This can be difficult in PHP because of the way data types are handled internally. Assuming that you don't mean complex types such as objects or resources, generic casting to strings may still result in incorrect conversion. In some cases pack/unpack may even be required, and then you still have the possibility of problems with string encoding. I know this might sound like a stretch but these are the type of cases where standard type juggling such as $myText = $my_var .''; and $myText = (string)$my_var; (and similar) may not work. Otherwise I would suggest a generic cast, or using serialize() or json_encode(), but again it depends on what you plan on doing with the string.

主要的区别在于，Java和. net在处理二进制数据和基本类型以及从特定类型转换到/从特定类型转换到字符串方面有更好的功能，即使对用户抽象了特定的情况也是如此。PHP的情况则完全不同，即使是处理十六进制也会让你摸不着头脑，直到你掌握了它。

我想不出更好的方法来回答这个问题，这是可比Java/。NET中的_toString()和此类方法通常以特定于对象或数据类型的方式实现。在这种情况下，神奇的方法__toString()和__serialize()/__unserialize()可能是最好的比较。

还要记住，PHP没有基本数据类型的相同概念。从本质上讲，PHP中的每一种数据类型都可以被认为是一个对象，它们的内部处理程序试图使它们具有某种通用性，即使这意味着在将float转换为int时失去准确性。你不能像在Java中那样处理类型，除非你在本地扩展中使用它们的zvals。

While PHP userspace doesn't define int, char, bool, or float as an objects, everything is stored in a zval structure which is as close to an object that you can find in C, with generic functions for handling the data within the zval. Every possible way to access data within PHP goes down to the zval structure and the way the zend vm allows you to handles them without converting them to native types and structures. With Java types you have finer grained access to their data and more ways to to manipulate them, but also greater complexity, hence the strong type vs weak type argument.

这些链接可能会有帮助:

https://www.php.net/manual/en/language.types.type-juggling.php https://www.php.net/manual/en/language.oop5.magic.php

您还可以使用var_export PHP函数。

您总是可以创建一个名为.ToString($in)的方法来返回

$in . '';  

__toString方法或(string)强制转换

$string=(string)$variable;  //force make string 

可以将对象视为字符串

class Foo
{

  public function __toString()
  {
     return "foo";
  }

}

echo new Foo(); //foo

还有另一个技巧，假设你有int变量，你想把它变成字符串

$string=''.$intvariable;

你可以使用强制转换操作符:

$myText = (string)$myVar;

PHP手册的字符串部分有关于字符串强制转换的更多细节，包括对布尔值和空值的特殊处理。

你可以使用类型转换:

$var = (string)$varname;

或StringValue:

$var = strval($varname);

或SetType:

$success = settype($varname, 'string');
// $varname itself becomes a string

它们在类型杂耍方面的作用都是一样的。