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change() is not a function


change() is not a function

By : user2185078
Date : November 20 2020, 04:01 AM
it helps some times You have to remove html, because you need to attach a change event handler to the input element.
Also, you need to change your selector to
code :
$('.person-range-value input')
$(function (ready) {
    var txtInput = $('.person-range-element input');
    txtInput.change(function () {
        console.log($(this).val());
    });
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<div class="person-range-element">
        4
       <input class="person-range slider-range" id="nombre" type="range" min="4" max="12" value="4"/> 
       12
</div>


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How to change the value of a dropdown box from within it's own change function so that the change function is called aga

How to change the value of a dropdown box from within it's own change function so that the change function is called aga


By : omveer singh
Date : March 29 2020, 07:55 AM
I wish this helpful for you I run some code when a value is selected from a drop-down box: , This should work methinks:
code :
$('#temp_id').change( function() {
    if ($('#temp_id').val() == "temp_val") {
        $('#temp_id').val("temp_val2");    
        $('#temp_id').change(); // or $('#temp_id').trigger('change');
    }
}):
Whether to go for a member function or friend function when the function is supposed to change state of object?

Whether to go for a member function or friend function when the function is supposed to change state of object?


By : Javier López G.
Date : March 29 2020, 07:55 AM
around this issue To be honest, I think the only reasons to make such a decision are syntactic convenience and tradition. I'll explain why by showing what are (not) the differences between the two and how these differences matter when making a decision.
What differences are there between non-member friend functions and public member functions? Not much. After all, a member function is just a regular function with a hidden this parameter and access to the class's private members.
code :
// what is the difference between the two inv functions?
// --- our code ---
struct matrix1x1 { // this one is simple :P
private:
    double x;
public:
    //... blah blah
    void inv() { x = 1/x; }
    friend void inv(matrix1x1& self) { self.x = 1/self.x; }
};
matrix1x1 a;

// --- client code ---

// pretty much just this:
a.inv();
// vs this:
inv(a);

void lets_try_to_break_encapsulation(matrix1x1& thingy) {
    thingy.x = 42; // oops, error. Nope, still encapsulated.
}
Do function arguments of an outer function change for an asynchronous inner function?

Do function arguments of an outer function change for an asynchronous inner function?


By : Roland Moroni Lafort
Date : March 29 2020, 07:55 AM
Hope that helps I am confused in this topic which is important to me in terms of asynchronous function calls and closures. I must have missed the most important part of javascript function invocation and wasn't able to find an answer to my question up to now, hence I hope for your help! , Function arguments are locally scoped variables.
code :
function foo (x) {

}
foo(1);
function foo () {
    var x = 1;
}
foo();
function foo () {
    x = 1;
}
foo();
Is it safe to change a function pointer (std::function) inside a called function?

Is it safe to change a function pointer (std::function) inside a called function?


By : pedro barbosa dos sa
Date : March 29 2020, 07:55 AM
should help you out This is legal with function pointers.
When you assign or construct a std::function with a target, it creates a copy of the target. In the case of assigning a function to the std::function, this in effect stores a function pointer as the target object.
code :
std::function<void()> fun;
struct bob {
  std::string name;
  bob* next = 0;
  void operator()() const {
    std::cout << name << "\n";
    if (next) fun = *next;
    // undefined behavior:
    // std::cout << name << "\n";
  }
};
bob foo = {"foo"};
bob bar = {"bar", &foo};

int main() {
  fun = bar;
  fun();
  fun();
}
How can I fix the repetitive root.mainloop() calls in the change() function? I want to change an image with a change in

How can I fix the repetitive root.mainloop() calls in the change() function? I want to change an image with a change in


By : merkaba
Date : March 29 2020, 07:55 AM
Does that help You do not need to call mainloop in your function. It should be called exactly once for the life of your program.
The fact that the image doesn't show when you remove the call to mainloop is because you aren't storing a reference to the image so it gets deleted by the garbage collector when the function goes out of scope. By running mainloop in the function you prevent that from happening, but cause even worse problems (namely, the one you're writing about)
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