Home > Backend Development > PHP Tutorial > php array_flip() deletes duplicate elements from array_PHP tutorial

php array_flip() deletes duplicate elements from array_PHP tutorial

WBOY
Release: 2016-07-21 15:47:57
Original
1193 people have browsed it

The method is as follows:
$arr = array(…………);//Suppose there is an array of 10,000 elements with repeated elements in it.
$arr = array_flip(array_flip($arr)); //This will remove duplicate elements.

What on earth is going on? Let’s take a look at the role of array_flip(): array_flip() is used to exchange the key and value of each element of an array, such as:
$arr1 = array ("age" => 30, "name" => ; "Happy Garden");
$arr2 = array_flip($arr1); //$arr2 is array(30 => "age", "Happy Garden" => "name");
In In PHP arrays, different elements are allowed to take the same value, but the same key name is not allowed to be used by different elements, such as:
$arr1 = array ("age" => 30, "name" = > "Happy Garden", "age" => 20); "age" => 20 will replace "age" => 30
$arr1 = array ("name" => "Happy Garden" ", "age" => 45);
Here $arr1 and $arr2 are equal.
So, we can know why array_flip(array_flip($arr)) can delete duplicate elements in the array. First, the value in $arr will become a key name, because the value is repeated. After becoming a key name, these repeated values ​​will become duplicate key names. The PHP engine will delete the duplicate key names and only keep the last one. . For example:
$arr1 = array ("age" => 30, "name" => "Happy Park", "age" => 20);
$arr1 = array_flip($arr1); //$arr1 becomes array("Happy Garden" => "name", 20 => "age");
//Restore the key name and value of $arr1:
$ arr1 = array_flip($arr1);

The above code can be written more concisely: $arr1 = array_flip(array_flip($arr1));

www.bkjia.comtruehttp: //www.bkjia.com/PHPjc/319822.htmlTechArticleThe method is as follows: $arr = array(…………);//Assume an array with 10,000 elements , which contains repeated elements. $arr = array_flip(array_flip($arr)); //This will remove duplicate elements. ...
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template