打开/关闭花括号而不是打开/关闭双引号在 PHP 中解释字符串的区别

Question

我阅读了 PHP 手册，但并不清楚其中有什么区别。我很困惑使用这个有什么意义：

echo "Print this {$test} here.";

与此相比：

echo "Print this $test here.";

Answer 1

在您的示例中，没有区别。但是，当变量表达式更复杂时，它会很有帮助，比如带有字符串索引的数组。例如：

$arr['string'] = 'thing';

echo "Print a {$arr['string']}";
// result: "Print a thing";

echo "Print a $arr['string']";
// result: Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE

Answer 2

来自PHP.net

复杂（卷曲）语法

这不是因为语法复杂，而是因为 它允许使用复杂的表达式。

任何标量变量、数组元素或带有字符串的对象属性 可以通过此语法包含表示。简单地写 表达方式与它出现在字符串之外的方式相同，并且 然后将其包装在 { 和 } 中。由于 { 无法转义，因此此语法将 仅当 $ 紧跟 { 时才被识别。使用 {\$ 来 得到一个字面量 {$.

例子：

<?php
// Show all errors
error_reporting(E_ALL);

$great = 'fantastic';

// Won't work, outputs: This is { fantastic}
echo "This is { $great}";

// Works, outputs: This is fantastic
echo "This is {$great}";
echo "This is ${great}";

// Works
echo "This square is {$square->width}00 centimeters broad."; 


// Works, quoted keys only work using the curly brace syntax
echo "This works: {$arr['key']}";


// Works
echo "This works: {$arr[4][3]}";

// This is wrong for the same reason as $foo[bar] is wrong  outside a string.
// In other words, it will still work, but only because PHP first looks for a
// constant named foo; an error of level E_NOTICE (undefined constant) will be
// thrown.
echo "This is wrong: {$arr[foo][3]}"; 

// Works. When using multi-dimensional arrays, always use braces around arrays
// when inside of strings
echo "This works: {$arr['foo'][3]}";

// Works.
echo "This works: " . $arr['foo'][3];

echo "This works too: {$obj->values[3]->name}";

echo "This is the value of the var named $name: {${$name}}";

echo "This is the value of the var named by the return value of getName(): {${getName()}}";

echo "This is the value of the var named by the return value of \$object->getName(): {${$object->getName()}}";

// Won't work, outputs: This is the return value of getName(): {getName()}
echo "This is the return value of getName(): {getName()}";
?>

有关更多示例，请参见该页面。