,
ManageSubwebs = 24,
CreateGroups = 25,
ManagePermissions = 26,
BrowseDirectories = 27,
BrowseUserInfo = 28,
AddDelPrivateWebParts = 29,
UpdatePersonalWebParts = 30,
ManageWeb = 31,
AnonymousSearchAccessWebLists = 32,
UseClientIntegration = 37,
UseRemoteAPIs = 38,
ManageAlerts = 39,
CreateAlerts = 40,
EditMyUserInfo = 41,
EnumeratePermissions = 63
As you can see, some bits don't mean anything (like bit 11, 33, 34 etc)
So, we need to look at bit number 3 (for editing) from the first 32 bits to tell if user can edit this file/item/folder. We can just ignore the High bits. We must make a bitwise comparison with the integer which in binary format has only the 3rd bit set to 1 (the rest being 0). In this case this is 100 (this is the binary representation of 4). There is a formula for this, actually: binaryNr = 2^(bitIndex - 1). For our example, bitIndex is 3.
Now that we have the binaryNr we use it as a mask to find out if the third bit in the low sequence is 0 or 1:
hasEditingPermission = (binaryNr | LowSequence) == LowSequence
and here's a handy pseudocode function for the whole thing (^ is the power operator):
function hasPermission (low, high, bitIndex){
var sequence = low;
if (bitIndex >= 32){
sequence = high;
bit -= 32;
}
return ((2^(bitIndex - 1)) | sequence) == sequence
}
Radu made a great point in his answer for a different use case than I had given in my answer. If you're dredging for all permissions had by all users, use my approach. However, you'll need sufficient rights to get that information. If you want to know the base permissions of the user you're using to make the request, use his approach. I have both use cases in my work so I, in fact, do both.
However, as I commented to Radu in his answer, I had some trouble using his hasPermission() function. I'm adding a new answer here to provide an example of why it didn't work for me:
I'm definitely not a expert bit twiddler but, in Java at least, 1 << (bitIndex-1) is not equivalent to 2 ^ (bitIndex-1) as Radu asserted in his comment. Perhaps the overall expressions were intended to accomplish the same thing so here's an example where I discovered the XOR approach didn't work for me.
Example:
The permissions had by the user correspond to the permission list "Limited Access" (low = 134287360). I want to check if the user has the "Open" permission, bit 17. In binary, the low value (which becomes "sequence" in hasPermission()) is 1000000000010001000000000000. As you will see, bit 17 is set. Running the expression ((2^(bitIndex - 1)) | sequence) yields 1000000000010001000000010010 which obviously does not equal sequence as required to get a true response from hasPermission().
So, not understanding or knowing for sure exactly what was intended by Radu's XOR approach, I decided to use a more straightforward, less obtuse way for testing for the presence of a bit. Like so: return (sequence & (1 << (bitIndex - 1))) != 0; Taking 1 and shifting it bitIndex-1 spaces to the left and then doing a bitwise AND not only makes it obvious what I'm trying to accomplish but it also works in every case I've tested.
Not being a bit twiddler as I said, I briefly considered whether there might be problems with signed values etc. (I'm using ints for high and low) but I don't think my approach really would suffer from any of that since I'm not shifting the ints themselves and the remainder of my logic is simply bitwise.
$.ajax({
url: _spPageContextInfo.webAbsoluteUrl + "/_api/web/getFileByServerRelativeUrl('<your file or folder url>')/ListItemAllFields/effectiveBasePermissions",
type: "GET",
headers: { Accept: "application/json;odata=verbose" },
success: function(data){
var permissions = new SP.BasePermissions();
permissions.initPropertiesFromJson(data.d.EffectiveBasePermissions);
var hasPermissions = permissions.has(SP.PermissionKind.<any level to check>);
//for example: var bool_has_editListItems = permissions.has(SP.PermissionKind.editListItems);
}
});
to check more levels, see SP.PermissionKind list
I'm attempting to hash the values
10, 100, 32, 45, 58, 126, 3, 29, 200, 400, 0
I need a function that will map them to an array that has a size of 13 without causing any collisions.
I've spent several hours thinking this over and googling and can't figure this out. I haven't come close to a viable solution.
