如何從字符串中替換％符號？

Question

foreach($data_features as $feature){ 
    echo $feature['feature'].'<br>'; //string contains % symbol 
    $featur = str_replace("%", "'", $feature['feature']); 
    $featur = str_replace("!", '"', $feature['feature']);      
    echo '<li>'.$featur.'</li>'; // string still contains % symbol. 
} 

這裏$ feature ['feature']是一個包含％符號的字符串，我想用str_replace替換％符號，但％符號仍然存在於輸出中。

Answer 1

這可能幫助你： -

<?php 
    error_reporting(E_ALL); 
    ini_set('display_errors',1); 
    $data_features = Array ( 
    '0' => Array ( 
      'id' => 1, 
      'product_id' => 1, 
      'feature' => 'Customisation-With an open OS, you%re in control.', 
      'time_added' => '2016-09-28 15:33:28', 
      'product_asin' => 'B014UUQUAO' 
     ), 
    '1' => Array ( 
      'id' => 2, 
      'product_id' => 1, 
      'feature' => 'Enhanced Expirences- We believe core experiences like audio, email, and calling can be better. That%s we%re re.', 
      'time_added' => '2016-09-28 15:33:28', 
      'product_asin' => 'B014UUQUAO' 
     ), 
    '2' => Array ( 
      'id' => 3, 
      'product_id' => 1, 
      'feature' => 'Privacy & Securitye.', 
      'time_added' => '2016-09-28 15:33:28', 
      'product_asin' => 'B014UUQUAO' 
     ) 
); 

echo "<pre/>";print_r($data_features); 
foreach($data_features as $feature){ 
    $featur = $feature['feature']; 
    $find = array('/%/', '/!/'); 
    $replace = array("'", '"'); 
    $result = preg_replace($find, $replace, $feature['feature']); 

    echo '<li>'.$result.'</li>'; 
} 
?> 

輸出： - https://eval.in/651753

注： -

而是多個str_replace()，單preg_replace()與$find和$replace陣列是好主意。

在不久的將來，如果需要更多的重新補充，那麼只需要更改$find和$replace就可以了。

Answer 2

我發現我的錯誤必須放入一個變量，然後應用str_replace操作。不知何故，不知道原因。

foreach($data_features as $feature){ 
    $featur = $feature['feature']; 
    $featur = str_replace("%", "'", $featur); 
    $featur = str_replace("!", '"', $featur);      
    echo '<li>'.$featur.'</li>'; 
}