>假设我有一个名为"translations"的类。我有一个名为"chinese_translations"的类实例
$chinese_translations = new translations;
之后,我想从动态构建的字符串调用此实例,并且尝试了几种方法:
$part1 = 'chinese_';
$part2 = 'translations';
$instance_of_translations = ${$part1.$part2};
$instance_of_translations->getMsg();//Doesn't work
也像这样:
$part1 = 'chinese_';
$part2 = 'translations';
$instance_of_translations = $part1.$part2;
$$instance_of_translations->getMsg();//Doesn't work
我总是收到"在非对象上调用成员函数 getMsg()"消息。我做错了什么?
*采取的一些行动和取得的成果:
//Lets see if the var is in scope:
echo $chinese_translations->getMsg(get_locale());//It works
$instance_of_translations = ${$part1.$part2};//Let's try to build the name dynamically
echo $chinese_translations->getMsg(get_locale()); //Call to a member function getMsg() on a non-object
echo $$chinese_translations->getMsg(get_locale()); //Object of class internal_message could not be converted to string in
var_dump($instance_of_translations);//It throws the following:
//object(internal_message)#1956 (1) { ["message"]=> array(2) { ["es_ES"]=> string(19) "The expected result" ["it_IT"]=> string(19) "The expected result" } }NULL
$part1.$part2
是一个字符串。因此,$instance_of_translations
是变量名称,而不是变量本身。
试试这个:
$part1 = 'chinese_';
$part2 = 'translations';
$varName = $part1.$part2;
$instance_of_translations = $$varName;
var_dump($instance_of_translations);
它应该显示类型 translations
的对象。
阅读有关变量的更多信息。
试试这个:
$className = $part1.$part2;
$instance = new $$className;
$instance->getMsg();
最后,括号版本起到了作用,但存在范围问题,因此两个答案都可能是正确的。这个:
$part1 = 'chinese_';
$part2 = 'translations';
$instance_of_translations = ${$part1.$part2};
$instance_of_translations->getMsg();
目前正在为我工作。