回显一个数组值，给出另一个数组的键

Question

我知道这很简单，但我无法理解它...

我有 2 个数组。两者都是从数据库查找中填充的。

数组 1

Array ( 
[sailID] => 7 
[sailTag] => 100004 
[assigneduser] => Jason Ellmers 
[assigneddate] => 2018-05-30 17:48:57 
[cutuser] => Jason Ellmers 
[cutdate] => 2018-05-30 20:31:23 
[stickuser] => Jason Ellmers 
[stickdate] => 2018-05-30 20:38:24 
[corneruser] => Jason Ellmers 
[cornerdate] => 2018-05-30 20:38:54 
[finishuser] => Jason Ellmers 
[finishdate] => 2018-05-30 20:39:53 
[checkuser] => 
[checkdate] => 0000-00-00 00:00:00 
[DesignRef] => 420abcdefg 
[OrderingLoft] => 1 
[ClassRef] => 1 
[ClothType] => Bainbridge 
[ClothColour] => White 
[ClothWeight] => 12oz 
[SailNo] => GB342398 )

数组 2

Array ( 
[0] => Array ( 
      [id] => 1 
      [name] => 420 ) 
[1] => Array ( 
      [id] => 2 
      [name] => J24 ) )

我之后要做的是能够回显到屏幕 $array1['Where the ClassRef is a lookup of the ID in Array2' 并显示 Array2 中的名称]

因此对于上面的示例，Echo 将是“420”

我想我可以使用 foreach 或 while 循环来完成，但这似乎有点麻烦？？？

Answer 1

我不得不将一些测试数据放在一起，但是从评论来看，我的想法是使用 array_column() 以 id 作为索引重新索引第二个数组，所以代码（如您已经解决了）是...

$array1 =[
    "sailID" => 7,
    "sailTag" => "100004",
    "ClassRef" => 1 ];

$array2 = [["id" => 1, "name" => "420"],
    ["id" => 2, "name" => "J24"]];

$array2 = array_column($array2, "name", "id");

echo $array2[$array1["ClassRef"]];

回显一个数组值，给出另一个数组的键

echoing an array value giving a key from another array

php

arrays

matching