JSON文件:无法从JSON返回值
{
"stock": [
{
"symbol": "AVGO",
"name": "Broadcom Ltd.",
"sector": "Technology",
"industry": "Semiconductors",
"SCTR": "97.8",
"delta": "2.5",
"close": "154.97",
"vol": "2297509"
},
{
"symbol": "CBS",
"name": "CBS Corp.",
"sector": "Cyclicals",
"industry": "Entertainment",
"SCTR": "92.3",
"delta": "-3.6",
"close": "53.58",
"vol": "4045416"
}
]
}
PHP:
$jsona = json_decode($json, true);
echo $jsona['stock']['symbol'];
为什么我不能得到的符号? 如何获得所有符号值? 谢谢!
做'的print_r($ jsona);'然后就看到结构和你有多个符号值。 – Rizier123
原始JSON在这里,但我无法得到我的价值http://stockcharts.com/j-sum/sum?cmd=sctr&view=L&timeframe=W – Fireghost
@FelixKling它是有效的json。将它放入jsonlint中可能只是有一个复制错误。 – Rizier123