[英]PHP: Set the values of one array to another array
我想創建一個新數組,並將其鍵設置為等於另一個數組的值。
開始:
$names = array("don","james","jennifer","paul");
$ages = array("don" => 25, "paul" => 32);
完:
$name_age_map = array(
"don" => 25,
"james" => null,
"jennifer" => null,
"paul" => 32,
);
應該如何在PHP中完成? 這是我到目前為止所取得的最好成績:
$name_age_map = array();
foreach ($names as $name) {
$name_age_map[$name] = $name_map[$name];
}
理想情況下,我什至不創建新的數組,我只是給$ names中的元素指定年齡。
只是為了好玩,您可以這樣做:
$names = array("don","james","jennifer","paul");
$ages = array("don" => 25, "paul" => 32);
$names = array_merge(array_fill_keys($names, null), $ages);
var_dump($names);
產量:
array(4) {
["don"]=>
int(25)
["james"]=>
NULL
["jennifer"]=>
NULL
["paul"]=>
int(32)
}
$name_map = array();
foreach ($names as $name) {
$name_map[$name] = isset($ages[$name])?$ages[$name]:null;
}
問題在於$names
中的值就是-值。 為了使它們“相等”一個年齡,您需要將它們作為關鍵。 因為PHP不支持通過引用傳遞鍵,所以這意味着創建一個新數組:
$name_age_map = array();
foreach($names as $name) {
if(isset($ages[$name])) {
$name_age_map[$name] = $ages[$name];
}
else {
$name_age_map[] = $name;
}
}
或者,如果您始終希望名稱成為鍵(可能更有意義):
$name_age_map = array();
foreach($names as $name) {
$name_age_map[$name] =
isset($ages[$name]) ? $ages[$name] : null;
}
$name_age_map = array();
foreach ($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;
}
}
要操作原始數組,您需要將其翻轉,然后添加關聯的年齡。
嘗試這個:
$names = array_flip($names);
foreach($names as $key => $value)
{
$names[$key] = (array_key_exists($key,$ages)) ? $ages[$key] : null;
}
比較兩個數組中的鍵時, array_key_exists()
可以幫助您。 應用循環並在找到匹配項的地方分配值,否則返回null。 這應該為您解決。
$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)
}
array_merge(
array_combine(
$names,
array_fill(0,count($names),NULL)
),
$ages);
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