使用 php mysql 创建嵌套 json

Posted 2023-02-24

【中文标题】使用 php mysql 创建嵌套 json

【英文标题】：Create nested json using php mysql

【发布时间】：2017-07-18 01:58:35

【问题描述】：

我有以下三个表

讲座（lec_id、名称、描述） 测试（test_id、test_name、lec_id、日期） 问题（q_id、q_name、q_desc、test_id）

我想生成这样的 json 响应

"lec_name": "Math",
"description": "Can you identify these brands by the background color?",
"test": [
    
        "name": "Algebra",
        "date": "10-6-2017",
        "question": [
            
                "q_name": "question 1",
                "description": "Lorem Ipsum is simply dummy text of the printing",
            ,
            
                "q_name": "question 2",
                "description": "Lorem Ipsum is simply dummy text of the printing",
            ,
            
                "q_name": "question 3",
                "description": "Lorem Ipsum is simply dummy text of the printing",
            
        ]
    


] 

但我越来越喜欢这个了

[
    [
        
            "algebra": "2017-02-28"
        
    ],
    
        "question 1": "Lorem Ipsum is simply dummy text of the printing"
    ,
    
        "0": "Math",
        "1": "1",
        "name": "Math",
        "lec_id": "1"
    ,
    [
        
            "trigonometry": "2017-02-28"
        
    ],
    
        "question 2": "Lorem Ipsum is simply dummy text of the printing"
    ,
    
        "0": "Chemistry",
        "1": "2",
        "name": "Chemistry",
        "lec_id": "2"
    ,
    [
        
            "Bio test 1": "2017-02-26"
        
    ],
    
        "question 3": "Lorem Ipsum is simply dummy text of the printing"
    ,
    
        "0": "Physics",
        "1": "3",
        "name": "Physics",
        "lec_id": "3"
    ,
    [
        
            "Bio test 2": "2017-02-28"
        
    ],
    
        "question 4": "Lorem Ipsum is simply dummy text of the printing"
    ,
    
        "0": "Biology",
        "1": "4",
        "name": "Biology",
        "lec_id": "4"
    
]

这是我的代码，

    $sql = "SELECT name, lec_id FROM lecture";

$sqlRun = mysqli_query($conn , $sql);
//var_dump($sqlRun);
//echo $sqlRun;     
$json = array();
$total_records = mysqli_num_rows($sqlRun);

if($total_records > 0)
    while($row = mysqli_fetch_array($sqlRun))
        $row_array= array();


        $qus_pk = $row['lec_id'];
        $lec_desc = '';
        $lec_name = '';

        $option_qry = mysqli_query($conn, "SELECT test_name, date, test_id FROM test WHERE test_id= $qus_pk");
        //$option_qry = mysqli_query($conn, "SELECT t.name");
        while($opt_fet = mysqli_fetch_array($option_qry))
            $row_array[]= array(
                $opt_fet['test_name'] => $opt_fet['date'],

            );
            $quest_array = array();
            $quest_pk = $opt_fet['test_id'];
            $test_query = mysqli_query($conn, "SELECT q_name, q_desc FROM question WHERE q_id = $quest_pk");
            while($test_fet = mysqli_fetch_array($test_query))
                $quest_array= array(
                    $test_fet['q_name'] => $test_fet['q_desc'],
                );
            
        
        array_push($json, $row_array, $quest_array);
        $json[] = $row;
    



echo json_encode($json);

【参考方案1】：

以此更改您的代码。

if($total_records > 0)
    $i = 0;
    while($row = mysqli_fetch_array($sqlRun))
        $row_array= array();
        $qus_pk = $row['lec_id'];
        $json[$i]['lec_name'] = $row['name'];    
        $json[$i]['description'] = $row['description'];    

        $option_qry = mysqli_query($conn, "SELECT test_name, date, test_id FROM test WHERE test_id= $qus_pk");
        //$option_qry = mysqli_query($conn, "SELECT t.name");
        while($opt_fet = mysqli_fetch_array($option_qry))
            $json[$i]['test']['name'] = $opt_fet['name'];
            $json[$i]['test']['date'] = $opt_fet['date'];

            $quest_array = array();
            $quest_pk = $opt_fet['test_id'];
            $test_query = mysqli_query($conn, "SELECT q_name, q_desc FROM question WHERE q_id = $quest_pk");
            $j = 0;
            while($test_fet = mysqli_fetch_array($test_query))
                $json[$i]['test']['question'][$j] = array('q_name' => $test_fet['q_name'], 'description' => $test_fet['q_desc']);
                $j++;
            
        
        $i++;
    



echo json_encode($json);

【讨论】：

#naincy 感谢您的回答，我在 while 循环中遇到错误，它的“解析错误：语法错误，意外 '['” while($test_fet = mysqli_fetch_array($test_query)) $json [$i]['test']['question'][] = ['q_name' => $test_fet['q_name'], 'description' => $test_fet['q_desc']];

@BilalCh 它不应该给出错误...... :(你能确保你复制粘贴正确......这个错误在哪一行和那一行写的你可以分享跨度>

#naincy 请看一下这些屏幕截图prnt.sc/edytlnprnt.sc/edytrv

做或不做。没有“尝试”。 好的答案将始终解释所做的事情以及为什么以这种方式完成，不仅是为了 OP，也是为了 SO 的未来访问者。跨度>

@BilalCh 是第 33 行吗？