搜索元素，並刪除它從PHP的

Question

我有這樣下面的陣列$days：

$days= array ("Monday 1 April 2013","Tuesday 2 April 2013","Wednesday 3 April"); 

我會通過搜索它來刪除所有與週一和週日開始元素。

Answer 1

您可以使用preg_grep（）來獲取與正則表達式匹配的數組元素。要獲得元素不匹配的表達，使表達negative lookahead（?!）的

$days = array ("Monday 1 April 2013", "Tuesday 2 April 2013", "Wednesday 3 April"); 
$result = preg_grep('/^(?!Monday|Sunday)/', $days); 
print_r($result); 

結果

Array 
(
    [1] => Tuesday 2 April 2013 
    [2] => Wednesday 3 April 
) 

正則表達式^(?!Monday|Sunday)意味着「找到^任何發生（開始主題字符串）後面跟着「星期一」或「星期天」。

Answer 2

$days = array_filter($days, function($el){ 
    return !(strpos($el, 'Monday') === 0 || strpos($el, 'Sunday') === 0); 
}); 

Codepad Example

Answer 3

嘗試這種情況：

for($i=0;$i<count($days);$i++) { 
    $words = explode(' ',$days[$i]); 
    if($words[0]==='Monday' || $words[0]==='Sunday') unset($days[$i]);  
} 

Answer 4

$new_array=array(); 

$days= array ("Monday 1 April 2013","Tuesday 2 April 2013","Wednesday 3 April"); 

foreach($days as $key=>$val) 
{ 

$words = explode(' ',$val); 
    if($words[0]!='Monday' || $words[0]!='Sunday') 
{ 

$new_array[]=$val; 

} 


}  


}