通过php访问多维JSON数据。

Ok I'm realtively new to PHP, so hang in there with me, trying hard to learn this stuff. Anyways, so pretty much I have a php page that get's the JSON data contents and echo's it on the page just fine. So my problem is that my JSON is notated multidimensionally like so:

{"product":[{"brandId":"632","brandName":"Sam Edelman","productId":"7515478","productName":"Gigi","styles":[{"price":"$64.95","styleId":"1788226","imageUrl":"http:\/\/www.zappos.com\/images\/z\/1\/7\/8\/8\/2\/2\/1788226-p-DETAILED.jpg","originalPrice":"$64.95","productUrl":"http:\/\/www.zappos.com\/product\/7515478\/color\/25","percentOff":"0%","color":"Almond"},

So how would I access say the percentOff in the first element? Here's my javascript function that I use to ideally grab elements and make decisions based around grabbing the "percentOff" element.

<!DOCTYPE html>
<html>
<head>
<script type="text/javascript">
function ajax_get_json(){
var results = document.getElementById("results");
var hr = new XMLHttpRequest();
hr.open("GET", "my_json_list.php", true);
hr.setRequestHeader("Content-type", "application/json", true);
hr.onereadystatechange == function() {
    if(hr.readyState == 4 && hr.status == 200){
        var data = JSON.parse(hr.responseText);
        results.innerHTML = "";
        for(var obj in data){
        results.innerHTML += data[obj].styles[obj].percentOff+""; //Or like this?
           results.innerHTML += data[0].styles[obj].percentOff+""; //Would it look something like this?
        }
    }
}
hr.send(null);
results.innerHTML = "requesting...";

}   

</script>

</head>
<body>
<div id="results"></div>
<script type="text/javascript">ajax_get_json();</script>
</body>
</html>

Appreciate any help ahead of time!

写回答
好问题 0 提建议
追加酬金
关注问题
分享
邀请回答
编辑收藏删除结题
收藏举报

1条回答默认最新

dongtang5057 2014-02-17 02:09

关注

{"product":[{"brandId":"632","brandName":"Sam Edelman","productId":"7515478","productName":"Gigi","styles":[{"price":"$64.95","styleId":"1788226","imageUrl":"http://www.zappos.com/images/z/1/7/8/8/2/2/1788226-p-DETAILED.jpg","originalPrice":"$64.95","productUrl":"http://www.zappos.com/product/7515478/color/25","percentOff":"0%","color":"Almond"}]}]}

I copied and pasted it into here: http://jsoneditoronline.org/

<?php

$json = '{"product":[{"brandId":"632","brandName":"Sam Edelman","productId":"7515478","productName":"Gigi","styles":[{"price":"$64.95","styleId":"1788226","imageUrl":"http:\/\/www.zappos.com\/images\/z\/1\/7\/8\/8\/2\/2\/1788226-p-DETAILED.jpg","originalPrice":"$64.95","productUrl":"http:\/\/www.zappos.com\/product\/7515478\/color\/25","percentOff":"0%","color":"Almond"}]}]}';

$data = json_decode($json,true);

echo $data['product'][0]['styles'][0]['percentOff'];

returns '0%';

BTW "foreach" comes in handy....

e.g.

foreach ($data['product'][0]['styles'][0] as $key => $value) {
    echo "$key: $value<br>";
}

returns:

price: $64.95
styleId: 1788226
imageUrl: http://www.zappos.com/images/z/1/7/8/8/2/2/1788226-p-DETAILED.jpg
originalPrice: $64.95
productUrl: http://www.zappos.com/product/7515478/color/25
percentOff: 0%
color: Almond

foreach ($data as $key1 => $obj1) {
    echo "Level 1: $key1: $obj1<br>";
    foreach ($obj1 as $key2 => $obj2) {
        echo "Level 2: $key2: $obj2<br>";
        foreach ($obj2 as $key3 => $obj3) {
            echo "Level 3: $key3: $obj3<br>";
            if (is_array($obj3)) {
                foreach ($obj3 as $key4 => $obj4) {
                    echo "Level 4: $key4: $obj4<br>";
                    if (is_array($obj4)) {
                        foreach ($obj4 as $key5 => $obj5) {
                            echo "Level 5: $key5: $obj5<br>";
                            if ($key5 == 'percentOff') {
                                echo 'Percent Off is: '.$obj5.' or '.$obj4['percentOff'].'<br>';
                            }
                        }
                    }
                }
            }
        }
    }
}

returns:

Level 1: product: Array
Level 2: 0: Array
Level 3: brandId: 632
Level 3: brandName: Sam Edelman
Level 3: productId: 7515478
Level 3: productName: Gigi
Level 3: styles: Array
Level 4: 0: Array
Level 5: price: $64.95
Level 5: styleId: 1788226
Level 5: imageUrl: http://www.zappos.com/images/z/1/7/8/8/2/2/1788226-p-DETAILED.jpg
Level 5: originalPrice: $64.95
Level 5: productUrl: http://www.zappos.com/product/7515478/color/25
Level 5: percentOff: 0%
Level 5: color: Almond

本回答被题主选为最佳回答 , 对您是否有帮助呢?

