javascript - JSONP/Ajax 脚本出现问题 - 返回 'Undefined'

Question

我正在尝试从我无法控制的网络服务器检索数据。 我尝试过谷歌，但似乎没有任何解决办法。我尝试了一些不同的代码，包括带数据表和不带数据表的代码。

如果有人能指出我正确的方向，我将不胜感激。

 <script>
$(document).ready(function() {
$('#example').DataTable( {
    "ajax": {
        "url": 'https://Example.com',
        "dataSrc": "",
        "dataType": "jsonp"
    },
    "columns": [
        { "data": "name" },
        { "data": "description" }
    ]
} );
} );
</script>

当我在 Chrome 网页上运行代码时，在开发人员工具/网络中，我得到的 url 请求结果

[{"ID":10,"VIRTUAL":"No","QTY":2,"NAME":"Alto TS Sub 15\"","ALT_NAME":"Alto TS15","DESCRIPTION":"","MEMO":"","PRICEA":0,"PRICEB":0,"PRICEC":0,"PRICE_TYPE":"per Day","TAX_RATE":"0%","STATUS":"Active","CATEGORY":"Speakers","CATEGORY_ID":9,"CATEGORY_LFT":18,"CATEGORY_RGT":19,"IMAGE_ID":101,"IMAGE_URL":"https:\/\/www.hirehop.info\/uploads\/Alto-TS-Sub-15in\/1758_101.png","REPLACE_COST":300,"WEIGHT":"31 kg","WIDTH":0.451,"LENGTH":0.513,"HEIGHT":0.591,"SERVICE_INTERVAL":"P6M","PAT_INTERVAL":"P1Y","TEST_INTERVAL":"P6M","LOCATION":"Greenway Avenue","PART_NUMBER":"","BARCODE":"","LAST_UPDATE":"2019-09-30 21:42:15","MARKETING":1,"NOMINAL_CODE":"","LINKS":[],"CUSTOM_FIELDS":[]},{"ID":9,"VIRTUAL":"No","QTY":2,"NAME":"Electro-Voice ZLX-12P","ALT_NAME":"EV-ZLX12P","DESCRIPTION":"12\" Active PA Speaker","MEMO":"","PRICEA":0,"PRICEB":0,"PRICEC":0,"PRICE_TYPE":"per Day","TAX_RATE":"0%","STATUS":"Active","CATEGORY":"Speakers","CATEGORY_ID":9,"CATEGORY_LFT":18,"CATEGORY_RGT":19,"IMAGE_ID":100,"IMAGE_URL":"https:\/\/www.hirehop.info\/uploads\/Electro-Voice-ZLX-12P\/1758_100.png","REPLACE_COST":350,"WEIGHT":"15.6 kg","WIDTH":0.35,"LENGTH":0.35,"HEIGHT":0.65,"SERVICE_INTERVAL":"P6M","PAT_INTERVAL":"P1Y","TEST_INTERVAL":"P3M","LOCATION":"Greenway Avenue","PART_NUMBER":"","BARCODE":"3","LAST_UPDATE":"2019-09-30 21:27:27","MARKETING":1,"NOMINAL_CODE":"","LINKS":[],"CUSTOM_FIELDS":[]}]

Answer 1

你说:

"dataType": "jsonp"

...但 URL 返回 JSON，而不是 JSONP。

您必须使用与您获取的数据类型相匹配的数据类型(或者完全忽略它并相信响应中的 Content-Type header 是正确的)。