Question

我正在创建一个API，但是最终服务器以JSON格式向我发送回信息，其中包括RECORD属性：

{
    "RECORD": [{
        "@ID": "1",
        "FULLNAME": "*\"* **** ****",
        "PHONE": "*******",
        "CELLULAR": "********",
        "LOGIN_STATUS": "*",
        "LOGIN_STATUS_TEXT": "****",
        "STUDENT_ACADEMIC_YEAR": "",
        "STUDENT_DEPARTMENT": "",
        "STUDENT_SPECIALITY": "",
        "STUDENT_PHONE": "",
        "STUDENT_CELLULARPHONE": "",
        "STUDENT_ADDRESS": " *",
        "STUDENT_EMAIL": "",
        "STUDENT_STATUS": "",
        "STUDENT_ID": "*",
        "TEACHER_ID": "*******",
        "CURRENTYEAR": "****",
        "TOKEN": "*************",
        "CURRENTFULLYEAR": "****"
    }]
}

如何从内部属性中提取数据？我使用以下命令进行了解码：

$jsonRestData=json_decode($jsonRestData2, true);

我尝试过：

$request_json["attributes"] = array( 
        "userid" => str_replace(" ","",$user_uid),
        "fullname" => $jsonRestData->FULLNAME,
        "email" => $jsonRestData->STUDENT_EMAIL,
        "role" => $jsonRestData->STUDENT_STATUS,
        "year" => $jsonRestData->STUDENT_ACADEMIC_YEAR,
        "department" => $jsonRestData->STUDENT_DEPARTMENT,
        "speciality" => $jsonRestData->STUDENT_SPECIALITY
    );

我也尝试过：

 $request_json["attributes"] = array( 
            "userid" => str_replace(" ","",$user_uid),
            "fullname" => $jsonRestData->RECORD->FULLNAME,
            "email" => $jsonRestData->RECORD->STUDENT_EMAIL,
            "role" => $jsonRestData->RECORD->STUDENT_STATUS,
            "year" => $jsonRestData->RECORD->STUDENT_ACADEMIC_YEAR,
            "department" => $jsonRestData->RECORD->STUDENT_DEPARTMENT,
            "speciality" => $jsonRestData->RECORD->STUDENT_SPECIALITY
        );

对于第一个示例，我遇到一个错误：未定义的属性：stdClass :: $ FULLNAME

第二次出现错误：试图获取非对象的属性

Answer 1

更新：

如果对“”进行了大写，则代码有效，请检查以下内容：http://sandbox.onlinephpfunctions.com/code/9b1192db07cb3876a9c80e9ca25f88406117c0b5

您的代码中的问题出在JSON字符串中，您必须全名将字符串固定在其中。（修复结束字符串“”）

$json_string='{
"RECORD": [{
    "@ID": "1",
    "FULLNAME": "*",
    "PHONE": "*******",
    "CELLULAR": "********",
    "LOGIN_STATUS": "*",
    "LOGIN_STATUS_TEXT": "****",
    "STUDENT_ACADEMIC_YEAR": "",
    "STUDENT_DEPARTMENT": "",
    "STUDENT_SPECIALITY": "",
    "STUDENT_PHONE": "",
    "STUDENT_CELLULARPHONE": "",
    "STUDENT_ADDRESS": " *",
    "STUDENT_EMAIL": "",
    "STUDENT_STATUS": "",
    "STUDENT_ID": "*",
    "TEACHER_ID": "*******",
    "CURRENTYEAR": "****",
    "TOKEN": "*************",
    "CURRENTFULLYEAR": "****"
}]
}';

$json_array=json_decode($json_string,true);
$records=$json_array['RECORD'][0];

//example of use
echo "FULL NAME : ".$records['FULLNAME'];

您可以使用此链接来检查或验证JSON STRING：https://jsonformatter.curiousconcept.com/

Answer 2

您的原始代码（第二部分，带有RECORD的代码）只有一个问题：假设RECORD是一条记录。但是显然，这是一个包含单个记录的数组。

关于将true放在json_decode中，在如此多的上下文中这并不重要，因为它没有给出明确的利弊。但是，如果您确实在此处使用true，则需要相应地调整代码，因为使用true时，输出是嵌套数组，而没有使用true时，输出是嵌套对象和数组。

这里有一个示例PHP，展示了两种实现方式-使用<?php $jsonRestData2 = '{ "RECORD": [{ "@ID": "1", "FULLNAME": "*\"* **** ****", "PHONE": "*******", "CELLULAR": "********", "LOGIN_STATUS": "*", "LOGIN_STATUS_TEXT": "****", "STUDENT_ACADEMIC_YEAR": "", "STUDENT_DEPARTMENT": "", "STUDENT_SPECIALITY": "", "STUDENT_PHONE": "", "STUDENT_CELLULARPHONE": "", "STUDENT_ADDRESS": " *", "STUDENT_EMAIL": "", "STUDENT_STATUS": "", "STUDENT_ID": "*", "TEACHER_ID": "*******", "CURRENTYEAR": "****", "TOKEN": "*************", "CURRENTFULLYEAR": "****" }] }'; $jsonRestData = json_decode($jsonRestData2); $request_json = []; $request_json["attributes"] = array( "fullname" => $jsonRestData->RECORD[0]->FULLNAME, "email" => $jsonRestData->RECORD[0]->STUDENT_EMAIL, "role" => $jsonRestData->RECORD[0]->STUDENT_STATUS, "year" => $jsonRestData->RECORD[0]->STUDENT_ACADEMIC_YEAR, "department" => $jsonRestData->RECORD[0]->STUDENT_DEPARTMENT, "speciality" => $jsonRestData->RECORD[0]->STUDENT_SPECIALITY, ); print_r($request_json); $jsonRestData = json_decode($jsonRestData2, true); $request_json = []; $request_json["attributes"] = array( "fullname" => $jsonRestData['RECORD'][0]['FULLNAME'], "email" => $jsonRestData['RECORD'][0]['STUDENT_EMAIL'], "role" => $jsonRestData['RECORD'][0]['STUDENT_STATUS'], "year" => $jsonRestData['RECORD'][0]['STUDENT_ACADEMIC_YEAR'], "department" => $jsonRestData['RECORD'][0]['STUDENT_DEPARTMENT'], "speciality" => $jsonRestData['RECORD'][0]['STUDENT_SPECIALITY'], ); print_r($request_json);和不使用它。

Array
(
    [attributes] => Array
        (
            [fullname] => *"* **** ****
            [email] =>
            [role] =>
            [year] =>
            [department] =>
            [speciality] =>
        )

)
Array
(
    [attributes] => Array
        (
            [fullname] => *"* **** ****
            [email] =>
            [role] =>
            [year] =>
            [department] =>
            [speciality] =>
        )

)

这是示例代码输出的内容：

public class client {

private static final AtomicLong counter = new AtomicLong();

private final long id;

public client() {
    this.id = counter.getAndIncrement();
}

public long getId() {
    return id;
}
}

如您所见，输出是相同的，即两种方法都可以正常工作。

Answer 3

您必须使用$data = json_decode($jsondata, true);

之后，您可以将数据作为数组访问

如何在php中提取内部JSON变量？

3 个答案: