在PHP中的字符串中添加字符串

Question

我在PHP中有一个基本的字符串问题。

假设我有一个变量$story：

$story = 'this story is titled and it is really good';

如何在'标题'之后和'和'之前添加字符串？ 如果我在另一个变量中有标题，那么就说

$title = 'candy';

我可以用什么功能或方法来做这件事？

$story = 'this story is titled and it is really good';
$title = 'candy';
// do something
var_dump($story === 'this story is titled candy and it is really good'); // TRUE

Answer 1

有几个选择。

$title = 'candy';
$story = 'this story is titled '.$title.' and it is really good';
$story = "this story is titled $title and it is really good";
$story = sprintf('this story is titled %s and it is really good', $title);

请参阅：

• http://www.php.net/manual/en/language.types.string.php
• http://www.php.net/manual/en/function.sprintf.php

如果您使用带有html的php并且想要打印字符串（在php标签之外）

this story is titled <?php echo $title ?> and it is really good

Answer 2

你只需要使用双引号并将变量放在字符串中，如下所示：

$title = 'candy';
$story = "this story is titled $title and it is really good";

Answer 3

我建议您在原始字符串中使用占位符，然后将占位符替换为您的标题。

所以，修改你的代码是这样的：

$story = "this story is titled {TITLE} and it is really good";

然后，您可以使用str_replace将占位符替换为实际标题，如下所示：

$newStory = str_replace("{TITLE}", $title, $story);

Answer 4

简单的方法是：

$story="this story is titled $title and it is really good".

如果你问如何找到插入位置，可以这样做：

$i=stripos($story," and");
$story=substr($story,0,$i)." ".$title.substr($story,$i);

第三个是放置一个不太可能出现在文本中的标记，如|| TITLE ||。搜索并将其替换为标题文本，如：

$i=stripos($story,"||TITLE||");
$story=substr($story,0,$i).$title.substr($story,$i+9);

Answer 5

利用您的朋友字符串插值（GreenWevDev said）。

或者，如果您需要将单词title替换为字符串，并且只能单独替换，则可以使用正则表达式。

$story = preg_replace('/\btitle\b/', $title, $story);