Question

我想创建一个新数组，并将其键设置为等于另一个数组的值。

开始：

$names = array("don","james","jennifer","paul");
$ages = array("don" => 25, "paul" => 32);

完成：

$name_age_map = array(
 "don" => 25,
 "james" => null,
 "jennifer" => null,
 "paul" => 32,
);

如何在PHP中完成此操作？这是我迄今为止最好的：

$name_age_map = array();
foreach ($names as $name) {
  $name_age_map[$name] = $name_map[$name]; 
}

理想情况下，我甚至不会创建一个新数组，我只是给$ names中的元素赋予年龄值。

Answer 1

只是为了好玩，你可以这样做：

$names = array("don","james","jennifer","paul");
$ages  = array("don" => 25, "paul" => 32);
$names = array_merge(array_fill_keys($names, null), $ages);

var_dump($names);

收率：

array(4) {
  ["don"]=>
  int(25)
  ["james"]=>
  NULL
  ["jennifer"]=>
  NULL
  ["paul"]=>
  int(32)
}

Answer 2

$name_map = array();
foreach ($names as $name) {
  $name_map[$name] = isset($ages[$name])?$ages[$name]:null; 
}

Answer 3

问题是$names中的值只是 - 值。为了使它们“平等”成为一个时代，你需要它们成为关键。因为PHP不支持通过引用传递键，这意味着创建一个新数组：

$name_age_map = array();
foreach($names as $name) {
    if(isset($ages[$name])) {
        $name_age_map[$name] = $ages[$name];
    }
    else {
        $name_age_map[] = $name;
    }
}

或者如果你总是希望名字成为键（这可能更有意义）：

$name_age_map = array();
foreach($names as $name) {
    $name_age_map[$name] =
            isset($ages[$name]) ? $ages[$name] : null;
}

Answer 4

$name_age_map = array();

foreach ($names as $name){
    // this can also be replaced with
    // if (array_key_exists($name, $ages)){
    if (in_array($name, array_keys($ages))){
        $name_age_map[$name] = $ages[$name];
    }
    else {
        $name_age_map[$name] = null;
    }
}

Answer 5

要操作原始数组，您需要将其翻转，然后添加相关的年龄。

试试这个：

$names = array_flip($names);
foreach($names as $key => $value)
{
    $names[$key] = (array_key_exists($key,$ages)) ? $ages[$key] : null;
}

Answer 6

在比较两个数组中的键时，

array_key_exists()可以帮助您。应用循环并分配找到匹配项的值，否则返回null。这应该适合你。

$names = array("don","james","jennifer","paul");
$ages = array("don" => 25, "paul" => 32);

$merged = array();
foreach($names as $n) {
    if(array_key_exists($n, $ages)) {
        $merged[$n] = $ages[$n];
    } else {
        $merged[$n] = null;
    }
}

var_dump($merged);

//Produces

array(4) {
  ["don"]=>
  int(25)
  ["james"]=>
  NULL
  ["jennifer"]=>
  NULL
  ["paul"]=>
  int(32)
}

Answer 7

array_merge(
   array_combine(
       $names,
       array_fill(0,count($names),NULL)
   ),
   $ages);

PHP：将一个数组的值设置为另一个数组

7 个答案: