Custom data collector with regular expression problem


I am looking to use Scrapebox to remove all mentions of domain names in a list that barely reaches 4000 URLs of web pages.

Domain names are formatted on pages like this:

Domain names are plain text. They are not hyperlinks.

If it helps, they are also always in the middle



I already have my list of almost 4000 URLs that I want to scan.

I am using 5 private proxies that have been tested and saved.
I think they are being applied when using Custom Data Grabber, but honestly, I struggle with Scrapebox.

I created inbound and outbound rules for Scrapebox in Windows Firewall.
I can do other things using Scrapebox that work. How to get internal links in the domain from which I get the URLs.

I created a custom data capture module and below that a module mask:

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I tried several examples of regular expressions and found this:




I tried it using the tool at and 3 sample URLs appear as matches (as far as I can tell):

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However, when I run my Module, all I get is this:

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The module data folder contains a csv for each time I run the Module, which contains two odd characters in the first cell:

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I ran several of the URLs through and the domain names in those URLs are readable according to that tool.

Does anyone know where I am going wrong here?
Or is there a better way to remove domain name MENTIONS from a URL list?

Thanks in advance!

Solving equations: why can we only find the asymptotic expression of the solution of the first implicit function?

Here are three equations of implicit functions

AsymptoticSolve((a*y^2 + Sin(x))^2 == x, y, {x, 1, 3})
AsymptoticSolve((a*Sin(y)^2 + Sin(x))^2 == x, y, {x, 1, 3})
AsymptoticSolve((a*y^2 + Sin(x*y))^2 == x, y, {x, 1, 3})

Why can we only find the asymptotic expression of the solution of the first implicit function?

c # – Cannot convert lambda expression to type 'Delegate' because it is not a delegate type

using (Linq) DataSet – I group and totalize it from one datatable to another, but I just want to grab a feature of an element as if adding, or anything but this error appears to me the code where I get an error is

DataRow row2 = dt.NewRow();
   row2("Nombre") = item.Key;
row2("pesoCOMPO") = item.Sum(x => Convert.ToInt32(x("pesoCOMPO")));
row2("CantidadoCOMPO") = item.Sum(x => Convert.ToInt32(x("CantidadoCOMPO")));
row2("Tipo") = item.First(x => x("Tipo").ToString());

calculation and analysis: how to simplify an expression with implicit vector function and cross product operation?

oneThe pattern is established to calculate the integral of a derivative of a function (to return the original function, the first fundamental theorem of the calculation)

integrateByIndefiniteSub = 
Integrate(arg_, {x_, xLow_, xHigh_}) :> 
Subtract @@ (Integrate(arg, x) /. {{x -> xHigh}, {x -> xLow}});

two. This is the equation. The load is a constant. The rest are all vector functions. Sorry to paste an image. I tried several times but still did not know how to paste correctly. Note: x is cross product operation
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3. The expected result should be like this. How to achieve this in a general way? I have about 100 equations like this.
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syntax: the exact dimension of an expression cannot be found

I have defined a variable called y, it contains a huge equation, but when I look for the dimension of and it shows 3. How does a single expression have dimensions of 3? It's supposed to be 1 right?

y=39.25 ω^2 (1.00519 Subscript(a,4)(1)^2+1.00102 Subscript(a,4)(2)^2+2.19056 Subscript(a,4)(1) Subscript(a,4)(3)-1.07645 Subscript(a,4)(2) Subscript(a,4)(3)+1.5 Subscript(a,4)(3)^2+0.55507 Subscript(a,4)(1) Subscript(a,4)(4)+1.54742 Subscript(a,4)(2) Subscript(a,4)(4)+1.5 Subscript(a,4)(4)^2);;

flow: how to use a variable within a Power Automate expression

I'm trying to work with him setProperty To edit a JSON body during the execution of a flow, here is the expression:

setProperty(X, 'Y','bla bla 123456789' )


  • X is the json
  • And the property that I want to modify
  • & # 39; bla bla 123456789 & # 39; is the value

My question is: how can I replace & # 39; 123456789 & # 39; for a variable (& # 39; ID & # 39;) inside it, if I do the following?

setProperty(X, 'Y','bla bla variable('ID')' )

It will take exactly what is written "variable (& # 39; ID & # 39;) and not the value of this variable!

some help ?

java – Replacing the negative look behind the regular expression

This method should return all occurrences of strings in single quotes. However, a single quote with escape ( & # 39;) should be treated as a normal single quote character (just like a double quote with escape in Java). Example: "This is not easy", the method should return the unique string "is not".
My code:

public static List findSingleQuotedTextWithEscapes(String input) {
        Pattern pattern = Pattern.compile ("(?

How would you replace the negative aspect behind this?

c ++ – What is a lambda expression in C ++ 11?

