我无法使用PHP从JSON结果中提取数据

这是JSON查询

的结果
{
  "ACSExecution_HasError": false,"ACSExecutionErrorMessage": "","ACSOutputResponce": {
    "ACSValueOutput": [
      {
        "ACSObjectOutput": [
          {
            "GeoID": 1329068,"Resolved_Street": "kolopetinitsas","Resolved_Street_Num": "48","Resolved_Zip": "56432","Resolved_Area": "tsalonika","Resolved_Long": 32.9179268,"Resolved_Lat": 30.6484871,"Resolved_GeoDataType": 1,"Resolved_GeoDataID": 1329068,"Resolved_Station_ID": "ΘΟ","Resolved_Branch_ID": 1,"Resolved_As_Inaccesible_Area_With_Cost": 0,"Resolved_As_Inaccesible_Area_WithOut_Cost": 0,"Resolved_Confidence": 87,"Resolved_GeoRegionType": 2,"Resolved_Providence": "N.Thessalonikis","Resolved_Correction": "000","Resolved_Station_Descr": "Monasthriou (2310-567462)","AddressID": ""
          }
        ]
      }
    ],"ACSTableOutput": {}
  }
}

我要提取值Resolved_Station_ID 我尝试过

$json = json_decode($response1,true);// decode the JSON feed 
echo "STATION ID=".$json->ACSValueOutput[0]->ACSObjectOutput[0]->Resolved_Station_ID;

但是它无法正常工作,我得到了这些错误

注意:尝试获取/home/deorum/public_html/pveshop.gr/wp-content/plugins/insert-php-code-snippet/shortcode-handler.php(65)中非对象的属性'ACSValueOutput' :eval()在第116行上的代码

注意:尝试获取/home/deorum/public_html/pveshop.gr/wp-content/plugins/insert-php-code-snippet/shortcode-handler.php(65)中非对象的属性'ACSObjectOutput' :eval()在第116行上的代码

注意:尝试获取/home/deorum/public_html/pveshop.gr/wp-content/plugins/insert-php-code-snippet/shortcode-handler.php(65)中非对象的属性“ Resolved_Station_ID” :eval()在第116行上的代码

lilyweiwei 回答:我无法使用PHP从JSON结果中提取数据

应该是:

$json["ACSOutputResponce"]["ACSValueOutput"][0]["ACSObjectOutput"][0]["Resolved_Station_ID"]

您不能将->与数组一起使用,而只能与对象一起使用。由于true中的json_decode()参数,您拥有的是一个关联数组。如果更改为json_decode(...,false),则可以通过以下方式获取它:

$json->ACSOutputResponce->ACSValueOutput[0]->ACSObjectOutput[0]->Resolved_Station_ID
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