LibroHoyosLouisSacharpdf [Extra Quality]

by unybeth
Published: September 12, 2022 (2 weeks ago)
Category

LibroHoyosLouisSacharpdf [Extra Quality]



 
 
 
 
 
 
 

LibroHoyosLouisSacharpdf

A:

Here’s a small solution:
I don’t know if it can help but:
$documento = $_GET[‘documento’];
$archivo = $_GET[‘archivo’];
$ruta = “”. $_GET[‘documento’]. “&archivo=”. $_GET[‘archivo’] . “&autor=”. $documento. “&tipo=”. $documento;

if (file_exists(‘../uploads/doc_master/’.$documento.’.docx’)) {
$archivo = ‘../uploads/doc_master/’.$documento.’.docx’;
}
header(‘Location: ‘. $ruta);
exit;

And if $documento is “Günter Grass”, $archivo is “Wallenstein” (The Memoirs of General Field Marshal Günter Grass).

Q:

Excel VBA vs Access VBA

I have a current project in Access that is written in Access and has many subroutines built into it.
This application also has some VBA code written in Excel that calls a couple of these subroutines.
Recently they have added a large chunk of VBA code written in Excel that pulls data from a database and places it into another database.
I am not the original developer of this application, but I wanted to know if this code will work when I move it to Access.
Can someone tell me what the differences are between Access VBA and Excel VBA?

A:

VBA uses a very basic common specification that is the same in VBA for Access and VBA for Excel.
The primary difference between VBA for Access and VBA for Excel is that Microsoft Access VBA uses an internal code form, which is different than Excel VBA.
The common specification and the specifications of the Access VBA code are not compatible with any of the specifications of the Excel VBA code. (It’s not because the specifications aren’t the same, it’s just that they are different.) For example, the Access VBA code can make use of the following standard declarations, though that doesn’t mean that code for Excel can. The Access VBA and Excel VBA code is compatible with the same data types, and

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Scala case class constructor calling

I have the following code :
case class A(param: String, q: Option[String])
case class B(param: String) extends A(param, null)

I am getting the following exception :

java.lang.IllegalStateException: The constructor B in class A cannot
call itself

I know that constructor that I am calling is private and I am not allowed to call it in another constructor. But is there any workaround to create B?
My requirement is I am working on an existing program, and I cannot modify it to call some other constructor. It has been compiled in Scala 2.9.x.
Any workaround here?

A:

I think the problem is actually in the fact that class B extends the protected class A, and constructor B is calling the same constructor A, as constructor A(param: String, q: Option[String]).
What you can do, without touching the sourcecode of the program, is a

Extension of A, A subclass or mixin, A implementation from library A, giving you your own A instance. (but this is, as you say, not an option because this is not in the source)
Writing a method, that creates an instance of A from it’s parameters, called in the constructor. (looks like a hack, but if you don’t want to touch the source)

Otherwise, you could try to find a reference to the constructor A and call it from constructor B. This will, of course, only work if you can guarantee, that this constructor is the one used (you use the same compiler version and such…)
Regarding the problem of extending A from library A: I know, that you can’t extend (from libraries) protected classes. But regarding this, my answer would be:

Use the same library. If it can solve your problem, I am not sure (it shouldn’t, but I’m not sure).
Modify it
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