douzhi4830 2017-08-21 09:45
浏览 44

ajax错误,数据被埋没了

this is my code

main.js

var save_method, table;
//Menerapkan plugin datatables
$(function(){
   table = $('.table').DataTable({
      "processing" : true,
      "ajax" : {
         "url" : "ajax/ajax_user.php?action=table_data",
         "type" : "POST"
      }
   });

});

ajax_user.php

<?php 

include "../config/database.php";
include "../library/view.php";

if ($_GET['action'] == "table_data") {
    $query = mysqli_query($mysqli, "SELECT p.*, v.nama_provinsi, k.nama_kota, j.nama_tagihan FROM pelanggan P 
            INNER JOIN provinsi v ON p.id_provinsi = v.id_provinsi 
            INNER JOIN kota k ON p.id_kota = k.id_kota 
            INNER JOIN jenis_tagihan j ON p.id_jenis_tagihan = j.id_jenis_tagihan 
            ORDER BY p.id DESC");
    $data = array();
    $no = 1;
    while($p = mysqli_fetch_assoc($query)) {
        $row = array();
        $row[] = $no;
        $row = $p['nama_masjid'];
        $row = $p['nama_pengurus'];
        $row = $p['id_pelanggan'];
        $row = $p['tagihan'];
        $row = $p['nama_provinsi'];
        $row = $p['nama_kota'];
        $row = $p['nama_tagihan'];
        $data[] = $row;
        $no++;
    }

    $output = array("data" => $data);
    echo json_encode($output);
}

 ?>

and this is my error

enter image description here

The last data in the array appears but the previous data disappears what's wrong ?

  • 写回答

1条回答 默认 最新

  • dounue1965 2017-08-21 10:01
    关注

    try this :

    $data = [];
    $i=0;
    while($p = mysqli_fetch_assoc($query)) {
    
        $data[$i]['nama_masjid']  = $p['nama_masjid'];
        $data[$i]['nama_pengurus']= $p['nama_pengurus'];
    
        $i++;
    }
    
    echo json_encode($data);
    
    评论

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