使用 php mysql 创建嵌套的 json 对象答案

【问题标题】：Create nested json object using php mysql使用 php mysql 创建嵌套的 json 对象
【发布时间】：2015-07-21 05:21:26
【问题描述】：

我有两个表，表 1 有 2 个字段（question_pk、question_name），表 2 有 4 个字段（ans_pk、options、question_fk 和 right_answer）。 我想创建如下结构的json

{
    "type": "quiz",
    "name": "Brand Colors",
    "description": "Can you identify these brands by the background color?",
    "questions": [
        {
            "name": "Can you identify this color?",
            "description": "#ea4c89",
            "answers": [
                {
                    "name": "Dribbble",
                    "description": "dribbble.png",
                    "weight": 1
                },
                {
                    "name": "Amazon",
                    "description": "amazon.png",
                    "weight": 0
                },
                {
                    "name": "Apple",
                    "description": "apple.png",
                    "weight": 0
                }
            ]
        },
        {
            "name": "Can you identify this color?",
            "description": "#3d9ae8",
            "answers": [
                {
                    "name": "Youtube",
                    "description": "youtube.png",
                    "weight": 0
                },
                {
                    "name": "Dropbox",
                    "description": "dropbox.png",
                    "weight": 1
                },
                {
                    "name": "Wordpress",
                    "description": "wordpress.png",
                    "weight": 0
                }
            ]
        },
        {
            "name": "Can you identify this color?",
            "description": "#c4302b",
            "answers": [
                {
                    "name": "Youtube",
                    "description": "youtube.png",
                    "weight": 1
                },
                {
                    "name": "Twitter",
                    "description": "twitter.png",
                    "weight": 0
                },
                {
                    "name": "Vimeo",
                    "description": "vimeo.png",
                    "weight": 0
                }
            ]
        }

    ]
}

我的 PHP 代码

<?php
include '../config/config.php';
if(isset($_GET['sub_cat_id']))
{
         $sub_cat_id = $_GET['sub_cat_id']; 
        $result = mysql_query("select * from $questions where sub_cat='$sub_cat_id' order by level_fk asc"); 
        $json_response = array(); //Create an array
        $i=1;
                        while ($row = mysql_fetch_array($result))
                        {
                        $row_array['qus_pk'] = $row['qus_pk'];        
                        $row_array['question'] = $row['question'];
                        $qus_pk = $row['qus_pk'];  


                        $option_qry = mysql_query("select * from $qus_ans where qus_pk=$qus_pk");
                        while ($opt_fet = mysql_fetch_array($option_qry))
                        {

                        $row_array['options'] = $opt_fet['options'];  
                        $row_array['right_ans'] = $opt_fet['right_ans'];  
                        array_push($json_response,$row_array); //push the values in the array
                            }                       


                        $i++;
                        }
        echo json_encode($json_response);
}

?>

而我的结果我得到如下 json 响应

[
    {
        "qus_pk": "1",
        "question": "Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?",
        "options": "45",
        "right_ans": "0"
    },
    {
        "qus_pk": "1",
        "question": "Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?",
        "options": "40",
        "right_ans": "0"
    },
    {
        "qus_pk": "1",
        "question": "Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?",
        "options": "35",
        "right_ans": "1"
    },
    {
        "qus_pk": "1",
        "question": "Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?",
        "options": "50",
        "right_ans": "0"
    },
    {
        "qus_pk": "2",
        "question": "Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age?",
        "options": "4 times",
        "right_ans": "0"
    },
    {
        "qus_pk": "2",
        "question": "Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age?",
        "options": "1 times",
        "right_ans": "0"
    },
    {
        "qus_pk": "2",
        "question": "Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age?",
        "options": "3 times",
        "right_ans": "1"
    },
    {
        "qus_pk": "2",
        "question": "Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age?",
        "options": "5 times",
        "right_ans": "0"
    }
]

