我是PHP的新手,请原谅这个问题。
我想知道PHP是否具有字符串格式函数(例如Python的f-strings函数)而不是str.format()。我已经看到了有关该主题的几篇文章,但是作为答案而被接受的大多数示例都涉及Python处理格式化字符串str.format()
的较旧方法。就我而言,我想使用一个格式化的字符串来构建变量,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, $location);
但是如果我想使用$format
作为变量怎么办?主要思想是基于其他变量创建动态变量,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, $db_password);
注意:我知道根据PHP文档,有几种方法可以进行字符串连接,但是我真的找不到类似的东西。
答案 0 :(得分:3)
您可以使用点符号将string concatenation的任何变量附加到其他变量:
.=
或$a = "Hello ";
$b = "World";
$a .= $b; // $a now contains "Hello World"
表示法:
$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
您还可以使用双引号中包含的单个字符串,该字符串将自动评估变量。请注意,单引号不会不计算变量:
$num = 5;
$location = 'tree';
$output = "There are $num monkeys in the $location"; // There are 5 monkeys in the tree
与分配变量时的作用相同:
$output = "There are {$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
这可以用大括号进一步阐明:
{{1}}