报告相同问题？

关注问题

访问多维JSON对象中的数据 json php
2015-04-03 17:55

回答 1 已采纳 Try something like this <?php $result = '{"status":"success","data":{"lists":[{"id":"0032","n
PHP中JSON多维数组的访问值 json php
2018-07-18 00:43

回答 1 已采纳 but appear to be unable to access values (typically seeing nothing/null), through something lik
PHP - 将新的JSON数据附加到现有的多维JSON json php
2016-03-20 22:27

回答 2 已采纳 This should help you out. If you have more than one channel you'll need the index of the channel
PHP数组通过JSON格式转换传递到前端Javascript
2021-07-28 17:52

weixin_45018995的博客 php $name1="test1"; $age1=60; $age2=38; $name2="test2"; $a=array(); //创建一个空数组a $b=array("name"=>"$name1", "age"=>"$age1"); //定义关联数组b $c=array("name"=>"$name2", "age"=>"$age2")...
如何从PHP中的多维JSON字符串数组中获取元素的值 json php
2017-05-13 08:49

回答 5 已采纳 Hope this simple foreach will be helpful for you. Solution 1: Try this code snippet here $result
使用MySQL数据构建JSON多维数组 json mysql php
2019-03-01 17:33

回答 2 已采纳 I'd do it with a single query: $sql = " SELECT `cuname` AS `name`, `date`, count(`
循环通过多维数组 - PHP json php
2019-07-13 07:24

回答 1 已采纳 Here I think is a similar minimal data structure that is equivalent to yours: <?php $data = [
PHP的serialize序列化数据以及JSON格式化数据分析
2020-10-23 12:33

然而，如果数据需要跨语言交互，特别是前端JavaScript环境，或者对数据的可读性有要求，那么JSON格式化是更合适的选择。值得注意的是，JSON格式化也有其局限性，比如它不支持PHP中的资源类型（resource），并且在...
新手，关于PHP操作JSON数据的问题。不胜感激！ json php
2015-07-26 08:08

回答 2 已采纳 ``` $s='{"status":"1","Msg":"登陆成功"}'; $dom=new DomDocument; $dom->loadXML($s); $s=$dom
PHP上未显示JSON多维数组 json php
2018-08-21 08:59

回答 1 已采纳 header('Content-Type: application/json'); $return = array(); $conn = mysqli_connect("l
使用PHP解析多维嵌套JSON json php
2014-08-06 11:15

回答 2 已采纳 The problem is that you have both objects and arrays in your json, but are using array syntax in y
php mysql数据库json存储数据_php将json格式的数据直接存入mysql数据库
2021-01-30 06:43

UXOFFER的博客直接使用php函数 json_encode ，数据存入mysql时会报错：$imgs_str = json_encode($imgs_array);$imgs_str = addslashes($imgs_str);$update_query = "updatearticleset imageList = $imgs_str where id = $targetid...
PHP 二维array转换json的实例讲解
2020-10-18 04:35

在进行PHP编程的过程中，将二维数组转换为JSON格式是常见的操作，特别是涉及到Web开发时，经常需要把数组数据以JSON格式返回给前端。JSON（JavaScript Object Notation）作为一种轻量级的数据交换格式，易于人阅读和...
php json_encode()函数返回json数据实例代码
2020-10-25 09:23

通过本知识点的学习，我们了解了如何在PHP中使用`json_encode()`函数将数组或对象转换为JSON格式，并通过实例代码演示了如何结合数据库操作来返回JSON数据给前端。同时，我们也学习了如何对数组进行深层次的处理以及...
使用js对php返回的json数据进行解析
2019-03-08 15:39

simon_xing的博客想的是接收php传来的 json 数据的话，就更好解析数据，并获取里面的字段名和数据。所以在php文件里，将返回的数据解析成json格式，使用 json_encode()。问题接收后发现，接收到的数据是json格式的字符串，...
没有解决我的问题, 去提问

悬赏问题

¥15 R语言卸载之后无法重装，显示电脑存在下载某些较大二进制文件行为，怎么办
¥15 java 的protected权限，问题在注释里
¥15 这个是哪里有问题啊？
¥15 关于#vue.js#的问题：修改用户信息功能图片无法回显，数据库中只存了一张图片（相关搜索：字符串）
¥15 texstudio的问题，
¥15 spaceclaim模型变灰色
¥15 求一份华为esight平台V300R009C00SPC200这个型号的api接口文档
¥15 字符串比较代码的漏洞
¥15 欧拉系统opt目录空间使用100%
¥15 ul做导航栏格式不对怎么改？

码龄粉丝数原力等级 --

通过php访问多维JSON数据。

1条回答默认最新

码龄粉丝数原力等级 --

悬赏问题

通过php访问多维JSON数据。

1条回答 默认 最新

悬赏问题

1条回答默认最新