在PHP中从数组中移除键值

Question

这是我的数据库查询结果，

Array ([0] => Array ( [shopname] => Shop name [fueltype] => Pertol [amount] => 1000 ) 
       [1] => Array ( [shopname] => dfsdfsd [fueltype] => Pertol [amount] => 54456 )
       [2] => Array ( [shopname] => dfsdfsd [fueltype] => Disel [amount] => 54456 )
)

我需要像这样的结果

[["Shop name", "Pertol", 1000],["dfsdfsd", "Pertol", 54456],["Shop name", "Disel", 54456]]

我该如何做到这一点，我不知道？

- Ajin

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首先进行从键值到仅值的转换(array_values())，然后使用json_encode()函数... 这些现在是你的好朋友了... ;) - Raphioly-San

Ajin，检查答案并标记最合适的答案是你的责任。这将有助于未来的访问者更轻松地找到解决方案。谢谢。（如果其他答案有用，您也可以给它们投票） - Alive to die - Anant

Answer 1

array_map() 和 array_values() 可以一起使用，适用于你的情况：

<?php
$array = Array ( '0' => Array ( 'shopname' => 'Shop name','fueltype' => 'Pertol','amount' => 1000 ), 
        '1' => Array ( 'shopname' => 'dfsdfsd' ,'fueltype' => 'Pertol','amount' => 54456 ),
        '2' => Array ( 'shopname' => 'dfsdfsd','fueltype' => 'Disel','amount' => 54456 )
);


$values_data_only = array_map('array_values', $array);
$desire_result = json_encode($values_data_only);
echo $desire_result;

?>

输出：- https://eval.in/395344

同时使用简单的foreach()也是可能的：

<?php
$array = Array ( '0' => Array ( 'shopname' => 'Shop name','fueltype' => 'Pertol','amount' => 1000 ), 
        '1' => Array ( 'shopname' => 'dfsdfsd' ,'fueltype' => 'Pertol','amount' => 54456 ),
        '2' => Array ( 'shopname' => 'dfsdfsd','fueltype' => 'Disel','amount' => 54456 )
);

$new_array = array();

foreach ($array as $k=> $arr){
    $new_array[$k][] = $arr['shopname'];
    $new_array[$k][] = $arr['fueltype'];
    $new_array[$k][] = $arr['amount'];
}
echo "<pre/>";print_r($new_array);
$desired_result_2 = json_encode($new_array);
echo $desired_result_2;
?>

Output:-https://eval.in/395354

- Anant - Alive to die

Answer 2

$mapped = array_map('array_values', $input_array); // apply filter so we dont get the keys
$json = json_encode($mapped);

- viral

Answer 3

尝试使用：

$input  = array( /* your input data */ );
$output = array();

foreach ($input as $data) {
  $output[] = array_values($data);
}

- hsz

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我认为你不需要在 foreach 外初始化 $output，这样可以节省一行代码！ - viral

@Viral 那个主意不太好。 - hsz