Question

我来这里寻求帮助，我现在已经绞尽脑汁锻炼了3个小时。我有两个数组，$ authors和$ books。

$作者：

$authors = array( "Steinbeck", "Kafka", "Tolkien", "Dickens", "Milton", "Orwell" );

$书：

$books = array(
            array(
                "title" => "The Hobbit",
                "authorId" => 2,
                "pubYear" => 1937
                ),
            array(
                "title" => "The Grapes of Wrath",
                "authorId" => 0,
                "pubYear" => 1939
                ),
            array(
                "title" => "A Tale of Two Cities",
                "authorId" => 3,
                "pubYear" => 1859
                ),
            array(
                "title" => "Paradise Lost",
                "authorId" => 4,
                "pubYear" => 1667
                ),
            array(
                "title" => "Animal Farm",
                "authorId" => 5,
                "pubYear" => 1945
                ),
            array(
                "title" => "The Trial",
                "authorId" => 1,
                "pubYear" => 1925
                ),
        );

如您所见，$ authors是一个二维索引数组，而$ books是一个多维关联数组。我的任务是创建一个新密钥（我认为一个密钥？甚至是数组的词汇表让我感到困惑......），这些名为＆＃34; authorName＆＃34;的书籍，并且与$ authors数组中的作者填充密钥。问题是$ authors数组中作者的索引对应于＆＃34; authorId＆＃34;在$ books数组中，但ID无序。

换句话说，我的任务是从$ authors数组中提取数据，以便books数组最终得到以下数据：

$books = array(
            array(
                "title" => "The Hobbit",
                "authorId" => 2,
                "pubYear" => 1937
                "authorName" => "Tolkien"
                ),
            array(
                "title" => "The Grapes of Wrath",
                "authorId" => 0,
                "pubYear" => 1939
                "authorName" => "Steinbeck"
                ),
            array(
                "title" => "A Tale of Two Cities",
                "authorId" => 3,
                "pubYear" => 1859
                "authorName" => "Dickens"
                ),

......等等。任何帮助将不胜感激，因为我完全不知道如何做到这一点。

Answer 1

以下代码段可以解决问题。迭代books数组，将authorName设置为等于迭代书authorId的值。

foreach($books as $key => $book) {
    $books[$key]['authorName'] = $authors[$book['authorId']];
}

Answer 2

恭喜学习PHP！我会从像词汇这样的小事开始;我认为在编程中做正确的事情非常重要; - ）

$authors数组不是二维
你说它出了故障，但我认为不是。请记住，数组索引是从0开始的。
是，＆＃34;键＆＃34;是一种描述你想要添加的内容的好方法。在语义上稍微合适的可能是＆＃34; entry＆＃34;，其中＆＃34; entry＆＃34;由一个＆＃34;键＆＃34;组成。它的价值＆＃34;

这是一些代码。我选择比我通常编写的代码稍微冗长一些，以便清楚发生了什么。

foreach($books as $book_index => $book_array) {
    // Get the index of the author in the $authors array.
    // With this value, $authors[$authorId] will be the name of the author of this book
    $authorId = intval($book_array['authorId']);

    // this line adds an entry to the current book in the $books array.
    $books[$book_index]['authorName'] = $authors[$authorId];
}

Answer 3

您只需在$books上运行循环即可。

foreach($books as $index => $book){
    $books[$index]['authorName'] = $authors[$book['authorId']];
}

如何将索引数组合并为无序关联数组PHP

3 个答案: