mysqli_fetch_array没有返回任何内容

Question

我已经在phpMyAdmin检查了我的Sql语句，它返回了多个结果，但是当我将它作为php脚本运行时，它不会返回任何内容。 谁能告诉我我犯的是什么错误？

<?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

<?php

/* This file allows you to connect to a database */


$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 "true";
} 
*/

?>

Answer 1

改变你的代码如下，并检查一次： -

conn.php （检查文件名并自行更正）： -

<?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();
    }
?>

和其他页面： -

<?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

更改MYSQL_NUM到MYSQLI_NUM ; -

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


}