I want to make a Spinner to select the age ofthe users of my app. I wanted to know how I could create an array of all integers between 0 and 10 in the string resource file .xml in android Studio.
This is what I thought I was going to do
<integer-array name="AgeArray">
<item>1</item>
<item>2</item>
<item>3</item>
<item>4</item>
<item>5</item>
<item>6</item>
<item>7</item>
<item>8</item>
<item>9</item>
<item>10</item>
<item>12</item>
<item>13</item>
<item>14</item>
<item>15</item>
<item>16</item>
<item>17</item>
<item>18</item>
<item>19</item>
<item>20</item>
</integer-array>
But I thought that there must be a more elegant, more efficient way of creating such an array, or am I wrong?
Thanks in advance
You cannot write to a resource file (at runtime), resource files are read-only.
While developing you can create a scratch file in Android Studio, execute the following code and then copy-paste the result from the output panel to the resource file:
val list = (1..100).joinToString("\n") { " <item>$it</item>" }
val xml = "<integer-array name=\"AgeArray\">\n$list\n</integer-array>\n"
println(xml)
Related
I am new to Oracle Apex 5.1, and I have been asked to implement a button that when clicked, the user gets (downloads) a .doc file of an Interactive report.
I have noticed that the Interactive Report gives you the option to download it as .pdf, .xls, and so, but I need it to be a Word (.doc) file.
In addition, the file must be in a specific format (with heading, indentation, font, etc.) that I was given (as a template) in a Word file.
Any help would be appreciated.
Additional Information: I was able to open the template (.doc) file in NotePad++ and get the <html> version of it, so I could edit it in both NotePad++ and Word.
One of the best actually to do that is APEX OFFICE PRINT(AOP) but isn't free licence.
otherwise you can check this solution
How do we export a ms-word (or rtf) document (from a web browser) to generated by pl/sql?
I end up finding information in this page: http://davidsgale.com/apex-how-to-download-a-file/ and I wrote this code:
declare
l_clob clob;
begin
l_clob := null;
sys.htp.init;
sys.owa_util.mime_header('application/vnd.ms-word', FALSE,'utf-8');
sys.htp.p('Content-length: ' || sys.dbms_lob.getlength( l_clob ));
sys.htp.p('Content-Disposition: inline; filename="test_file.doc"' );
sys.owa_util.http_header_close;
sys.htp.p(SET_DOC_HEADER);
sys.htp.p(SET_TABLE_HEADER);
sys.htp.p(ADD_TABLE_ENTRY("arguments"));
sys.htp.p(SET_TABLE_FOOTER);
sys.wpg_docload.download_file(l_clob);
apex_application.stop_apex_engine;
exception when others then
--sys.htp.prn('error: '||sqlerrm);
apex_application.stop_apex_engine;
end;
It works, but I had to create functions in the SQL Workshop because writting a table in html is really long.
The LabelMe database can be downloaded from http://www.cs.toronto.edu/~norouzi/research/mlh/data/LabelMe_gist.mat
However, there is another link http://labelme.csail.mit.edu/Release3.0/
The webpage has a toolbox but I could not find any database to download. So, I was wondering if I could use the LabelMe_gist.mat which has the following fields. The field names contins the labels for the images, and img perhaps contains the images. How do I display the training and test images? I tried
im = imread(img)
Error using imread>parse_inputs (line 486)
The filename or url argument must be a string.
Error in imread (line 336)
[filename, fmt_s, extraArgs, msg] = parse_inputs(varargin{:});
but surely this is not the way. Please help
load LabelMe_gist.mat;
load('LabelMe_gist.mat', 'img')
Since we had no idea from your post what kind of data this is I went ahead and downloaded it. Turns out, img is a collection of 22019 images that are of size 32x32 (RGB). This is why img is a 32 x 32 x 3 x 22019 variable. Therefore, the i-th image is accessible via imshow(img(:,:,:,i));
Here is an animation of all of them (press Ctrl+C to interrupt):
for iImage = 1:size(img,4)
figure(1);clf;
imshow(img(:,:,:,iImage));
drawnow;
end
I am new to swift but I have made an android app where a string array is selected from an xml file. This is a large xml file that contains a lot of string arrays and the app gets the relevant string array based on a user selection.
I am now trying to develop the same app for iOS using swift. I would like to use the same xml file but I can not see and easy way to get the correct array. For example, part of the xml looks like this
<string-array name="OCR_Businessstudies_A_Topics">
<item>1. Business objectives and strategic decisions</item>
<item>2. External influences facing businesses</item>
<item>3. Marketing and marketing strategies</item>
<item>4. Operational strategy</item>
<item>5. Human resources</item>
<item>6. Accounting and financial considerations</item>
<item>7. The global environment of business</item>
</string-array>
<string-array name="OCR_Businessstudies_AS_Topics">
<item>1. Business objectives and strategic decisions</item>
<item>2. External influences facing businesses</item>
<item>3. Marketing and marketing strategies</item>
<item>4. Operational strategy</item>
<item>5. Human resources</item>
<item>6. Accounting and financial considerations</item>
</string-array>
If I have the string "OCR_Businessstudies_A_Topics" how do i get the "OCR_Businessstudies_A_Topics" array from the xml file.
