在PHP中提取JSONP结果集

Question

我希望能够获取此url的返回数据。我什至可以在PHP中做到这一点？

    <?php
    $yahooSS = "http://d.yimg.com/autoc.finance.yahoo.com/autoc?query=yahoo&callback=YAHOO.Finance.SymbolSuggest.ssCallback";

    $yss = fopen($yahooSS,"r");
    ....

我相信这会返回Javascript回调函数，但是我不知道从哪里开始。

以下是返回的结果集的示例。

YAHOO.Finance.SymbolSuggest.ssCallback({"ResultSet":{"Query":"yahoo","Result":[{"symbol":"YHOO","name": "Yahoo! Inc.","exch": "NMS","type": "S","exchDisp":"NASDAQ","typeDisp":"Equity"},{"symbol":"YAHOY.PK","name": "YAHOO JAPAN CORP","exch": "PNK","type": "S","exchDisp":"Pink Sheets","typeDisp":"Equity"},{"symbol":"ETD","name": "Citigroup Inc. ELKS On Yahoo","exch": "PCX","type": "S","typeDisp":"Equity"},{"symbol":"YOJ.BE","name": "YAHOO JAPAN","exch": "BER","type": "S","exchDisp":"Berlin","typeDisp":"Equity"},{"symbol":"YHO.SG","name": "YAHOO","exch": "STU","type": "S","exchDisp":"Stuttgart","typeDisp":"Equity"},{"symbol":"YAHOF.PK","name": "YAHOO JAPAN CORP","exch": "PNK","type": "S","exchDisp":"Pink Sheets","typeDisp":"Equity"},{"symbol":"YHO.HM","name": "YAHOO","exch": "HAM","type": "S","exchDisp":"Hamburg","typeDisp":"Equity"},{"symbol":"YOJ.DE","name": "YAHOO JAPAN","exch": "GER","type": "S","exchDisp":"XETRA","typeDisp":"Equity"},{"symbol":"YHO.DU","name": "YAHOO","exch": "DUS","type": "S","exchDisp":"Dusseldorf Stock Exchange","typeDisp":"Equity"},{"symbol":"YHOO.BA","name": "YAHOO  INC.","exch": "BUE","type": "S","exchDisp":"Buenos Aires","typeDisp":"Equity"}]}})

任何帮助是极大的赞赏。

Answer 1

正确，它是 带有填充的JSON。
您必须删除函数名称（和括号），然后可以使用解析JSON json_decode。

我曾经为此写过一个函数：

function jsonp_decode($jsonp, $assoc = false) { // PHP 5.3 adds depth as third parameter to json_decode
    if($jsonp[0] !== '[' && $jsonp[0] !== '{') { // we have JSONP
       $jsonp = substr($jsonp, strpos($jsonp, '('));
    }
    return json_decode(trim($jsonp,'();'), $assoc);
}

用法：

$data = jsonp_decode($response);