在我每次重复问题时的回答中，如何避免和如果我想获得第一个 json 结构， 在我的 PHP 代码中我需要在哪里进行更改？如果有人知道帮助我。

【问题讨论】：

标签： php mysql json recursive-query

【解决方案1】：

你好，试试这个，

<?php
include '../config/config.php';
if(isset($_GET['sub_cat_id']))
{
    $sub_cat_id = $_GET['sub_cat_id']; 
    $result = mysql_query("SELECT * FROM $questions WHERE sub_cat='$sub_cat_id' ORDER BY level_fk ASC"); 
    $json_response = array(); //Create an array
    while ($row = mysql_fetch_array($result))
    {
        $row_array = array();
        $row_array['qus_pk'] = $row['qus_pk'];        
        $row_array['question'] = $row['question'];
        $row_array['answers'] = array();
        $qus_pk = $row['qus_pk'];  

        $option_qry = mysql_query("SELECT * FROM $qus_ans WHERE qus_pk=$qus_pk");
        while ($opt_fet = mysql_fetch_array($option_qry))
        {
            $row_array['answers'][] = array(
                'options' => $opt_fet['options'],
                'right_ans' => $opt_fet['right_ans'],
            );

        }
        array_push($json_response, $row_array); //push the values in the array
    }
    echo json_encode($json_response);
}
?>

【讨论】：

谁能给出运行此代码的表结构？我们可以用它来构建 D3 树 json 吗？？

【解决方案2】：

我认为这段代码更容易理解，而且它使用 mysqli 的方式...

这是基于我自己的数据结构，我在做某事，我没有时间上午。使其适应问题，但应该很容易弄清楚如何使其适应其他结构：

$usersList_array =array();
$user_array = array();
$note_array = array();

$fetch_users = mysqli_query($mysqli, "SELECT 
        ID, 
        Surname, 
        Name 
    FROM tb_Users 
    WHERE Name LIKE 'G%' 
    ORDER BY ID") or die(mysqli_error($mysqli));
while ($row_users = mysqli_fetch_assoc($fetch_users)) {
    $user_array['id'] = $row_users['ID'];
    $user_array['surnameName'] = $row_users['Surname'].' '.$row_users['Name'];
    $user_array['notes'] = array();

    $fetch_notes = mysqli_query($mysqli, "SELECT 
        id, 
        dateIns, 
        type, 
        content 
     FROM tb_Notes 
     WHERE fk_RefTable = 'tb_Users' AND 
         fk_RefID = ".$row_users['ID'].""
   ) or die(mysqli_error($mysqli));
    while ($row_notes = mysqli_fetch_assoc($fetch_notes)) {
        $note_array['id']=$row_notes['id'];
        $note_array['dateIns']=$row_notes['dateIns'];
        $note_array['type']=$row_notes['type'];
        $note_array['content']=$row_notes['content'];
        array_push($user_array['notes'],$note_array);
    }

    array_push($usersList_array,$user_array);
}

$jsonData = json_encode($usersList_array, JSON_PRETTY_PRINT);


echo $jsonData;

生成的 JSON：

[
{
    "id": "1",
    "surnameName": "Xyz Giorgio",
    "notes": [
        {
            "id": "1",
            "dateIns": "2016-05-01 03:10:45",
            "type": "warning",
            "content": "warning test"
        },
        {
            "id": "2",
            "dateIns": "2016-05-18 20:51:32",
            "type": "error",
            "content": "error test"
        },
        {
            "id": "3",
            "dateIns": "2016-05-18 20:53:00",
            "type": "info",
            "content": "info test"
        }
    ]
},
{
    "id": "2",
    "cognomeNome": "Xyz Georg",
    "notes": [
        {
            "id": "4",
            "dateIns": "2016-05-20 14:38:20",
            "type": "warning",
            "content": "georg warning"
        },
        {
            "id": "5",
            "dateIns": "2016-05-20 14:38:20",
            "type": "info",
            "content": "georg info"
        }
    ]
}
]

【讨论】：

欢迎来到 Stack Overflow。虽然您提供的链接可能会回答问题，但最好将解决方案的基本部分直接放在您的答案中，以防链接上的页面将来消失。
谢谢，我认为避免重复是首选。我会输入我的代码:-)

【解决方案3】：

一个将表格嵌套到 php 数组中的基本类。

PHP 类

// chain data into a php array, filtering by relation to parent, based on a structure definition array
// nest child data by relation to parent data
// assign a array label "arr_label" to child definition to define what key the filtered data will use 
// assign a parent key "p_key" and a child key "c_key" to child definition to assign connection points from child to parent
// load array data to filter into "arr" key on child definition

class class_chain_filter
{
    var $return_arr;
    
    function __construct()
    {
    } // CONSTRUCTOR
    
    // input a defined filter tree array and output a processed result
    function chain_filter($filter_tree)
    {
        // can feed either a single record a set of rows...
        if(!$this->is_assoc($filter_tree['arr']))
            $this->return_arr = $filter_tree['arr']; // root for return array
        else
            $this->return_arr[] = $filter_tree['arr']; // force a numeric array so return is consistent.
            
