Question

这是我的回复

Array ([code] => 202 [message] => Accepted [data] => Array ( [resultMap] => Array ( [D3856~H158] => Array ( [AppDay] => * [HosTown] => Colombo 06 [SpecName] => Physiotherapist/Sports Physiotherapist [HosName] => Revival Healthcare Services (Pvt)-Colombo [SpecializationId] => 333 [HosCode] => H158 [AppDate] => Any [DocName] => MR CHAMARA MATHANGAWEERA [DoctorNo] => D3856 ) ) ) [detailMessage] => Success )

现在我想为这个值分配变量并以正确的方式回声我尝试这个但它给出了一个错误消息

未定义的索引：DocName

这是我的代码 ////////////////////////////////////

if( $response ){
    if ( isset($result->error) )die( $result->error_message );
    /* Convert json data to array */
    $arr=json_decode( $response, true );
    //print_r($arr);
    foreach($arr['data'] as $data)
    {
        $output="Doctor".$data['DocName']."<br/>";
        $output="Doctor".$data['SpecName']."<br/>";
        $output="Doctor".$data['HosName']."<br/>";
        $output="Doctor".$data['Day']."<br/>";
        $output="Doctor".$data['Date']."<br/>";
    }

Answer 1

查看数组的结构。您要查找的值嵌套了几个数组：

$data = array (
    'code' => 202,
    'message' => 'Accepted', 
    'data' => array ( 
        'resultMap' => array ( 
            'D3856~H158' => array ( 
                'AppDay' => '*', 
                'HosTown' => 'Colombo 06',
                'SpecName' => 'Physiotherapist/Sports Physiotherapist',
                'HosName' => 'Revival Healthcare Services (Pvt)-Colombo',
                'SpecializationId' => 333,
                'HosCode' => 'H158',
                'AppDate' => 'Any',
                'DocName' => 'MR CHAMARA MATHANGAWEERA',
                'DoctorNo' => 'D3856',
            ) 
        ) 
    ),
    'detailMessage' => 'Success'
);

打印resultMap中第一条记录的所有值：

foreach(current($data['data']['resultMap']) as $key => $value) {
    echo $key . " => " . $value . "<br />";
}

您需要一个额外的循环才能在resultMap中打印所有记录。

Answer 2

你可以试试这个工作正常对于这个固定的数组值意味着给定

foreach($arr['data']['resultMap']['D3856~H158'] as $data)
 {
        $output="Doctor".$data['DocName']."<br/>";
        $output="Doctor".$data['SpecName']."<br/>";
        $output="Doctor".$data['HosName']."<br/>";
        $output="Doctor".$data['Day']."<br/>";
        $output="Doctor".$data['Date']."<br/>";


 }

或更改的数组值给出了答案

foreach($arr['data']['resultMap'] as $data) { $output="Doctor".$data['DocName']." "; $output="Doctor".$data['SpecName']." "; $output="Doctor".$data['HosName']." "; $output="Doctor".$data['Day']." "; $output="Doctor".$data['Date']." "; }

用于json回复的php变量

2 个答案: