php - 如何将多个mysql结果集转换为json数组

Question

我在我的 php 脚本中执行多个“选择”语句。然后我将它们添加到数组中$Output=array($out, $out1);，然后对其进行编码json_encode()

输出格式为[ [{},{}] , [{}, {}] ]

我如何将其转换为形式 -{"str1": [{},{}], "str2": [{},{}]}

php脚本：

<?php

require 'DbConnect.php';


$query = ("SELECT * FROM `cars`");
$query1 = ("SELECT * FROM `City`");

if ($query_run = mysql_query($query)){

while ($query_row = mysql_fetch_assoc($query_run)){

    $out [] = $query_row;

}

}else{
    echo 'Fail';
} 

//second query 

if ($query_run1 = mysql_query($query1)){

while ($query_row1 = mysql_fetch_assoc($query_run1)){

$out1 [] = $query_row1;

}

}else{
    echo 'Fail';
} 

$Output=array($out, $out1); 
echo(json_encode($Output));
?>

Answer 1

如果你得到结果 [ [{},{}] , [{}, {}] ]

然后尝试

$Output=array('str1'=>$out,'str2'=>$out1); 
  echo(json_encode($Output,JSON_FORCE_OBJECT));

Answer 2

试试这个

    $query = "SELECT 
    post_title, 
    guid 
    FROM testwp_posts
    WHERE 
    post_author = $user_id 
    AND
    post_status = 'publish' 
    ORDER BY ID DESC LIMIT $number_of_posts";

        $result = mysql_query($query) or die('Errant query:  '.$query);

        /* create one master array of the records */
        $posts = array();
        if(mysql_num_rows($result)) {
            while($post = mysql_fetch_assoc($result)) {
                $posts[] = array('post'=>$post);
            }
        }


            header('Content-type: application/json');
            echo json_encode(array('posts'=>$posts));

Answer 3

<?php

require 'DbConnect.php';


$query = ("SELECT * FROM `cars`");
$query1 = ("SELECT * FROM `City`");

if ($query_run = mysql_query($query))
{
    while ($query_row = mysql_fetch_assoc($query_run))
    {   
        $output['cars'][] = $query_row;
    }

}
else
{
    echo 'Fail';
} 

//second query 

if ($query_run1 = mysql_query($query1))
{
    while ($query_row1 = mysql_fetch_assoc($query_run1))
    {
        $output['city'][] = $query_row1;
    }
}
else
{
    echo 'Fail';
} 

echo(json_encode($output));
?>

可能有帮助吗

Answer 4

我不确定这是否有助于您的要求，但请尝试以下操作..

而不是使用

$out [] = $query_row;

采用

$out['str1'][] = $query_row;

和而不是

$out1 [] = $query_row1;

采用

$out['str2'][] = $query_row1;

现在试试这个

echo(json_encode($out));