To check if your website is blocked by Facebook, you can use the Facebook Sharing Debugger tool.

All you need to do is enter your site’s URL and click the Debug button. If your website is blocked by Facebook, you should see the following message (or something similar):

We can’t review this website because the content doesn’t meet our Community Standards. If you think this is a mistake, please let us know.

How to fix a website blocked by Facebook

If your website is indeed blocked by Facebook, here’s how to fix the problem so that you and others can share your content again.

1. Make sure you’re not violating any Community Standards

Before doing anything else, you need to make sure that your website doesn’t violate Facebook’s Community Standards.

If your website is legitimately violating any standards, you’re not going to be able to unblock it.

Read through the Facebook Community Standards in detail and think about how/if any of them could apply to your site.

2. Submit your website to Facebook for reconsideration

Once you’re 100% sure that your website doesn’t violate Facebook’s Community Standards, you can submit it for reconsideration.

To do this, go back to the Facebook Sharing Debugger tool and enter your website again. You should see the same message telling you that your website is blocked.

To submit your website for reconsideration, click the let us know link:

This will take you to a form where you can submit a short message explaining why your site should be unblocked:

• If your site never violated the Community Standards, you can explain that here.
• If your site did violate the Community Standards but you’ve since removed the offending content, you can also let them know about that.

Facebook doesn’t offer a specific timeline for when they’ll respond. However, you should expect it to take anywhere from a few weeks to a month (and you might not get a response at all – you can try resubmitting again if that’s the case).

That’s great! Since you’re looking for something that can help you develop an Android app with drag and drop facility, I think I have the solution you’re looking for. That’s right! I think It’ll be great if you try OnGoBuyo. It’s a great Android app maker. Just drag and drop and you can build an app for not only Android but for iOS platform also.

It offers some really exciting features. For example:

• Custom color selection
• Supports multiple languages and currencies
• Right-to-left language
• Homepage layout customization
• Maintenance and live mode
• Push notifications for orders
• Real-time cart management

It actually offers many more features than I have mentioned here. I think, maybe you should try OnGoBuyo and see how it works for you. I/m sure you will be able to get your app developed with it with great ease.

ব্লু লাইট ফিল্টার করে এইরকম চশমা ব্যবহার করতে পারেন। আমি যেটা ব্যবহার করি সেটা এই লিঙ্কে দেখতে পারেন।

Amazon.com: PROSPEK Computer Glasses, Blue Light Blocking Glasses – Professional (+0.00 (No Magnification) for Men and Women Regular Size, Red and Black): Office Products

When you declare a reference variable (i.e. an object) you are really creating a pointer to an object. Consider the following code where you declare a variable of primitive type int:

int x;
x = 10;

In this example, the variable x is an int and Java will initialize it to 0 for you. When you assign it the value of 10 on the second line, your value of 10 is written into the memory location referred to by x.

But, when you try to declare a reference type, something different happens. Take the following code:

Integer num;
num = new Integer(10);

The first line declares a variable named num, but it does not actually contain a primitive value yet. Instead, it contains a pointer (because the type is Integer which is a reference type). Since you have not yet said what to point to, Java sets it to null, which means “I am pointing to nothing“.

In the second line, the new keyword is used to instantiate (or create) an object of type Integer and the pointer variable num is assigned to that Integer object.

The NullPointerException occurs when you declare a variable but did not create an object and assign to the variable before trying to use the contents of the variable (called dereferencing). So you are pointing to something that does not actually exist.

Dereferencing usually happens when using . to access a method or field, or using [ to index an array.

If you attempt to dereference num BEFORE creating the object you get a NullPointerException. In the most trivial cases, the compiler will catch the problem and let you know that “num may not have been initialized,” but sometimes you may write code that does not directly create the object.

For instance, you may have a method as follows:

public void doSomething(SomeObject obj) {
   //do something to obj
}

In which case, you are not creating the object obj, but rather assuming that it was created before the doSomething() method was called. Note, it is possible to call the method like this:

doSomething(null);

In which case, obj is null. If the method is intended to do something to the passed-in object, it is appropriate to throw the NullPointerException because it’s a programmer error and the programmer will need that information for debugging purposes. Please include the name of the object variable in the exception message, like

Objects.requireNonNull(a, "a");

Alternatively, there may be cases where the purpose of the method is not solely to operate on the passed in object, and therefore a null parameter may be acceptable. In this case, you would need to check for a null parameter and behave differently. You should also explain this in the documentation. For example, doSomething() could be written as:

/**
  * @param obj An optional foo for ____. May be null, in which case 
  *  the result will be ____.
  */
public void doSomething(SomeObject obj) {
    if(obj == null) {
       //do something
    } else {
       //do something else
    }
}