PHP：當變量（b）包含文本時，如何在另一個變量（b）中顯示變量（a）

Question

我正在將文本存儲在我的數據庫中。 這裏是下面的文字：

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

查詢我給你這個文本到一個名爲$news變量的數據庫，然後呼應它之後。

但是，文本完全按照上面的方式輸出到屏幕上，而不會將變量$teamName*和$sector替換爲相應的值。

我向你保證在我查詢數據庫之前定義了$teamName和$sector。

它甚至有可能做我想做的事情嗎？

Answer 1

在這裏，您最好使用sprintf()。

$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. 

在您的數據庫中，您存儲$string。使用sprintf()替代變量值。

Answer 2

嘗試使用此，我想：http://php.net/manual/en/function.sprintf.php

Answer 3

這是胡亂猜測，但你因爲也許他們在單引號和不評估存儲在數據庫中的變量名？

$foo = 'bar';echo '$foo'; //$foo

$foo = 'bar';echo "$foo"; //bar

Answer 4

這不是它的工作原理。如果要對$teamname進行評估，則需要先將其評估，然後再將其存儲在數據庫中。如果你需要改變它們，你可以爲所有變量做一些字符串替換。

SQL：INSERT INTO ... VALUES ('My team has won ##num_won## games this year.')

PHP：

$string = get_string_from_sql(); // substitute for whatever method you are using to get the string. 
$num_won = 16; 
$string = str_replace('##num_won##', $num_won, $string); 
echo $string; // Will echo My team has won 16 games this year. 

Answer 5

你應該存儲在數據庫中的以下字符串（略有不同，你的）：

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

然後，你可以做一個的兩件事：

$news = eval('return "'.$news.'";'); 

......或者......

$news = str_replace(array('$teamName','$sector'),array($teamName,$sector),$news); 

或者更好的是，使用sprintf()，其中字符串是：

%s, is the name of a recently formed company hoping to take over the lucrative hairdryer design %s. 

...你得到的實際值是這樣的：

$news = sprintf($news, $teamName, $sector); 

Answer 6

試試這個：

$teamName = 'foo'; 
$sector = 'bar'; 
$news = $teamname . ', is the name of a recently formed company hoping to take over the lucrative hairdryer design ' . $sector '.'; 
echo $news; 

如果你真的想顯示的雙引號，然後嘗試：

$news = '\"' . $teamname . '\", is the name of a recently formed company hoping to take over the lucrative hairdryer design \"' . $sector . '\".';