php 如何替换字符串中的占位符？
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how to replace a placeholder in a string?
提问by Alica Kindel
Not very sure how to word this question but I'll give an example.
不太确定如何表达这个问题，但我会举一个例子。
$string = 'Hey, $name. How are you?';
When I post this string, $namedoesn't change. How can I make it so I can put something like +name+so it changes to the name. I've tried seaching for it but I don't know what to search for so I don't have any luck. It's probably really simple but i'm just blanking out.
Thanks
当我发布此字符串时，$name不会改变。我怎样才能做到这一点，以便我可以将类似的+name+内容更改为名称。我试过搜索它，但我不知道要搜索什么，所以我没有任何运气。这可能真的很简单，但我只是空白。谢谢
回答by Brad Christie
You can use place-holders and str_replace. Or use PHP's built-in sprintfand use %s. (And as of v4.0.6 you can swap the ordering of arguments if you'd like).
您可以使用占位符和str_replace. 或者使用 PHP 的内置sprintf并使用%s. （从 v4.0.6 开始，您可以根据需要交换参数的顺序）。
$name = 'Alica';
// sprintf method:
$format = 'Hello, %s!';
echo sprintf($format, $name); // Hello, Alica!
// replace method:
$format = "Hello, {NAME}!";
echo str_replace("{NAME}", $name, $format);
And, for anyone wondering, I understood that is trouble templating a string, not PHP's integrated concatenation/parsing. I just assume keep this answer up though as I'm still not 100% sure this OP's intent
而且，对于任何想知道的人，我明白这是模板字符串的麻烦，而不是 PHP 的集成连接/解析。我只是假设保留这个答案，因为我仍然不是 100% 确定这个 OP 的意图
回答by Nick
I've always been a fan of strtr.
我一直是strtr.
$ php -r 'echo strtr("Hi @name. The weather is @weather.", ["@name" => "Nick", "@weather" => "Sunny"]);'
Hi Nick. The weather is Sunny.
The other advantage to this is you can define different placeholder prefix types. This is how Drupal does it; @indicates a string to be escaped as safe to output to a web page (to avoid injection attacks). The format_stringcommand loops over your parameters (such as @nameand @weather) and if the first character is an @, then it uses check_plainon the value.
另一个优点是您可以定义不同的占位符前缀类型。Drupal 就是这样做的；@指示要转义的字符串以安全输出到网页（以避免注入攻击）。的FORMAT_STRING命令遍历您的参数（如@name和@weather），并且如果所述第一字符是一个@，那么它使用check_plain上的值。
回答by Tim Cooper
You should be wrapping the string in double quotes (") if you want variables to be expanded:
"如果要扩展变量，则应将字符串括在双引号 ( ) 中：
$name = "Alica";
$string = "Hey, $name. How are you?"; // Hey, Alica. How are you?
回答by Josh
A third solution, no better or worse than the aforementioned, is concatenation:
第三种解决方案，并不比上述更好或更差，是串联：
echo 'Hello '. $name .'!!!';
echo 'Hello '. $name .'!!!';
