Question

我有一个AJAX，它用一个间隔更新自己并从SELECT PHP返回数据。它返回一个包含多个元素的数组，但我不知道如何访问各个元素。我想使用它，因为我需要在我的网页上显示刷新的项目和其他不同的值。

AJAX代码：

var a=0;
setInterval(function() {
    $.ajax({
        url: "http://localhost/Seyroku/extras/mostrarBote.php",
        dataType: "jsonp",
        jsonp: "callback",
        data: {"a": "a"},
        success: function (respJSON) {
            //I usually use this to access the properties of the object but doesn't work
            var ItemName = respJSON.itemName;
            console.log(ItemName);
        }
    });
}, 5000);

mostrarBote.php代码：

$sql="SELECT  itemName, itemPrice, itemUrl, nameUser, color FROM potitems";

$result=mysqli_query($conn, $sql);
$numero = mysqli_num_rows($result);

$num=1;
if ($result!=NULL) {
    if (mysqli_num_rows($result)>0) {
        while($row=mysqli_fetch_array($result)) {
            $resposta = '{"itemName":"'.$row['itemName'].'","itemPrice":"'.$row['itemPrice'].'","itemUrl":"'.$row['itemUrl'].'"}';

            if (isset($_GET['callback'])) {
                echo $_GET['callback'].'('. $resposta.')';
            } else {
                echo ($resposta); 
            }
        }
    } else {
    }
    mysqli_free_result($result);
}

AJAX返回此响应（来自控制台输出）：

jQuery21404145154874458823_1465128123711（{＆＃34; itemName＆＃34;：＆＃34; Sticker | wayLander | MLG Columbus 2016＆＃34;，＆＃34; itemPrice＆＃34; ：＆＃34; 0.33＆＃34;＆＃34; itemUrl＆＃34;：＆＃34; http://steamcommunity-a.akamaihd.net/economy/image/-9a81dlWLwJ2UUGcVs_nsVtzdOEdtW wKGZZLQHTxDZ7I56KU0Zwwo4NUX4oFJZEHLbXQ9QVcJY8gulReQ0DFSua4xJ2DAgs7Ng1Qia2kMgth3ffNY3MXtd3uwteJlvKjN-KHxDsGvMYl2byYrNrxjQ3n80Q-Nzz2JIaWIQ5rfxiOrWILvkNz＆＃34; }）jQuery21404145154874458823_1465128123711（{＆＃34; itemName＆＃34;：＆＃34; Shadow Case＆＃34;，＆＃34; itemPrice＆＃34;：＆＃34; 0.03＆＃34;，＆＃34; itemUrl＆＃34 ;：＆＃34; HTTP ：//steamcommunity-a.akamaihd.net/economy/image/-9a81dlWLwJ2UUGcVs_nsVtzdOEdtWwKGZZLQHTxDZ7I56KU0Zwwo 4NUX4oFJZEHLbXU5A1PIYQNqhpOSV-fRPasw8rsUFJ5KBFZv668FF4u1qubIW4Su4mzxYHbzqGtZ-KGlz8EuJcg3rnE9NiijVe3_UY-Zzr2JJjVLFEEeiQRtg＆＃34; }）jQuery21404145154874458823_1465128123711（{＆＃34; itemName＆＃34;：＆＃34; Revolver Case＆＃34;，＆＃34; itemPrice＆＃34;：＆＃34; 0.03＆＃34;，＆＃34; itemUrl＆＃34 ;：＆＃34; HTTP ：//steamcommunity-a.akamaihd.net/economy/image/-9a81dlWLwJ2UUGcVs_nsVtzdOEdtWwKGZZLQHTxDZ7I56KU0Zwwo 4NUX4oFJZEHLbXU5A1PIYQNqhpOSV-fRPasw8rsUFJ5KBFZv668FFYwnfKfcG9HvN7iktaOkqD1auLTxD5SvZYgiLvFpo7xjVLh-kdrYWnzcoGLMlhpsyM-5vg＆＃34; }）jQuery21404145154874458823_1465128123711（{＆＃34; itemName＆＃34;：＆＃34; Five-SeveN | Forest Night（破损不堪）＆＃34;，＆＃34; itemPrice＆＃34; ：＆＃34; 0.03＆＃34;＆＃34; itemUrl＆＃34;：＆＃34; http://steamcommunity-a.akamaihd.net/economy/image/-9a81dlWLwJ2UUGcVs_nsVtzdOEdtW wKGZZLQHTxDZ7I56KU0Zwwo4NUX4oFJZEHLbXH5ApeO4YmlhxYQknCRvCo04DEVlxkKgposLOzLhRlxfbGTjVb09q5hoWYg8j6OrzZglRc7cF4n-SP9o2gjQbhqRVla2GnJ46VIQA_ZlyD-VHvxuu808S7tc7NzXFm7iUh4mGdwULQPtyarQ＆＃34; }）

如何从JSON元素访问属性ItemName，ItemPrice和ItemUrl？

Answer 1

您的Ajax调用配置为使用'jsonp'而不是'json'，它不会为您提供您想要/期望的结果。您应该将dataType选项更改为'json'，然后删除jsonp选项（无论如何都会被忽略）。

您应该能够使用jQuery.each来处理Ajax success回调中JSON数组的每个元素。这样您就可以按预期访问ItemName，ItemPrice和ItemUrl。

以下是更新的Ajax代码：

setInterval(function() {
    $.ajax({
        url: "http://localhost/Seyroku/extras/mostrarBote.php",
        dataType: "jsonp",
        jsonp: "callback",
        data: {"a": "a"},
        success: function (respJSON) {
            $.each(respJSON, function(index, item) {
                var itemName = item.itemName;
                var itemPrice = item.itemPrice;
                var itemUrl = item.itemUrl;
                console.log("JSON array[", index, "]: ItemName=", itemName, " ItemPrice=", itemPrice, " ItemUrl=", itemUrl);
                //TODO: Do something useful with itemName, itemPrice, and itemUrl here
            });
        }
    });
}, 5000);

处理返回JSON对象数组的Ajax响应

1 个答案: