php-如何在函数中传递一些参数[复制]

I have a function with a known number of parameters like

    function myFunction($name, $id=1,$salary=1000) {
        echo $id;
        echo $name;
        echo $salary
    }

when i call this function

myFunction("name");

it displays

• 1 name 1000

But How I can ignore the id parameter of this function and want to give 1st and 3rd arguments only? like

myfunction("name",ignore 2nd parameter and give value of 3rd parameter)

so that it should display

• name 1 given salary

</div>

You can't with native code. Two solutions :

• Take an array of parameters like that :

  function myFunction($params) {
    if (!isset($params['name'])) throw new Exception();
    if (!isset($params['id'])) $params['id'] = 1;
    if (!isset($params['salary'])) $params['salary'] = 1000;
    echo $params['id'];
    echo $params['name'];
    echo $params['salary'];
  }
  

• Create a new function myFunction2 :

  function myFunction2($name, $salary) {
    return myFunction($name, 1, $salary);
  }
  

Variable argument function?

function myFunction() {
    switch (func_num_args()) {
    case 2:
        $name = func_get_arg(0);
        $salary = func_get_arg(1);
        break;
    case 3:
        $name = func_get_arg(0);
        $id = func_get_arg(1);
        $salary = func_get_arg(2);
        break;
    default:
        throw new Exception();
    }
}
myFunction('name', 1000);
myFunction('name', 1, 2000);

If you wanted to set defaults for these, then you would do that inside the switch statement.

I'm not advocating this as a good solution, but it is a solution.

A second function will do the job.

function second($name,$another){
  first( $name ,1,$another);
}