PHP：当变量（b）包含文本时，如何在另一个变量（b）中显示变量（a）

Question

I am storing text in my database. Here is the following text:

".$teamName.", is the name of a recently formed company hoping to take over the lucrative hairdryer design ".$sector."

After querying the database I assign this text to a variable called $news, then echo it.

However the text is outputted to the screen exactly as above without the variables $teamName*and $sector replaced by there corresponding values.

I assure you that both $teamName and $sector are defined before I query the database.

Is it even possible to do what I am trying to do?

Answer 1

You might be better off using sprintf() here.

$string = "%s is the name of a recently formed company hoping to take over the lucrative hairdryer design %s.";

$teamName = "My Company";
$sector = "sector";

echo sprintf($string, $teamName, $sector);
// My Company is the name of a recently formed company hoping to take over the lucrative hairdryer design sector.

In your database, you store $string. Use sprintf() to substitute the variable values.