已成功獲取JSON PHP數據，但無法訪問它

我正在使用PHP向另一個Web服務器發出HTTP請求以檢索JSON數據，該請求似乎可行，並且我可以使用var_dump查看數據，但我不確定我可以訪問數據。已成功獲取JSON PHP數據，但無法訪問它

這是我獲得的數據代碼：

$response = Requests::get($mydetails); 

$array1 = json_decode(json_encode($response), true);

幾乎所有的我都試過的東西（如回聲$ array1->操作 - > result->狀態）給我同樣的錯誤信息：

PHP Notice: Trying to get property of non-object

這是的var_dump輸出的前幾行：

array(9) { 
    ["body"]=> 
    string(7017) "{"operation":{"result":{"status":"Success","message":"Request details listed successfully"},"details":[{"fields":[{"name":"requesttype","value":"Incident","type":"Pick List","displayvalue":"Request Type","allowedvalues":["Incident","Request for Change","Request For Information","Request for Work"],"mandatory":"false"},{"name":"impact","value":"Affects User","type":"Pick List","displayvalue":"Impact","allowedvalues":["Affects Business","Affects Department","Affects Group","Affects User"],"mandatory":"false"},{"name":"status","value":"Closed","type":"Pick List","displayvalue":"Status","allowedvalues":["Awaiting Customer","Awaiting Third Party","Closed","More Information Required","Open","Resolved"],"mandatory":"false"},{"name":"impactdetails","value":"-","type":"Multi Line","displayvalue":"Impact Details","allowedvalues":[""],"mandatory":"false"},{"name":"mode","value":"Phone Call","type":"Pick List","displayvalue":"Mode","allowedvalues":["Customer Portal","E-Mail","Fax","On-Site Request","Phone Call"],"mandatory":"false"},{"name":"priority","value":"Severity 2","type":"Pick List","displayvalue":"Priority","allowedvalues":["National Level 1","National Level 2","Request for Change","Severity 1","Severity 2","Severity 3","Severity 4"],"mandatory":"false"},{"name":"department","value":"General Practice","type":"Not Editable","displayvalue":"Department","allowedvalues":[""],"mandatory":"true"},{"name":"requesteremail","value":"","type":"Not Editable","displayvalue":"E-mail","allowedvalues":[""],"mandatory":"false"},{"name":"requestercontactnumber","value":"","type":"Not Editable","displayvalue"

來源

2014-02-18 djh123

你爲什麼要做'json_decode（json_encode（$ response），true）;'？什麼是編碼，然後立即解碼？什麼是'$ response'？爲什麼你不能直接使用它？你想用這條線解決什麼問題？ –

P.S. '$ array1'不是一個對象，它是一個數組（即''json_decode（）'的第二個參數）。 'echo'array1 ['operation'] ['result'] ['status']' –

我設法通過執行以下操作來解決問題： $ response = Requests :: get（$ mydetails）; $ array = json_decode（$ response - > body）; echo $ array-> operation-> result-> status; – djh123

的DECOD後的結構ing是一個array，所以你應該使用像$array1['body']['operation']['result']['status']這樣的數組運算符來訪問它，而不是->--它是對象遍歷運算符。你也可以將它解碼作爲一個對象：

$response = Requests::get($mydetails); 

$array1 = json_decode(json_encode($response));

但你的反應似乎已經是一個array，所以只是訪問它，而無需重新編碼，也許？

編輯： 後做你的代碼中的一些假設，你也許應該這樣做：

$response = Requests::get($mydetails); 

$array1 = json_decode($response['body'], true); 
var_dump($array1); // looks ok?

來源

2014-02-18 15:32:06 ddinchev

感謝您的回覆，我刪除了'true'選項，並將我的代碼更改爲以下內容，但仍然無法使用echo $ jsondata-> operation-> result-> status訪問數據：： $ response =請求::得到（$ myMANAGEENGINEAddressfull）; $ array = json_decode（json_encode（$ response））; $ jsondata = $ array-> body; – djh123

你能解釋一下你爲什麼要做'json_decode（json_encode（$ response））'？這將從PHP數組（json_encode）創建一個JSON字符串，然後將從剛剛編碼的JSON字符串（'json_decode'）創建相同的數組。這不是你應該做的！你可以在獲得它之後立即顯示'var_dump（$ response）'，而不需要解碼或編碼任何東西？ – ddinchev

如果使用true屬性爲JSON解碼你有你自己的JSON關聯數組所以你會需要訪問示波器

$array1['operation']['result']['status'];

來源

2014-02-18 15:32:38 Fabio

謝謝您的回覆，但是當我嘗試給出錯誤「PHP Notice：Undefined index：operation」 – djh123

已成功獲取JSON PHP數據，但無法訪問它

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