Question

如何使用能够在调用函数时定义的节来组合函数？

<?php
class MyClass {
        function doSomething() {
        $conn = $this->connectToDatabase("database");
        $rows = $conn->query($query);

        if ($rows->num_rows > 0) {  
            while($row = $rows->fetch_assoc()) {
                $col_value = $row['name'];
                function defineMe() {
                   //i need to be defined when called
                }
}
        } else {
            echo "0 results";
        }   

        $conn->close();
    }
}
?>

这样，说我想从数据库中提取数据，我可以更改defineMe（）以匹配其吐出的信息的布局。如何覆盖此函数并仍然可以从doSomething（）函数访问数据？

像

$myClass = new MyClass;
$myClass->doSomething(){
    function defineMe() {
       echo $col_value;
    }
};

Answer 1

您可以将回调传递给您的函数：

<?php
class MyClass
{
    function doSomething(callable $callback)
    {
        $conn = $this->connectToDatabase("database");
        $rows = $conn->query($query);

        if ($rows->num_rows > 0)
        {  
            while($row = $rows->fetch_assoc())
            {
                $col_value = $row['name'];
                $callback($col_value);      // Call callback
            }
        } else {
            echo "0 results";
        }   

        $conn->close();
    }
}

$myClass = new MyClass;
$myClass->doSomething(function($col_value) {
    echo "Inside callback: ".$col_value;
});
?>

php函数，在调用时具有可定义的函数

1 个答案: