PHP Programming for Beginners. Key Programming Concepts. How to use PHP with MySQL and Oracle databases (MySqli, PDO). Sergey D Skudaev

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convertFtoC ($F) {

      $C = round (($F – 32) * 5/9);

      return $C;

      }

      Now to convert Fahrenheit to Celsius you must call your function and pass the temperature in Fahrenheit.

      $T= convertFtoC (80);

      echo $T.”<br>”;

      The Output: will be 27

      The round is a built-in PHP function. It will round 26.6666666667 to 27.

      Variable Scope

      Variables in the PHP code may have different scopes. They may be seen or accessed globally, anywhere on the current php page or only inside the function. Let us analyze the following code:

      <?php

      $greeting=“Hello!”;

      function greetme ($name) {

      echo $greeting.””. $name;

      }

      //function call

      greetme (“John’);

      ?>

      Output: John

      At the beginning of the code, we assigned the “Hello” string to the $greeting variable. Then we defined the function with one parameter $name. Inside the function we concatenated the $name variable to the $greeting variable. Then we called the function and passed the name “John’ as a parameter.

      You might expect the Output: would be “Hello John”, since $greeting equals “Hello”, but it’s “John”. “Hello” is not visible because the $greeting variable inside the function is not the same as outside. It’s local, and because we assigned no value to the local variable $greeting, the value we passed as a parameter is the only one visible.

      Let us modify the function.

      <?php

      $greeting=“Good morning!”;

      function greetme ($name) {

      $greeting=“Hi There”;

      echo $greeting.””. $name;

      }

      echo greetme (“John’);

      echo "<br>”;

      echo “Greeting=”. $greeting.”<br>”;

      ?>

      Output:

      Hi There John

      Greeting=Good Morning

      Inside the function we assigned the value “Hi There” to the local variable $greeting. As a result, the function will return “Hi There John”. However, because we printed the $greeting variable outside the function, the global variable $greeting value remains “Good Morning”.

      Let us modify our code again.

      <?php

      $greeting=“Good Morning”;

      function greetme ($name) {

      global $greeting=“Hi There”;

      echo $greeting.””. $name;

      }

      echo greetme (“John’);

      echo "<br>”;

      echo “Greeting=”. $greeting.”<br>”;

      ?>

      Output:

      Hi There John

      Greeting=Hi There

      In the above we’ve declared a global variable inside the function, using global. Then we assigned the value “Hi There” to the global $greeting variable.

      We previously assigned “Good Morning” to the global variable $greeting outside the function. Then, inside the function, we assigned a global variable with the same name to the value “Hi There”. Now the global variable $greeting no longer has the value “Good Morning”, but rather “Hi There”. So, when we print the value of the global variable outside the function it displays, “Hi There”.

      Static variable

      A static variable is the same as a local variable, but it is not destroyed when the function execution is ended.

      <?php

      function greetme ($name) {

      static $greeting=“Hi There”;

      $greeting=$greeting.””. $name;

      echo $greeting.”! <br>”;

      }

      greetme (“John’);

      greetme (“John’);

      greetme (“John’);

      ?>

      Output:

      Hi There John!

      Hi There John John!

      Hi There John John John!

      The value of the static $greeting variable is changing after each function call because the $name variable is added to it each time.

      Remove the static keyword and the Output: of the function will be the same in the second and the third call

      <?php

      function greetme ($name) {

      $greeting=“Hi There”;

      $greeting=$greeting.””. $name;

      echo $greeting.”! <br>”;

      }

      greetme (“John’);

      greetme (“John’);

      greetme (“John’);

      ?>

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