Question

我只是想让它{string}变成$string变量的值。

例如，考虑我有以下变量：

$car = 'Audi';
$speed = 'Fast';
$sentence = "The {car} goes {speed}";

我想要输出：The Audi goes Fast

这是我的代码无效：

$setting['leads_low_subject'] = '{dealer_name} has {leads_left} leads left!';
$dealer_name = 'My Test Dealer';
$leads_left = 6;

$subject = preg_replace('/\{([a-z]+)\}/e', "$$1", $setting['leads_low_subject']);

echo $setting['leads_low_subject'].'<br />';
echo "$subject";

这回应：

{dealer_name} has {leads_left} leads left!
{dealer_name} has {leads_left} leads left!

何时应该回应：

{dealer_name} has {leads_left} leads left!
My Test Dealer has 6 leads left!

为什么它无法正常工作？

Answer 1

我看到的最快的方法是使用strtr（）：

$trans = array( '{car}'   => 'Audi',
                '{speed}' => 'Fast');

$result = strtr($sentence, $trans);

Answer 2

好像你正在寻找一个穷人的模板引擎。您可以使用preg_replace_callback()：

$tpl_vars = array(
    'car' => 'Ford',
    'speed' => 'fast'
);


$tpl = 'The {car} goes {speed}';

echo preg_replace_callback('/\{(.*?)\}/', function($match) use ($tpl_vars) {
    return $tpl_vars[$match[1]];
}, $tpl);

变量字符串的简单preg_replace到值不起作用

2 个答案: