如何分离json数组并将其放在html表上？

Question

[“ cus_AKIT4Rz6NHlKMa”，“ cus_AKIP3muBIL95H0”，“ cus_AKI1l8vRoFhUyK”，“ cus_AKI0B1DHs2IBxB”，“ cus_AKHxuw8UvlwH8v”，“ cus_AKHx6iyvJJziw03”，

上面的数据是下面代码的结果，它来自我在Kuha.php上创建的PHP数组

function StartRetrieve()
{
var a = $('#selType').val();

    $.ajax({

        type:"POST",
        url:"Kuha.php",
        data:'&Type='+a,
        datatype:'json',
        cache:false,
        success:function(s){
        alert(s);

        }

    })
}

如何访问每个数据并将其分开？ 因为当我使用

for (var i = 0; i <= s.length; i++) {
$('#tb').append('<tr><td>'+s[i][0]+'</td></tr>');
}

包含[“”]，它像

[
"
c
u
s
A
K
I
T
4
R
z
...
"
]

我希望结果分开并水平打印

cus_AKIT4Rz6NHlKMa
cus_AKIP3muBIL95H0
cus_AKI1l8vRoFhUyK

这是Kuha.php

<?php
function httpGet($url) {
    $ch = curl_init();
    $headers = array('Authorization: Bearer xxxxxxxxxxxxxxxxxxxxxx');
    curl_setopt($ch, CURLOPT_URL,$url);
    curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
    curl_setopt($ch,CURLOPT_RETURNTRANSFER,true);
    curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
    $output=curl_exec($ch);
    curl_close($ch);
     $output = json_decode($output,true);//only necessary if the result of $output is a json encoded array
    return $output;
 }



$Type = $_POST['Type'];
$url = "https://api.stripe.com/v1/".$Type;

$array = httpGet($url);
$id = '';
 foreach($array["data"] as $a){
    $id .= $a['id'];
    $id .= ',';

 }
 //echo $id;
$a = rtrim($id,',');
$a = explode(',', $a);

echo json_encode($a);
?>

谢谢那些会回答的人。

Answer 1

 var s = ["cus_AKIT4Rz6NHlKMa", "cus_AKIP3muBIL95H0", "cus_AKI1l8vRoFhUyK", "cus_AKI0B1DHs2IBxB", "cus_AKHxuw8UvlwH8v", "cus_AKHx6iyvJJOwWe", "cus_AK03kmAOGCziJ8"]; var tr = $('<tr/>');//comment this for vertical for (var i = 0; i < s.length; i++) { //var tr = $('<tr/>');//comment this for horizontal tr.append('<td>' + s[i] + '</td>'); $('#tb').prepend(tr); } 

 td { background-color: blue; } 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <table id="tb"></table> 

在成功函数中使用此函数，只需删除var的初始化