This is very straight forward in android and although I have used online tutorials for swift it seems like I have to parse the xml file but do not seem to be getting anywhere.
Is there a better approach than trying to parse the whole xml fie?
Thanks
Barry
You can write your own XML parser, conforming to NSXMLParser or use a library like HTMLReader:
let fileURL = NSBundle.mainBundle().URLForResource("data", withExtension: "xml")!
let xmlData = NSData(contentsOfURL: fileURL)!
let topic = "OCR_Businessstudies_A_Topics"
let document = HTMLDocument(data: xmlData, contentTypeHeader: "text/xml")
for item in document.nodesMatchingSelector("string-array[name='\(topic)'] item") {
print(item.textContent)
}
I've done a few searches here, and while some issues are similar, they don't seem to be exactly what I need.
What I'm trying to do is import an Excel file into a SQL table via SSIS, but the problem is that I will never know the exact filename. We get files at no steady interval, and the file usually has a date/month in the name. For instance, our current file is "Census Data - May 2013.xls". We will only ever load ONE file at a time, so I don't need to loop through a directory for multiple Excel files.
My concept is that I can take this file, copy it to a "Loading" directory, and load it from there. At the start of the package, I will first clear out the loading directory, then scan the original directory for an Excel file, copy it to the loading directory and then load it into SQL. I suppose I may have to store the file names somewhere so I don't copy the same file into the loading directory in subsequent months, but I'm not really sure of the best way to handle that.
I've pretty much got everything down except the part that scans the directory for the Excel file and copies it to the loading directory. I've taken the majority of my info from this page, which (again) is close to what I want to do but not quite exactly the solution I need.
Can anyone get me over the finish line? I can't seem to get the Excel Connection Manager right (this is my first time using variables), and I can't figure out how to get the file into the Loading directory.
Problem statement
How do I dynamically identify a file name?
You will require some mechanism to inspect the contents of a folder and see what exists. Specifically, you are looking for an Excel file in your "Loading" directory. You know the file extension and that is it.
Resolution A
Use a ForEach File Enumerator.
Configure the Enumerator with an Expression on FileSpec of *.xls or *.xlsx depending on which flavor of Excel you're dealing with.
Add another Expression on Directory to be your Loading directory.
I typically create SSIS Variables named FolderInput and FileMask and assign those in the Enumerator.
Now when you run your package, the Enumerator is going to look in Diretory and find all the files that match the FileSpec.
Something needs to be done with what is found. You need to use that file name that the Enumerator returns. That's done through the Variable Mappings tab. I created a third Variable called CurrentFileName and assign it the results of the enumerator.
If you put a Script Task inside the ForEach Enumerator, you should be able to see that the value in the "Locals" window for #[User::CurrentFileName] has updated from the Design time value of whatever to the "real" file name.
Resolution B
Use a Script Task.
You will still need to create a Variable to hold the current file name and it probably won't hurt to also have the FolderInput and FileMask Variables available. Set the former as ReadWrite and the latter as ReadOnly variables.
Chose the .NET language of your choice. I'm using C#. The method System.IO.Directory.EnumerateFiles
using System;
using System.Data;
using System.IO;
using Microsoft.SqlServer.Dts.Runtime;
using System.Windows.Forms;
namespace ST_fe2ea536a97842b1a760b271f190721e
{
[Microsoft.SqlServer.Dts.Tasks.ScriptTask.SSISScriptTaskEntryPointAttribute]
public partial class ScriptMain : Microsoft.SqlServer.Dts.Tasks.ScriptTask.VSTARTScriptObjectModelBase
{
public void Main()
{
string folderInput = Dts.Variables["User::FolderInput"].Value.ToString();
string fileMask = Dts.Variables["User::FileMask"].Value.ToString();
try
{
var files = Directory.EnumerateFiles(folderInput, fileMask, SearchOption.AllDirectories);
foreach (string currentFile in files)
{
Dts.Variables["User::CurrentFileName"].Value = currentFile;
break;
}
}
catch (Exception e)
{
Dts.Events.FireError(0, "Script overkill", e.ToString(), string.Empty, 0);
}
Dts.TaskResult = (int)ScriptResults.Success;
}
enum ScriptResults
{
Success = Microsoft.SqlServer.Dts.Runtime.DTSExecResult.Success,
Failure = Microsoft.SqlServer.Dts.Runtime.DTSExecResult.Failure
};
}
}
Decision tree
Given the two resolutions to the above problem, how do you chose? Normally, people say "It Depends" but there only possible time it would depend is if the process should stop/error out in the case that more than one file did exist in the Loading folder. That's a case that the ForEach enumerator would be more cumbersome than a script task. Otherwise, as I stated in my original response that adds cost to your project for Development, Testing and Maintenance for no appreciable gain.
Bits and bobs
Further addressing nuances in the question: Configuring Excel - you'll need to be more specific in what isn't working. Both Siva's SO answer and the linked blogspot article show how to use the value of the Variable I call CurrentFileName to ensure the Excel File is pointing to the "right" file.
You will need to set the DelayValidation to True for both the Connection Manager and the Data Flow as the design-time value for the Variable will not be valid when the package begins execution. See this answer for a longer explanation but again, Siva called that out in their SO answer.
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.