这个问题不完全是一个编程问题。我有要求显示图片的JSON输出。我知道json_encode()
的用法,但问题是我没有得到如何逻辑我会显示这些细节。PHP中表格的JSON输出
我得到的输出是:
{"name":"ZXY","success":1,"subjects":["DIGITAL COMMUNICATION TECHNIQUE(2012-13)","DIGITAL SIGNAL PROCESSING(2012-13)","ANTENNAS AND WAVE PROPAGATION(2012-13)","DIGITAL SWITCHING THEORY AND NETWORKS(2012-13)","ACCOUNTING FOR MANAGERS(2012-13)","DIGITAL COMMUNICATION TECHNIQUE LAB(2012-13)","DIGITAL SIGNAL PROCESSING LAB(2012-13)"]}
。
请告诉我什么都可以的我有细节的有效JSON输出,
编辑:我得从MySQL数据输出JSON。我在图片中提供的学生的示例数据。我已经显示的JSON是PHP页面的结果,我不满意我得到的JSON输出。我正在寻求一种更好的方式来以JSON格式有效地表示数据。
这是我为获取JSON输出而编写的PHP代码。
<?php
$regno = $_GET['regno'];
$host = 'localhost';
$user = 'root';
$password = '';
$database = 'android_app_details';
$dbc = mysqli_connect($host,$user,$password,$database)
or die('Error in connecting the database');
$query =
"SELECT SD.REGNO, STD.FNAME, STD.MNAME, STD.LNAME, SD.SUBJECT_NAME, AD.PRESENT_COUNT, AD.TOTAL_COUNT ".
"FROM ATTENDANCE_DETAILS AS AD, SUBJECT_DETAILS AS SD, STUDENT_DETAILS STD ".
"WHERE AD.ENROLLMENT_ID = SD.ENROLLMENT_ID ".
"AND SD.REGNO = '".$regno."' ".
"AND AD.BATCH_SUBJECT_ID = SD.BATCH_SUBJECT_ID ".
"AND SD.STUDENT_ID = STD.STUDENT_ID ";
$result = mysqli_query($dbc, $query)
or die('Error in query');
$subject = Array();
$att = Array();
$total = Array();
$i = 0;
$success = 0;
while($row = mysqli_fetch_array($result)){
if($row['MNAME'])
$name = $row['FNAME'].' '.$row['MNAME'].' '.$row['LNAME'];
else
$name = $row['FNAME'].' '.$row['LNAME'];
$subject[$i] = $row['SUBJECT_NAME'];
$i++;
$success = 1;
}
if($success)
$response = array(
"name" => $name,
"success" => $success,
"subjects" => $subject
);
else
$response = array(
"success" => $success
);
echo json_encode($response);
?>
发表一些代码。 – Yogus
您当前的JSON输出究竟有什么问题? – Onheiron
那么,首先,主体没有任何'标记名',因此访问它们将会很困难.. – NewUser