通过在多维数组中搜索标题返回值

Posted 2023-02-24

【中文标题】通过在多维数组中搜索标题返回值

【英文标题】：return value by searching for title inside a multidimensional array

【发布时间】：2015-05-12 05:24:01

【问题描述】：

我希望能够搜索标题为 Seattle 的数组，该数组将由一个变量设置。然后返回该数组的 x 或 y 坐标。我已经尝试了 5 或 6 种不同的方法来尝试定位它，但没有任何运气。

这是我正在使用的查询以及我如何打印我的数组：

global $wpdb;
$myquery = $wpdb->get_row("SELECT * FROM wp_maps WHERE title = 'Test Map'"); 
$mymap =  $mylink->data;

print_r($mymap);

这是实际输出。

 "title":"USA", "location":"World", "levels":[  "id":"states", "title":"States", "locations":["id":"bhAAG","title":"Seattle","description":"The City of Goodwill","x":47.6097,"y":122.3331,"id":"biAAG","title":"Portland","description":"Portland, Maine. Yes. Life’s good here.","x":43.6667,"y":70.2667]  ] 

相同的输出（格式化以便于查看）。

    "title":"USA",
    "location":"World",
    "levels":[
        
            "id":"states",
            "title":"States",
            "locations":[
                
                    "id":"bhAAG",
                    "title":"Seattle",
                    "description":"The City of Goodwill",
                    "x":47.6097,
                    "y":122.3331
                ,
                
                    "id":"biAAG",
                    "title":"Portland",
                    "description":"Portland, Maine. Yes. Life’s good here.",
                    "x":43.6667,
                    "y":70.2667
                
            ]
        
    ]

任何帮助将不胜感激。

【问题讨论】：

使用 print_r($mymap) 的实际输出进行编辑

我已经添加了上面的实际输出。

【参考方案1】：

您的 myMap 数据采用 JSON 格式。您可以将json_decode 放入一个数组中，然后在所有位置中搜索具有指定标题的数组：

$myMap = ' "title":"USA", "location":"World", "levels":[  "id":"states", "title":"States", "locations":["id":"bhAAG","title":"Seattle","description":"The City of Goodwill","x":47.6097,"y":122.3331,"id":"biAAG","title":"Portland","description":"Portland, Maine. Yes. Life’s good here.","x":43.6667,"y":70.2667]  ] ';

// Convert JSON and grab array of locations
$array     = json_decode($myMap, true);
$locations = $array['levels'][0]['locations'];

// What we are looking for
$title = 'Seattle';

// Search locations
foreach ($locations as $location) 
    if ($location['title'] == $title) 
        $x = $location['x'];
        $y = $location['y'];
    


echo "x = $x, y = $y", php_EOL;

输出：

x = 47.6097, y = 122.3331

【讨论】：

效果很好，这正是我所要求的。你是男人，谢谢你的帮助。看起来我没有得到每个位置的正确位置，但你把它钉牢了。再次感谢。

【参考方案2】：

紧凑的解决方案PHP5 >= 5.3

$term = ''; // term being used to search
if(isset($mymap['levels']) && isset($mymap['levels']['locations']))
    $locations = $mymap['levels']['locations'];

    // filtered will be an array of x, y values
    $filtered = array_map(function($location)
        return [ 'x' => $location['x'], 'y' => $location['y']]; // transform into required format
    , array_filter($locations, function($location) use ($term) // filter by title
        return $location['title'] === $term;
    ));

array_filter() array_map()

【讨论】：

我已经尝试过了，但是我收到了这个错误，解析错误：语法错误，意外的''。里面有多余的角色吗？

@IYBITWC 哎呀，对不起。修复。如果你真的需要这个，请回来查看。

当我回到这里时将测试并发布更新。