php 从 JSON 输入中检索数组键
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Retrieving array keys from JSON input
提问by Nick
I have this array:
我有这个数组：
$json = json_decode('
{"entries":[
{"id": "29","name":"John", "age":"36"},
{"id": "30","name":"Hyman", "age":"23"}
]}
');
and I am looking for a PHP "for each" loop that would retrieve the key names under entries, i.e.:
我正在寻找一个 PHP“for each”循环，它可以检索 下的键名entries，即：
id
name
age
How can I do this?
我怎样才能做到这一点？
回答by Sena
Try it
尝试一下
foreach($json->entries as $row) {
foreach($row as $key => $val) {
echo $key . ': ' . $val;
echo '<br>';
}
}
In the $key you shall get the key names and in the val you shal get the values
在 $key 中，您将获得键名，而在 val 中，您将获得值
回答by Kar.ma
I was not satisfied with other answers so I add my own. I believe the most general approach is:
我对其他答案不满意，所以我添加了我自己的答案。我认为最通用的方法是：
$array = get_object_vars($json->entries[0]);
foreach($array as $key => $value) {
echo $key . "<br>";
}
where I used entries[0]because you assume that all the elements of the entriesarray have the same keys.
我使用的地方entries[0]是因为您假设entries数组的所有元素都具有相同的键。
Have a look at the official documentation for key: http://php.net/manual/en/function.key.php
看看官方文档key：http: //php.net/manual/en/function.key.php
回答by Emmanuel Okeke
You could do something like this:
你可以这样做：
foreach($json->entries as $record){
echo $record->id;
echo $record->name;
echo $record->age;
}
If you pass trueas the value for the second parameter in the json_decodefunction, you'll be able to use the decoded value as an array.
如果您true作为json_decode函数中第二个参数的值传递，您将能够将解码后的值用作数组。
回答by Inamur Rahman
$column_name =[];
foreach($data as $i){
foreach($i as $key => $i){
array_push($column_name, $key);
}
break;
}
回答by gingerCodeNinja
Alternative answer using arrays rather than objects - passing true to json_decodewill return an array.
使用数组而不是对象的替代答案 - 传递 truejson_decode将返回一个数组。
$json = '{"entries":[{"id": "29","name":"John", "age":"36"},{"id": "30","name":"Hyman", "age":"23"}]}';
$data = json_decode($json, true);
$entries = $data['entries'];
foreach ($entries as $entry) {
$id = $entry['id'];
$name = $entry['name'];
$age = $entry['age'];
printf('%s (ID %d) is %d years old'.PHP_EOL, $name, $id, $age);
}
回答by Mihai Stancu
foreach($json->entries[0] AS $key => $name) {
echo $key;
}
回答by karthik
You could try getting the properties of the object using get_object_vars:
您可以尝试使用以下方法获取对象的属性get_object_vars：
$keys = array();
foreach($json->entries as $entry)
$keys += array_keys(get_object_vars($entry));
print_r($keys);