How would I go about finding a hash function of this sort? I've played with gperf, but I don't really understand it and I couldn't get the results I was looking for.
if you know the exact keys then it is trivial to produce a perfect hash function -
int hash (int n) {
switch (n) {
case 10: return 0;
case 100: return 1;
case 32: return 2;
// ...
default: return -1;
}
}
Found One
I tried a few things and found one semi-manually:
(n ^ 28) % 13
The semi-manual part was the following ruby script that I used to test candidate functions with a range of parameters:
t = [10, 100, 32, 45, 58, 126, 3, 29, 200, 400, 0]
(1..200).each do |i|
t2 = t.map { |e| (e ^ i) % 13 }
puts i if t2.uniq.length == t.length
end
On some platforms (e.g. embedded), modulo operation is expensive, so % 13 is better avoided. But AND operation of low-order bits is cheap, and equivalent to modulo of a power-of-2.
I tried writing a simple program (in Python) to search for a perfect hash of your 11 data points, using simple forms such as ((x << a) ^ (x << b)) & 0xF (where & 0xF is equivalent to % 16, giving a result in the range 0..15, for example). I was able to find the following collision-free hash which gives an index in the range 0..15 (expressed as a C macro):
#define HASH(x) ((((x) << 2) ^ ((x) >> 2)) & 0xF)
Here is the Python program I used:
data = [ 10, 100, 32, 45, 58, 126, 3, 29, 200, 400, 0 ]
def shift_right(value, shift_value):
"""Shift right that allows for negative values, which shift left
(Python shift operator doesn't allow negative shift values)"""
if shift_value == None:
return 0
if shift_value < 0:
return value << (-shift_value)
else:
return value >> shift_value
def find_hash():
def hashf(val, i, j = None, k = None):
return (shift_right(val, i) ^ shift_right(val, j) ^ shift_right(val, k)) & 0xF
for i in xrange(-7, 8):
for j in xrange(i, 8):
#for k in xrange(j, 8):
#j = None
k = None
outputs = set()
for val in data:
hash_val = hashf(val, i, j, k)
if hash_val >= 13:
pass
#break
if hash_val in outputs:
break
else:
outputs.add(hash_val)
else:
print i, j, k, outputs
if __name__ == '__main__':
find_hash()
Bob Jenkins has a program for this too: http://burtleburtle.net/bob/hash/perfect.html
Unless you're very lucky, there's no "nice" perfect hash function for a given dataset. Perfect hashing algorithms usually use a simple hashing function on the keys (using enough bits so it's collision-free) then use a table to finish it off.
Just some quasi-analytical ramblings:
In your set of numbers, eleven in all, three are odd and eight are even.
Looking at the simplest forms of hashing - %13 - will give you the following hash values:
10 - 3,
100 - 9,
32 - 6,
45 - 6,
58 - 6,
126 - 9,
3 - 3,
29 - 3,
200 - 5,
400 - 10,
0 - 0
Which, of course, is unusable due to the number of collisions. Something more elaborate is needed.
Why state the obvious?
Considering that the numbers are so few any elaborate - or rather, "less simple" - algorithm will likely be slower than either the switch statement or (which I prefer) simply searching through an unsigned short/long vector of size eleven positions and using the index of the match.
Why use a vector search?
You can fine-tune it by placing the most often occuring values towards the beginning of the vector.
I assume the purpose is to plug in the hash index into a switch with nice, sequential numbering. In that light it seems wasteful to first use a switch to find the index and then plug it into another switch. Maybe you should consider not using hashing at all and go directly to the final switch?
The switch version of hashing cannot be fine-tuned and, due to the widely differing values, will cause the compiler to generate a binary search tree which will result in a lot of comparisons and conditional/other jumps (especially costly) which take time (I've assumed you've turned to hashing for its speed) and require space.
If you want to speed up the vector search additionally and are using an x86-system you can implement a vector search based on the assembler instructions repne scasw (short)/repne scasd (long) which will be much faster. After a setup time of a few instructions you will find the first entry in one instruction and the last in eleven followed by a few instructions cleanup. This means 5-10 instructions best case and 15-20 worst. This should beat the switch-based hashing in all but maybe one or two cases.
I did a quick check and using the SHA256 hash function and then doing modular division by 13 worked when I tried it in Mathematica. For c++ this function should be in the openssl library. See this post.
If you were doing a lot of hashing and lookup though, modular division is a pretty expensive operation to do repeatedly. There is another way of mapping an n-bit hash function into a i-bit indices. See this post by Michael Mitzenmacher about how to do it with a bit shift operation in C. Hope that helps.
Try the following which maps your n values to unique indices between 0 and 12
(1369%(n+1))%13