MySQL表的JSON API

Question

我有一個API請求，我需要它放入MySQL DB

API示例

"items": [
    {
        "market_name": "\u2605 Bayonet",
        "market_hash_name": "\u2605 Bayonet",
        "icon_url": "\/\/steamcommunity-a.akamaihd.net\/economy\/image\/-9a81dlWLwJ2UUGcVs_nsVtzdOEdtWwKGZZLQHTxDZ7I56KU0Zwwo4NUX4oFJZEHLbXU5A1PIYQh5hlcX0nvUOGsx8DdQBJjIAVHubSaKQZ53P3NZXMXvYmykdLSxqWkZ7-HkjMIvpIj3u2Y84733gzh_RU_MG_zIYLEdQ45fxiOrdJh0ExF",
        "name_color": "8650AC",
        "quality_color": "EB4B4B"
    },
    {
        "market_name": "\u2605 Bayonet | Blue Steel (Battle-Scarred)",
        "market_hash_name": "\u2605 Bayonet | Blue Steel (Battle-Scarred)",
        "icon_url": "\/\/steamcommunity-a.akamaihd.net\/economy\/image\/-9a81dlWLwJ2UUGcVs_nsVtzdOEdtWwKGZZLQHTxDZ7I56KU0Zwwo4NUX4oFJZEHLbXH5ApeO4YmlhxYQknCRvCo04DEVlxkKgpotLu8JAllx8zJYAJA4MmzkL-DkvbiKvXTkzNVucNzj7mX9tWk21Xkr0JvN231JYGcdA47NF3Y81Hoxebs1sftot2XnmcyW1u0",
        "name_color": "8650AC",
        "quality_color": "EB4B4B"
    },

和我的PHP文件

$json = file_get_contents('API Link');
$obj = json_decode($json,true);

$hostname="localhost";
$database="API";
$username="XxX";
$password="XxX";

$link = mysql_connect($hostname, $username, $password);
 mysql_select_db($database) or die('Could not select database');

我知道很熱，使用一個項目

$obj['items'][0]['market_name'];

我需要自動將數字從0更改為7226，然后將其添加到db。 你能幫我嗎 ？

Answer 1

嘗試這個 ：

$items = $obj['items'];
for($i=0; $i<count($items);$i++){
    //Do the stuff whatever you want to do inside the loop
    // $i will change the index from 0 to the last index of array $items

    $market_name = $items[$i]['market_name'];
}