php - 在服务器上生成一个 JSON 对象并在客户端迭代它

Question

我的 php 方面

   $i = 1;
   while ($row = mysql_fetch_array($query)) {
       $productarray[$i][] = $row['product_id'];
       $productarray[$i][] = $row['product_model'];
       $productarray[$i][] = $row['product_type'];
       $productarray[$i][] = $row['product_return'];
       $i++;
    }
    $jsonstring = json_encode($productarray);

这就是我得到的 这是一个有效的 json 字符串吗

{
 "1":["1","HFJ5G1.5","plat","graviteits"],
 "2":["2","HHJ5S2.5","holle plunjer","veer"],
 "3":["3","HTJ5S7.5","inbouw","veer"]
}

我不能遍历这个有人请告诉我如何迭代这个 json

从 json 站点我认为这是格式，如果是的话我如何将上面的 json 字符串更改为这种格式？

{
 "1":[{"1","HFJ5G1.5","plat","graviteits"}],
 "2":[{"2","HHJ5S2.5","holle plunjer","veer"}],
 "3":[{"3","HTJ5S7.5","inbouw","veer"}]
}

Answer 1

您需要稍微重构您的 PHP 代码。尝试做这样的事情:

$productarray = array();
while ($product = mysql_fetch_assoc($query)) { // NOTE: using associative array
    $productarray[] = $product;
}
echo json_encode($productarray);

// Output
// [{
//     "product_id": "1",
//     "product_model": "HFJ5G1.5",
//     "product_type": "plat",
//     "product_rturn": "graviteits"
// }, {
//     "product_id": "2",
//     "product_model": "HHJ5S2.5",
//     "product_type": "holle plunjer",
//     "product_rturn": "veer"
// }, {
//     "product_id": "3",
//     "product_model": "HTJ5S7.5",
//     "product_type": "inbouw",
//     "product_rturn": "veer"
// }]

请注意，它现在变得非常容易迭代。您可以按如下方式进行:

var o = [{
    "product_id": "1",
    "product_model": "HFJ5G1.5",
    "product_type": "plat",
    "product_rturn": "graviteits"
}, {
    "product_id": "2",
    "product_model": "HHJ5S2.5",
    "product_type": "holle plunjer",
    "product_rturn": "veer"
}, {
    "product_id": "3",
    "product_model": "HTJ5S7.5",
    "product_type": "inbouw",
    "product_rturn": "veer"
}];
for (var i = 0; i < o.length; i++) {
    console.log("Product " + (i + 1) + " has model:" + o[i]["product_model"]);
    console.log("Product " + (i + 1) + " has type:" + o[i]["product_type"]);
}