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我是 PHP 新手，所以请原谅这个问题。

我想知道 PHP 是否有字符串格式函数，例如 Python 的 f-strings 函数，而不是 str.format()。我看过一些关于这个主题的帖子，但大多数被接受为答案的例子都是指 Python 处理格式化字符串的旧方式str.format()。就我而言，我想使用格式化的字符串构建一个变量，例如（Python）：

f_name = "John"
l_name = "Smith"
sample = f`{f_name}'s last name is {l_name}.`
print(sample)

我知道我可以使用（PHP）：

 $num = 5;
 $location = 'tree';
 $format = 'There are %d monkeys in the %s';
 echo sprintf($format, $num, $location);

但是如果我想$format用作变量呢？主要思想是基于其他变量创建一个动态变量，例如：

$db_type = $settings['db_type'];  # mysql
$db_host = $settings['db_host'];  # localhost
$db_name = $settings['db_name'];  # sample

var $format = "%s:host=%s; dbname=%s";

# Not sure what to do after that, but I can use string concatenation:

var $format = $db_type + ":host=" + $db_host + "; dbname=" + $db_name;
var $connection = new PDO($format, $db_user, $db_password);

注意：我知道根据 PHP 文档有几种方法可以进行字符串连接，但是我真的找不到这样的东西。

4

1 回答 1

8

您可以使用点符号将任何变量附加到具有字符串连接的任何其他变量：

$num = 5;
$location = 'tree';
$output = 'There are ' . $num . ' monkeys in the ' . $location; // There are 5 monkeys in the tree

或.=符号：

$a = "Hello ";
$b = "World";
$a .= $b; // $a now contains "Hello World"

您还可以使用包含在double-quotes中的单个字符串，它会自动评估变量。请注意，单引号不评估变量：

$num = 5;
$location = 'tree';
echo 'There are $num monkeys in the $location'; // There are $num monkeys in the $location
echo "There are $num monkeys in the $location"; // There are 5 monkeys in the tree

这在分配给变量时是一样的：

$num = 5;
$location = 'tree';
$output = "There are $num monkeys in the $location"; // There are 5 monkeys in the tree

这可以用大括号进一步澄清：

$output = "There are {$num} monkeys in the {$location}"; // There are 5 monkeys in the tree
// OR
$output = "There are ${num} monkeys in the ${location}"; // There are 5 monkeys in the tree
于 2019-02-25T22:26:35.527 回答