PHP解析器:大括号围绕变量 [英] PHP parser: braces around variables

问题描述

我想知道大括号的语义是如何精确定义的 在PHP里面?例如，假设我们已定义:

I was wondering, how is the semantics of braces exactly defined inside PHP? For instance, suppose we have defined:

$a = "foo";

然后有什么区别:

echo "${a}";

echo "{$a}";

也就是说，在任何情况下 大括号外的美元符号与大括号内的美元符号相反 差异或结果始终相同(使用花括号) 几乎可以分组)?

that is, are there any circumstances where the placement of the dollar sign outside the braces as opposed to within braces makes a difference or is the result always the same (with braces used to group just about anything)?

推荐答案

花括号的可能性很多(例如省略它们)，并且在处理对象或数组时事情变得更加复杂.

There are a lot of possibilities for braces (such as omitting them), and things get even more complicated when dealing with objects or arrays.

相对于串联，我更喜欢插值，并且在不需要时，我更喜欢省略花括号.有时候是.

I prefer interpolation to concatenation, and I prefer to omit braces when not necessary. Sometimes, they are.

您不能使用${}语法使用对象运算符.调用方法或链接运算符时，必须使用{$...}(如果只有一个运算符来获取成员，则可以省略花括号).

You cannot use object operators with ${} syntax. You must use {$...} when calling methods, or chaining operators (if you have only one operator such as to get a member, the braces may be omitted).

${}语法可用于变量变量:

The ${} syntax can be used for variable variables:

$y = 'x';
$x = 'hello';
echo "${$y}"; //hello

$$语法不会在字符串中插值，因此需要${}进行插值.您还可以使用字符串(${'y'})，甚至可以在${}块内连接.但是，可变变量可能被认为是一件坏事.

The $$ syntax does not interpolate in a string, making ${} necessary for interpolation. You can also use strings (${'y'}) and even concatenate within a ${} block. However, variable variables can probably be considered a bad thing.

对于数组，任何一个都将在${foo['bar']}与{$foo['bar']}之间工作.我更喜欢$foo[bar](仅用于插值-在字符串bar之外的字符串在该情况下将被视为常量).

For arrays, either will work ${foo['bar']} vs. {$foo['bar']}. I prefer just $foo[bar] (for interpolation only -- outside of a string bar will be treated as a constant in that context).

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