JSON verilerini PHP'den nasıl çıkarabilirim ve bir veri tablosunda görüntüleyebilirim?

Question

Veritabanımdan raporları PHP kullanarak JSON biçimine dönüştürerek ve AJAX JQuery ile görüntüleyerek raporları almak istiyorum. Ancak verileri doğru şekilde iletemiyorum. Birisi bana hatamı söyler mi lütfen?

$(document).ready(function() { 
 

 
    $("#searchOverall").click(function() { 
 

 

 
    var ay = $("#Overall_acadyear").val(); 
 
    var year = $("#Overall_year").val(); 
 

 
    if (ay === undefined || ay == '') { 
 

 
     alert("Select Academic year."); 
 

 
    } else if (year === undefined || year == '') { 
 

 
     alert("Select year level."); 
 

 
    } else { 
 

 

 
     $.ajax({ 
 

 
     url: "js/overallreport.php", 
 
     dataType: "json", 
 
     data: "ay=" + ay + "&year=" + year, 
 
     success: function(data) { 
 
       //left this blank because I am not sure of what I am doing. 
 
       //I used $.getJSON and $.each 
 
     }, 
 

 
     error: function() { 
 
      alert('Cannot retrieve data from server.'); 
 
     } 
 

 
     }); //ajax 
 

 

 

 

 
    } //else 
 

 
    }); //btnOverall 
 

 

 

 
}); 
 
//This is the JS FILE

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> 
 
<div class="form-group col-sm-4"> 
 
    <select class="form-control" id="Overall_acadyear"> 
 
    <option value='' active>Select AY</option> 
 
    <?php $r->selectAcademicYear(); ?> 
 
    </select> 
 

 
    <select class="form-control col-sm-4" id="Overall_year" align="right"> 
 
    <option value='' active>Select year level</option> 
 
    <?php $r->selectYear(); ?> 
 
    </select> 
 
    <div class="form-group"> 
 
    <button type="button" class="control-label btn-success" id="searchOverall" name='createAccount'>Search</button> 
 
    <button type="reset" class="control-label btn-danger" id="reset">Clear Fields</button> 
 
    </div> 
 

 
    </th> 
 

 
    </tr> 
 
    <tr> 
 
    <th>Student Number</th> 
 
    <th>Student Name</th> 
 
    <th>Section</th> 
 
    <th>Status</th> 
 
    </tr> 
 
    </thead> 
 
    <tbody id="tableBody"> 
 

 
    </tbody> <!--This is the view file -->

<?php 
require($_SERVER['DOCUMENT_ROOT']."/finalsis/include/config.php"); 
include_once($_SERVER['DOCUMENT_ROOT']."/finalsis/include/class.utility.php"); 
header("Content-Type: application/json"); 
$ay = $_GET['ay']; 
$year = $_GET['year']; 

$ay = $obj->cleanString($ay); 
$year = $obj->cleanString($year); 


$conn = mysqli_connect(db_server,db_user,db_password,db_database); 
$ay = mysqli_escape_string($conn,$ay); 
$year = mysqli_escape_string($conn,$year); 

$selectSQL = "SELECT studentlevel_student, student_fname, student_mname, student_lname, 
section_name, student_status FROM tblstudentlevel inner join 
tblstudent on studentlevel_student = student_number inner join 
tblyearsection on studentlevel_ys = ys_id WHERE studentlevel_acadyear = '" .$ay."' AND year_name = '" .$year. "'"; 
$result = mysqli_query($conn,$selectSQL); 

$output = '{"student": ['; 
while($rs = mysqli_fetch_array($result)){ 
     $name = $obj->getFullName($rs['student_fname'],$rs['student_mname'],$rs['student_lname']); 
     $output .= '{"sno":"' .$rs['studentlevel_student']. '", '; 
     $output .= '"name":"' .$name. '", '; 
     $output .= '"section":"' .$rs['section_name']. '",'; 
     $output .= '"status":"' .$rs['student_status']. '"},'; 
    } 

$output .= "]}"; 
mysqli_close($conn); 
echo json_encode($output); 

//This is where my data gathering happens. ?> 

Answer 1

Sorun şu uyumludur:

echo json_encode($output); 

json_encode bir dizi ihtiyacı

İşte benim kod görünümüdür. Dize geçiyorsun.

$jsonData = array(); 
while($rs = mysqli_fetch_array($result)){ 
    $name = $obj->getFullName($rs['student_fname'],$rs['student_mname'],$rs['student_lname']); 

    $jsonData[] = array(
     'sno' => $rs['studentlevel_student'], 
     'name' => $name, 
    ); 
} 

echo json_encode($jsonData);