One of the best explanations of lambda expression is given by the author of C ++ Bjarne Stroustrup in his book ***The C++ Programming Language*** Chapter 11 (ISBN-13: 978-0321563842):

What is a lambda expression?

A lambda expression, sometimes also known as lambda
function or (strictly speaking incorrectly, but colloquially) as a
lambda, is a simplified notation to define and use a anonymous function object. Instead of defining a named class with an operator (), then make an object of that class and finally
invoking it, we can use a shorthand.

When would I use one?

This is particularly useful when we want to pass an operation as a
argument to an algorithm. In the context of graphical user interfaces
(and elsewhere), such operations are often referred to as callbacks.

What class of problem do they solve that wasn't possible prior to their introduction?

Here I suppose that every action performed with the lambda expression can be resolved without them, but with much more code and much greater complexity. Lambda expression: this is the way to optimize your code and a way to make it more attractive. So sad for Stroustup:

effective ways to optimize

Some examples

through the lambda expression

void print_modulo(const vector& v, ostream& os, int m) // output v(i) to os if v(i)%m==0
        (&os,m)(int x) { 
           if (x%m==0) os << x << 'n';

or through the function

class Modulo_print {
         ostream& os; // members to hold the capture list int m;
         Modulo_print(ostream& s, int mm) :os(s), m(mm) {} 
         void operator()(int x) const
             if (x%m==0) os << x << 'n'; 

or even

void print_modulo(const vector& v, ostream& os, int m) 
     // output v(i) to os if v(i)%m==0
    class Modulo_print {
        ostream& os; // members to hold the capture list
        int m; 
           Modulo_print (ostream& s, int mm) :os(s), m(mm) {}
           void operator()(int x) const
               if (x%m==0) os << x << 'n';

if you need you can name lambda expression as below:

void print_modulo(const vector& v, ostream& os, int m)
    // output v(i) to os if v(i)%m==0
      auto Modulo_print = (&os,m) (int x) { if (x%m==0) os << x << 'n'; };

Or assume another simple sample

void TestFunctions::simpleLambda() {
    bool sensitive = true;
    std::vector v = std::vector({1,33,3,4,5,6,7});

         (sensitive)(int x, int y) {
             printf("n%in",  x < y);
             return sensitive ? x < y : abs(x) < abs(y);

    for_each(v.begin(), v.end(),
             ()(int x) {
                 printf("x - %i;", x);

will generate the next

0 0


0 0


0 0


0 0


0 0


0 rated x - 1; x - 3; x - 4; x - 5; x - 6; x - 7; x - 33;

() - this is the capture list or lambda introducer: Yes lambdas Do not require access to your local environment, we can use it.

Book Quote:

The first character of a lambda expression is always (. A lambda
The presenter can take several forms:

(): An empty capture list. Is
implies that local names of the surrounding context cannot be used
in the lambda body For such lambda expressions, the data is obtained from
arguments or non-local variables.

(&): implicitly captured by
reference. All local names can be used. All local variables are
Accessed by reference.

(=): implicitly capture by value. All local
Names can be used. All names refer to copies of local variables.
taken at the call point of the lambda expression.

(capture list): explicit capture; The capture list is the list of local variable names that will be captured (that is, stored in the object) by reference or by value. Variables with names preceded by & are captured by
reference. Other variables are captured by value. A capture list can
It also contains this and names followed by ... as elements.

(&, capture list): implicitly captures by reference all local variables with names not mentioned in the list. The capture list may contain this. The names listed cannot be preceded by &. Variables named in the
The capture list is captured by value.

(=, capture list): implicitly captures by value all local variables with names not mentioned in the list. The capture list cannot contain this. The names listed must be preceded by &. Variables named in the capture list are captured by reference.

Note that a local name preceded by & is always captured by
reference and a local name not preceded by and is always captured by
value. Only the capture by reference allows the modification of variables in
The calling environment


Lambda expression Format

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Additional References:

mathematical optimization: I need help finding the minimum of an expression with multiple restrictions

I need help finding a value for "re"That the expr has the minimum value in that"re"value. Code restrictions are added

ClearAll[c, x1, y1, x2, y2, d, expr]
c = 5;
expr = -x1*y1*c + x2*y2*c + d*x1*x2;
  x2*y2 + 2*c*y2 - d*x1*x2 < 0 &&
  -x1*y1 + 2*c*y1 - d*x1*x2 < 0 && 
  x1 + x2 == 1 &&
  y1 + y2 == 1 && 
  x1 > 0 && 
  x2 > 0 && 
  y1 >= 0 &&
  y2 >= 0}, {d}]

The code above does not give me any value for "re"and I don't receive any errors. I don't know what the problem is with this code.

Do liberals believe in freedom of expression?

Saying "liberals" moves your question to a vague territory, since there is a broad spectrum of individuals that are classified as liberals.

Among today's young university liberals, the obvious answer is: No. They attack those who say things with which they disagree.