JQuery从PHP JSON字符串返回undefined


JQuery returning undefined from PHP JSON string

我正在尝试这个JQuery代码:

val = $(this).val();
var data = {
    "action": "test"
};
data = $(this).serialize() + "&" + $.param(data);
$.ajax({
    type: "POST",
    dataType: "json",
    url: "?getCustomer=1&sequence="+val+"",
    data: data,
    success: function(data) {
        alert(data["sequence"]);
    }
});

但是CCD_ 1正在返回CCD_ 2

如果我检查URL(?getCustomer=1&sequence=4),我会返回以下信息:

[{"sequence":"53"}]

因此序列值定义为

这就是控制台中显示的内容:

[Object]0: Objectsequence: "112"__proto__: Objectlength: 1__proto__: Array[0]concat: concat()constructor: Array()copyWithin: copyWithin()entries: entries()every: every()fill: fill()filter: filter()find: find()findIndex: findIndex()forEach: forEach()indexOf: indexOf()join: join()keys: keys()lastIndexOf: lastIndexOf()length: 0map: map()pop: pop()push: push()reduce: reduce()reduceRight: reduceRight()reverse: reverse()shift: shift()slice: slice()some: some()sort: sort()splice: splice()toLocaleString: toLocaleString()toString: toString()unshift: unshift()Symbol(Symbol.iterator): values()Symbol(Symbol.unscopables): Object__proto__: Object

您误解了json。更改

alert(data["sequence"]);

alert(data[0]["sequence"]);

或更好的

alert(data[0].sequence );

[{"sequence":"53"}]是一个数组,其第一个元素是具有序列成员的对象。你需要alert(data[0]['sequence'])