Question

我想创建一个我不知道名字的类。在PHP中完成以下场景的最佳方法是什么？

$class_name = 'SomeClassName';
$code = "class {$class_name}_Control extends WP_Customize_Control {}";
eval( $code );

感谢。

Answer 1

我刚试过这个，它运行正常。如果你只需要暂时上课，你总是可以在底部扔unlink()。不需要exec()。

<?php

// your dynamic classname
$fart = "poot";

// define your class, don't forget the "<?php"
$class = <<<YOYOYOHOMIE
<?php
class $fart{
    public \$poop = "poop";
}
YOYOYOHOMIE;

// write the class to a file.
$filename = "dynamicClass".time().".php";
$fh = fopen($filename,"w+");
fwrite($fh, $class);
fclose($fh);

// require the file
require $filename;

// now your dynamically generated class is available
$tird = new $fart;
echo "<pre>";
var_dump($tird);

使用变量动态定义PHP类

1 个答案: