Question

我的php文件创建了一个jsonp响应，它有4个独立的对象和自己的嵌套。我想知道如何通过一个嵌套来执行$ .each（），而不是一次检索所有值，因为我对每个值的处理方式不同。例如，我只想通过[SubmittedAddress]下的$ .each函数浏览所有数据。我通过jquery ajax函数调用它。这是非jsonp格式的数组，便于阅读：

Array
(
    [SubmittedAddress] => Array
        (
            [address1] => 850 BRYANT STREET
            [address2] => 
            [addresscombined] => 850 BRYANT STREET
            [city] => SAN FRANCISCO
            [state] => CA
            [zip] => 94106
        )

    [CorrectAddress] => Array
        (
            [address1] => SUPERIOR COURT
            [addresscombined] => SUPERIOR COURT 850 BRYANT ST STE 306
            [address2] => 850 BRYANT ST STE 306
            [city] => SAN FRANCISCO
            [state] => CA
            [zip] => 94103
            [zip4] => 4667
        )

    [PercentMatch] => Array
        (
            [addresscombined] => 39
            [city] => 100
            [state] => 100
            [zip] => 80
        )

    [Response] => success
)

以下是我为jsonp获取的每个值，而不仅仅是[SubmittedAddress]中的值

success: function(data, textStatus){
    $.each(data, function(index, object) {
        $.each(object, function(property, value) {
            alert(property + "=" + value);
        }); 
    });
},

提前谢谢。

Answer 1

如果你想处理一个属性，那么就这样做，例如

success: function(data, textStatus){
    $.each(data.SubmittedAddress, function(index, object) {
        alert(property + "=" + value);
    });
    $.each(data.CorrectAddress, function(index, object) {
        console.log(property + "=" + value);
    });
    //etc
},

仅从嵌套的Json数组中的一个对象读取数据

1 个答案: