Question

我有两个相同的PHP数组。只有一个值不同。我想找到这个值：

var_dump(array_diff(array(
  "a" => "1",
  "b" => "SomeString",
  "c" => 1, // <- different value, same key 
  "d" => "4521",
  "e" => "7546654241",
  "f" => "78",
  "g" => "99.999",
  "h" => "34",
  "i" => "http://google.com/"
), array(
  "a" => "1",
  "b" => "SomeString",
  "c" => "0", // <- different value, same key 
  "d" => "4521",
  "e" => "7546654241",
  "f" => "78",
  "g" => "99.999",
  "h" => "34",
  "i" => "http://google.com/"
)));

结果为array(0) { }，但应该有新的"c"值，但不是。{n}。当我删除所有其他值时：

var_dump(array_diff(array(
  "c" => 1
), array(
  "c" => "0"
)));

我得到了我想要的array(1) { ["c"]=> int(1) }。

我不会理解它。为什么会这样？

Answer 1

array_diff（）将值与第一个数组匹配到第二个

returns the values in array1 that are not present in any of the other arrays.

第一个数组 中的键1的值c 存在于第二个数组中作为键{{1}的值因此，array_diff（）将识别

没有区别

也许使用array_diff_assoc()会让你得到你真正想要获得的结果.... http://ideone.com/xHCVfF

Answer 2

我尝试了http://php.net/manual/en/function.array-diff-assoc.php，我得到了

array (size=1)
  'c' => int 1

至于使用array_diff，它会比较值...

PHP array_diff：奇怪的行为

2 个答案: