2013-10-17 45 views

回答

3

如果定義 「具有價值」 爲「價值== true「:

if (count(array_filter($array)) == count($array)) { 
    echo 'All elements have values'; 
} 
+0

請與$設置=陣列( '主機'=>」本地主機」, '用戶'=>爲真, '密碼'=>爲真, '數據庫'=> TRUE);並得到真實:P –

1
function hasValue($v) { 
    return strval($v) != ''; 
}  
$res_array = array_filter($my_array, 'hasValue'); 
// any key has value 
$any_key_has_value = 0 < sizeof($res_array); 
// all keys have values 
$all_keys_have_values = sizeof($my_array) == sizeof($res_array); 
1
$my_array = array(1 => 'has value', 2 =>'', 3 => ''); 

只是試試這個

if (array_filter($my_array)) { 
    // here first value `has value` 
}else{ 
// all values are empty 
} 
+0

需要'count($ array_filter)'否則它不會工作,謝謝 – user007

1

試試這個:

<?php 
     $array = array('key1' => null, 'key2' => null, 'key3' => null, 'key4' => null); 

     if (!array_filter($array)) { 

      echo "All keys have null values"; 
     } 
     else 
     { 
       // do something 
     } 
     ?> 

-