Question

有人可以帮我解决这个问题，我有以下json字符串

string(1223) "YAHOO.Finance.SymbolSuggest.ssCallback({"ResultSet":{"Query":"google","Result":[{"symbol":"GOOG","name": "Google Inc.","exch": "NMS","type": "S","exchDisp":"NASDAQ","typeDisp":"Equity"},{"symbol":"GOOG.MX","name": "GOOGLE-A","exch": "MEX","type": "S","exchDisp":"Mexico","typeDisp":"Equity"},{"symbol":"GGQ1.F","name": "GOOGLE-A","exch": "FRA","type": "S","exchDisp":"Frankfurt","typeDisp":"Equity"}]}})"

但我似乎无法随心所欲。基本上我想循环出结果

[{"symbol":"GOOG","name": "Google Inc.","exch": "NMS","type": "S","exchDisp":"NASDAQ","typeDisp":"Equity"},{"symbol":"GOOG.MX","name": "GOOGLE-A","exch": "MEX","type": "S","exchDisp":"Mexico","typeDisp":"Equity"},{"symbol":"GGQ1.F","name": "GOOGLE-A","exch": "FRA","type": "S","exchDisp":"Frankfurt","typeDisp":"Equity"}]

很抱歉，我的问题是如何循环甚至打印第一个结果，例如

{"symbol":"GOOG","name": "Google Inc.","exch": "NMS","type": "S","exchDisp":"NASDAQ","typeDisp":"Equity"}

Answer 1

您的字符串不是JSON，它是JSON-in-Script。注意片段说：

YAHOO.Finance.SymbolSuggest.ssCallback(...)

当浏览器收到上述脚本（实际上是一个javascript代码）时，它将调用YAHOO.Finance.SymbolSuggest.ssCallback函数，并将JSON数据作为参数传递。

您没有提及是否要访问服务器端或客户端上的JASON数据？它是服务器端（PHP），然后您可以使用正则表达式或字符串替换函数来提取您喜欢的部分。您可以使用json_decode()函数将结果字符串转换为关联数组。

编辑----

将JSONP转换为JSON的快速而肮脏的黑客行为：

<?php
    $text = 'YAHOO.Finance.SymbolSuggest.ssCallback({"ResultSet":{"Query":"google","Result":[{"symbol":"GOOG","name": "Google Inc.","exch": "NMS","type": "S","exchDisp":"NASDAQ","typeDisp":"Equity"},{"symbol":"GOOG.MX","name": "GOOGLE-A","exch": "MEX","type": "S","exchDisp":"Mexico","typeDisp":"Equity"},{"symbol":"GGQ1.F","name": "GOOGLE-A","exch": "FRA","type": "S","exchDisp":"Frankfurt","typeDisp":"Equity"}]}})';
    # //CONVERT JSONP to JSON\\
    $text = preg_replace('/.+?({.+}).+/', '$1', $text); 
    # \\CONVERT JSONP to JSON//
    $data = json_decode($text);
    var_dump($data);
    var_dump($data->ResultSet->Result[0]);
    var_dump($data->ResultSet->Result[0]->symbol);
    var_dump($data->ResultSet->Result[0]->name);
    # etc etc
?>

Answer 2

您的结果不仅仅是一个JSON字符串，它是一个JSON函数调用前面的JSON字符串。这肯定是一个JSONP电话。您必须编写YAHOO.Finance.SymbolSuggest.ssCallback(data) javascript函数并在那里获取Json。检查JSONP查询，如果你想要另一个名字，你应该能够改变这个反向引用函数的名称，它通常是在GET查询中的参数上。

现在你可能直接从PHP调用它而你不在js环境中。因此，您必须在PHP代码中编写一些内容，以删除YAHOO.Finance.SymbolSuggest.ssCallback(部分以及之前的) 将其解析为JSON数据..

php和json的问题

2 个答案: