以json格式动态传递表头引导数据表

Question

我从json数据创建数据表。表格数据正在变好。现在我想用这些json数据传递表头。

我data.php

<?php 
include('config/db_i.php'); 
$sql_sel = mysqli_query($con,"SELECT * FROM `tbl`"); 
$array = array(); 
$array['data'] = array(); 

while($res_sel = mysqli_fetch_row($sql_sel)){ 
    $array['data'][] = $res_sel; 
} 
echo json_encode($array); 
?> 

现在输出来像

{"data":[["1","John","USA"],["2","BOB","INDIA"],["3","James","UK"],["4","Dani","CHINA"]]} 

我想转移列中的值太...即ID，姓名，国家

Answer 1

您需要从mysqli_fetch_assoc()取代数据库中的关联数组而不是mysqli_fetch_row()。然后，您将收到一个JSON对象，该属性是列名称，而不是数组。此外，我建议使用JSON_NUMERIC_CHECK option，以确保数字通过这样。

<?php 
include("config/db_i.php"); 
$array = array("data"=>array()); 
$sql_sel = $con->query("SELECT * FROM tbl"); 
if ($sql_sel) {  
    while($res_sel = $sql_sel->fetch_assoc()){ 
     $array['data'][] = $res_sel; 
    } 
} 

header("Content-Type:application/json"); 
echo json_encode($array, JSON_NUMERIC_CHECK); 

这将导致这样的事情：

{"data":[{"id":1,"name":"John","country":"USA"},{"id":2,"name":"BOB","country":"INDIA"},{"id":3,"name":"James","country":"UK"},{"id":4,"name":"Dani","country":"CHINA"}]}