        $this->do_chain_filter( $filter_tree['next_arrs'], $this->return_arr );
        
        return $this->return_arr;
    } // $this->chain_filter($filter_tree) // public
    
    
    function is_assoc($arr)
    {
        return array_keys($arr) !== range(0, count($arr) - 1);
    }

    
    function do_chain_filter(&$tree_arr, &$final_arr)
    {
        $cur_final_node = &$final_arr;
        
        if( !is_array($cur_final_node) )
            return false;
        
        // send the next_arrs
        foreach($final_arr as $f_key => $f_arr)
        {
            $cur_final_node = &$final_arr[$f_key];
            
            foreach($tree_arr as $n_key => $n_arr)
            {
                $cur_tree_node = $tree_arr[$n_key];
                // $final_cur_el['arr_label'] = 'true';
                $next_final_node = &$cur_final_node[$cur_tree_node['arr_label']];
                
                // data up hombre 
                // filter out array elements not related to parent array
                $result = $this->children_of_parent(
                    $cur_final_node, 
                    $cur_tree_node['arr'], 
                    $cur_tree_node['p_key'], 
                    $cur_tree_node['c_key']
                );

                $next_final_node = $result;

            
                // now recurse if we have more depths to travel...
                if(!empty($cur_tree_node['next_arrs']))
                    $this->do_chain_filter($cur_tree_node['next_arrs'], $next_final_node);

            }
        }
    } // this->function chain_filter(&$tree_arr, &$final_arr)




    // take 2 arrays
    // first array is an associative array. 
    // second array is an array of associative arrays.
    // return children of second array that belong to parent array
    function children_of_parent($arr_parent, $arr_children, $key_parent, $key_child )
    {
        // parent   = a record
        // child    = multiple records
        // filter out children that don't apply to parent.
        // return the result
        $parent_id = $arr_parent[$key_parent];
        
        foreach($arr_children as $arr_child)
        {
            $child_id = $arr_child[$key_child];
            
            if($child_id == $parent_id)
                $return_arr[] = $arr_child;
        }
        
        if(!empty($return_arr))
            return $return_arr;
    } // this->children_of_parent($arr_parent, $arr_children, $key_parent, $key_child )


} // end. class class_chain_filter

加载一些表（使用您自己的首选数据库类）

$areas = $db->get("SELECT * FROM areas");
$rooms = $db->get("SELECT * FROM rooms");
$exits = $db->get("SELECT * FROM exits");

定义我们返回的数组树结构

// predefine tree structure for generation
// structure definition array example...
$tree_arr = array (

    "arr"   => $areas, // root (can be multiple rows or a single record)
    
    "next_arrs" => array ( // children
    
        0 => array(
        
            "arr"           => $rooms,          // array to filter against parent
            "arr_label"     => "rooms",         // for the php array label
            "p_key"         => "id",            // field name of parent // eg) id
            "c_key"         => "areaid",        // this array's field name that links it to parent

            "next_arrs" => array( // children

                0 => array(
                
                    "arr"           => $exits,              // array to filter against parent
                    "arr_label"     => "exits",             // for the php array label
                    "p_key"         => "id",                // field name of parent / blank if root / eg) id
                    "c_key"         => "roomid"             // this array's field name that links it to parent
                )
            )
        )
    )

); // $tree_arr

现在将对象和进程创建到目标数组中

$c = new class_chain_filter();
$return_arr = $c->chain_filter($tree_arr);
print_r($return_arr);

...输出应该看起来像...

)

然后你可以json_encode 数组将它从 PHP 数组转换为 JSON 字符串

【讨论】：