我有一个JSON格式的@Javascript，并通过AJAX传递给PHP，问题是我无法为JSON数组中的每个JSON对象分配键。JSON数组如下所示：[{"jobId":"90","cname":"Subhasish","removal_id":101,"quantity":"3"},{"jobId":"90","cname":"Subhasish","removal_id":102,"quantity":"2"},{"jobId":"90","cname":"Subhasish","removal_id":103,"quantity":"4"},{"jobId":"90","cname":"Subhasish","removal_id":104,"quantity":"4"},{"jobId":"90","cname":"Subhasish","removal_id":105,"quantity":0},{"jobId":"90","cname":"Subhasish","removal_id":106,"quantity":"5"},{"jobId":"90","cname":"Subhasish","removal_id":107,"quantity":0},{"jobId":"90","cname":"Subhasish","removal_id":108,"quantity":0},{"jobId":"90","cname":"Subhasish","removal_id":109,"quantity":"4"}]
，现在，在PHP中，如何迭代数据，我是PHP的新手，所以如果我问了一个非常愚蠢的问题，请原谅我。
你提供的链接有带密钥的例子，我的没有密钥。
感谢
$json = '[{"jobId":"90","cname":"Subhasish","removal_id":101,"quantity":"3"},{"jobId":"90","cname":"Subhasish","removal_id":102,"quantity":"2"},{"jobId":"90","cname":"Subhasish","removal_id":103,"quantity":"4"},{"jobId":"90","cname":"Subhasish","removal_id":104,"quantity":"4"},{"jobId":"90","cname":"Subhasish","removal_id":105,"quantity":0},{"jobId":"90","cname":"Subhasish","removal_id":106,"quantity":"5"},{"jobId":"90","cname":"Subhasish","removal_id":107,"quantity":0},{"jobId":"90","cname":"Subhasish","removal_id":108,"quantity":0},{"jobId":"90","cname":"Subhasish","removal_id":109,"quantity":"4"}]';
$array = json_decode($json,true);
foreach($array as $item) {
echo $item['jobId']."-".$item['cname']."-".$item['removal_id']."-".$item['quantity']."'n";
}
Output
90-Subhasish-101-3
90-Subhasish-102-2
90-Subhasish-103-4
90-Subhasish-104-4
90-Subhasish-105-0
90-Subhasish-106-5
90-Subhasish-107-0
90-Subhasish-108-0
90-Subhasish-109-4
示例
使用json_decode（$json_object）。
$json = json_decode($json_object);
foreach($json as $j){
print_r($j)
}