Question

我有一个JSON形成@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数组看起来像这样： [{"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，所以如果我问一个非常愚蠢的问题，请原谅。

您提供的链接包含密钥的示例，我的密钥没有密钥。

由于

Answer 1

$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";
}

<强> 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

Example

Answer 2

使用json_decode（$ json_object）。

$json = json_decode($json_object);
foreach($json as $j){
   print_r($j)
}

如何在没有任何密钥的情况下在php中读取JSON数组？

2 个答案: