[英]How to delete a record in array without using for loop
I have this array collection which is formatted in this way: 我有此数组集合,其格式是这样的:
$collection = array(
'0' => array(
'user_id' => 12345
'name' => 'test'
),
'1' => array(
'user_id' => 6789
'name' => 'test2'
),
)
My problem is that I don't want to create a for loop to search for the id:6789. 我的问题是我不想创建for循环来搜索id:6789。
Currently, what I'm doing is this: 目前,我正在做的是:
$user_id = 6789
foreach($collection as $key => $collect)
{
if($collect['user_id'] == $user_id)
{
unset($collection[$key]);
}
}
I just want to ask if there is a much efficient way of deleting the data instead of doing a for loop. 我只想问一下是否有一种有效的方法来删除数据,而不是执行for循环。
Thanks. 谢谢。
there's a function in php called array_walk_recursive()
you can find the documentation here: php中有一个名为
array_walk_recursive()
的函数,您可以在这里找到文档:
http://www.php.net/manual/en/function.array-walk-recursive.php http://www.php.net/manual/zh/function.array-walk-recursive.php
im not sure if it will be more efficient than using loops but it should do what your asking 我不确定它是否比使用循环更有效,但它应该按照您的要求进行
I can't think of a better way of finding the key as the items are stored in a multi-dimensional array. 由于项目存储在多维数组中,我想不出找到密钥的更好方法。 However if you could build your initial array differently by putting the user_id in the key eg:
但是,如果您可以通过将user_id放在键中来不同地构建初始数组,例如:
$collection = array(
'12345' => array(
'user_id' => 12345
'name' => 'test'
),
'6789' => array(
'user_id' => 6789
'name' => 'test2'
),
)
Then you can just do 那你就可以做
$user_id = 6789;
unset($collection[$user_id ]);
You can also use array_filter to accomplish this. 您也可以使用array_filter完成此操作。 Given your input of:
根据您的输入:
$collection = array(
'0' => array(
'user_id' => 12345,
'name' => 'test',
),
'1' => array(
'user_id' => 6789,
'name' => 'test2',
),
);
$user_id = 6789;
You can use: 您可以使用:
$new_collection = array_filter( $collection, function($item) use ($user_id) {
return $item['user_id'] != $user_id;
});
Hope that helps 希望能有所帮助
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.