mysqli_fetch_array沒有返回任何東西

Question

我檢查了我的phpMyAdmin中的Sql語句，它返回多個結果，但是當我將它作爲php腳本運行時，它不返回任何東西。任何人都可以告訴我我犯了什麼錯誤嗎？

<?php 
require "conn.php"; 
/*$sqlQry = $_POST["sqlQry"];*/ 

$sqlQry = "SELECT bad.id, Name, Address, Latitude, Longitude, AvAge FROM baraddresses bad inner join barlivedata bld on bad.id = bld.id inner join bar_data bdt on bdt.id = bad.id"; 

$result = mysqli_query($conn ,$sqlQry); 

$json = array(); 
while ($row = mysqli_fetch_array($result , MYSQL_NUM)){ 
$json[] = array('id' => $row[0], 
     'Name' => $row[1], 
     'Address' => $row[2], 
     'Latitude'=> $row[3], 
     'Longitude' => $row[4],   
     'AvAge' => $row[5]   
      ); 
} 

$jsonstring = json_encode($json); 
    echo $jsonstring; 

?> 

Conn.php

​​

Answer 1

更改你的代碼像下面檢查一次： -

conn.php（檢查文件名，並糾正自己）： -

<?php 
    //comment these two lines when everything started working fine 
    error_reporting(E_ALL); 
    ini_set('display_errors',1); 

    $db_name = "bdata"; 
    $mysql_username = "root"; 
    $mysql_password = "" ; 
    $server_name = "localhost"; 
    $conn = mysqli_connect($server_name , $mysql_username , $mysql_password , $db_name ); 

    if(!$conn){ 
     echo "connection error:-".mysqli_connect_error(); 
    } 
?> 

和其他頁面： -

<?php 
    //comment these two lines when everything started working fine 
    error_reporting(E_ALL); 
    ini_set('display_errors',1); 
    require "conn.php"; 

    $sqlQry = "SELECT bad.id, bad.Name, bad.Address, bad.Latitude, bad.Longitude, bad.AvAge FROM baraddresses bad JOIN barlivedata bld on bad.id = bld.id JOIN bar_data bdt on bdt.id = bad.id"; 

    $result = mysqli_query($conn ,$sqlQry) or die(mysqli_error($conn)); 

    $json = array(); 
    if(mysqli_num_rows($result)>0){ 
     /* Either use this */ 
     while ($row = mysqli_fetch_assoc($result)){ 
      $json[] = array('id'=>$row['id'],'Name'=>$row['Name'],'Address'=>$row['Address'],'Latitude'=>$row['Latitude'],'Longitude'=>$row['Longitude'],'AvAge'=>$row['AvAge']); 
     } 
     /* Or use this 

     $json = mysqli_fetch_all($result,MYSQLI_ASSOC); 

     mysqli_free_result($result); */ 

     mysqli_close($conn); 

     echo "<pre/>";print_r($json); 

     $jsonstring = json_encode($json); 

     echo $jsonstring; 
    }else{ 
     die('No record exist'); 
    } 
?> 

Answer 2

變化MYSQL_NUM到MYSQLI_NUM; -

while ($row = mysqli_fetch_array($result , MYSQLI_NUM)){ 


}