Question

我有一个这样的数组：

Array
(
    [0] => Array
    (
        [objectid] => 197
        [adresse] => D554
        [city] => NEW-YORK
        [lat] => 12,545484654687
        [long] => 12,545484654687
    )

    [1] => Array
    (
        [objectid] => 198
        [adresse] => D556
        [city] => WASHINGTON
        [lat] => 12,545484654687
        [long] => 12,545484654687
    )
    ...
    ...
)

我想用0,1,2 ......等标识符更改城市名称。

实际上，我是通过这段代码完成的：

foreach ($big_array as $key => $value){
    if ($value['city'] == "NEW-YORK"){
        $big_array[$key] = str_replace("NEW-YORK", 0, $value);
    } elseif($value['city'] == "WASHINGTON") {
        $big_array[$key] = str_replace("WASHINGTON", 1, $value);
    } etc...
}

我认为这不是最好的方法，我有很多城市。是否可以定义如下数组：

$replacements = array(
    "NEW-YORK" => 0,
    "WASHINGTON" => 1,
    etc...
)

并使用函数简单地执行更改？

Answer 1

如果通过引用

传递数组，则可以直接替换数组的值

foreach ($big_array as &$value) {
    $city = $value['city'];

    // for cities that we don't have a replacement
    if (! isset($replacements[$city])) {
        continue;
    }

    $value['city'] = $replacements[$city];
}

// just to be sure we don't keep any reference to the $value variable
unset($value);

Answer 2

具有$replacements数组的逻辑应该类似于：

foreach ($big_array as $key => $value)
{
    if (!isset($replacements[$value['city']])
    {
        echo "Could not find city '".$value['city']."' in replacements array, skipping";
        continue;
    }

    $value['city'] = $replacements[$value['city']];

    $big_array[$key] = $value;
}

Answer 3

$find = array(0, 1, etc...);

$replace = array("NEW-YORK", "WASHINGTON", ect ...);

foreach ($big_array as $key => $value){
     $big_array[$key]['city'] = str_replace($find, $value['city']);
}

替换多维数组中的字符串

3 个答案: