我有这两个数组$fonts['google']和$data['value']，内容如下：
var_dump ($fonts['google']) outputs
array(4) {
[0]=> array(3) { ["family"]=> string(7) "ABeeZee" ["variants"]=> array(2) { [0]=> string(7) "regular" [1]=> string(6) "italic" } ["subsets"]=> array(1) { [0]=> string(5) "latin" } }
[1]=> array(3) { ["family"]=> string(4) "Abel" ["variants"]=> array(1) { [0]=> string(7) "regular" } ["subsets"]=> array(1) { [0]=> string(5) "latin" } }
[2]=> array(3) { ["family"]=> string(13) "Abril Fatface" ["variants"]=> array(1) { [0]=> string(7) "regular" } ["subsets"]=> array(2) { [0]=> string(5) "latin" [1]=> string(9) "latin-ext" } }
[3]=> array(3) { ["family"]=> string(8) "Aclonica" ["variants"]=> array(1) { [0]=> string(7) "regular" } ["subsets"]=> array(1) { [0]=> string(5) "latin" } }
}
var_dump ($data['value']) outputs
array(4) {
["size"]=> int(17)
["family"]=> string(3) "Exo"
["style"]=> string(3) "200"
["subsets"]=> string(5) "latin"
}
现在我从数据库中得到$data['value']['family']='Abel'。
问题：
- 如何获得给定$data的["value"]["family"]值的["variations"]
- 如何获取$data['value']['family']值所在的子数组的$fonts['gogle']索引
PHP支持关联数组，允许您为每个元素使用（字符串）键而不是数字索引。这些数组类似于javascript对象、Objective-C字典、javaHashMaps等。这使得这样的场景变得容易。您是否可以控制构建原始数据阵列？如果你可以重构你的存储，那么设置这样的阵列：
$fonts['google'] = [
["ABeeZee"] => [
["variants"]=>["regular", "italic"],
["subsets"]=>["latin"]
],
["Abel"] => [
["variants"]=>["regular"],
["subsets"]=>["latin"]
],
["Abril Fatface"] => [
["variants"]=>["regular"],
["subsets"]=>["latin", "latin-ext"]
],
["Aclonica"] => [
["variants"]=>["regular"],
["subsets"]=>["latin"]
]
]
额外的信用：如果你有帖子中的原始数据，你可以将其转换为
$newArray = array(); // or just [] in PHP >= 5.3 I believe
foreach($fonts['google'] as $index=>$fontArray) {
$newArray[$fontArray['family']] = $fontArray;
// this leaves a redundant copy of the family name in the subarray
unset $newArray[$fontArray['family']]['family']; // if you want to remove the extra copy
}
然后它变得微不足道。给定字体系列名称，您只需使用该系列名称作为数组索引来访问$fonts['google'][$fontFamilyName]（或$newArray[$fontFamilyName]）。