Question

我有一个JSON feed我通过PHP解析。我有问题得到一些嵌套元素回应我会很感激一些帮助..我已经看了很多相关的帖子，但似乎无法得到逻辑，以处理我的特定JSON提要。有人可以告诉我做错了吗？

JSON Feed在这里＆gt; https://api.lever.co/v0/postings/leverdemo?skip=1&limit=3&mode=json

我正在努力解析的元素是＆＃34;类别＆＃34; ＆＃34; team＆＃34;，＆＃34; location＆＃34;的父节点和子节点和＃34;承诺＆＃34;。

我以为这会起作用 - 但它不会......

<?php

$url = 'feed.json'; 
$data = file_get_contents($url); 
$characters = json_decode($data, true);

?>

<table>
<tbody>
    <tr>
        <th>Job title</th>
        <th>Team</th>
        <th>Location</th>
        <th>Commitment</th>
        <th>DescriptionPlain</th>
        <th>applyUrl</th>
    </tr>
    <?php foreach ($characters as $character) : ?>
    <tr>
        <td> <?php echo $character['text']; ?> </td>



        <td> <?php echo $character['categories'][2]['team'] ?></td>
        <td> <?php echo $character['categories'][2]->team ?></td>
        <td> <?php echo $character['categories'][1]->location ?></td>
        <td> <?php echo $character['categories'][0]->commitment ?></td>

        <td> <?php echo $character['descriptionPlain']; ?> </td>
        <td> <?php echo $character['applyUrl']; ?> </td>

    </tr>
    <?php endforeach; ?>
</tbody>

注意，它只是那些未能回应的孩子？还注意到，如果我在$ url变量中使用完整的url，它都会失败？但是从当地它起作用了吗？

任何想法???谢谢！

Answer 1

应该是：

    <td> <?php echo $character['categories']["team"] ?></td>
    <td> <?php echo $character['categories']["location"] ?></td>
    <td> <?php echo $character['categories']["commitment"] ?></td>

代替。数据中的数字键不存在数字键。另外＆＃34;类别＆＃34;不是对象，因此您不能使用箭头（ - ＆gt;）表示法。

Answer 2

修改

你得到一个错误，因为你试图访问一个对象，实际上你有一个数组，这是解决方案希望它有所帮助：

<?php $url = 'https://api.lever.co/v0/postings/leverdemo?skip=1&limit=3&mode=json'; $data = file_get_contents($url); $characters = json_decode($data, true); $nb = count($characters); ?> <table> <tbody> <tr> <th>Job title</th> <th>Team</th> <th>Location</th> <th>Commitment</th> <th>DescriptionPlain</th> <th>applyUrl</th> </tr> <?php while($nb > 0){ $nb--; $nb_lists = count($characters[$nb]['lists']); ?> <tr> <?php while($nb_lists > 0){ $nb_lists--; ?> <td> <?php if(isset($characters[$nb]['lists'][$nb_lists]['text'])){ echo $characters[$nb]['lists'][$nb_lists]['text'];} ?> </td> <?php } ?> <td> <?php if(isset($characters[$nb]['categories']['team'])){echo $characters[$nb]['categories']['team'];} ?></td> <td> <?php if(isset($characters[$nb]['categories']['team'])){echo $characters[$nb]['categories']['team'];} ?></td> <td> <?php if(isset($characters[$nb]['categories']['location'])) {echo $characters[$nb]['categories']['location'];} ?></td> <td> <?php if(isset($characters[$nb]['categories']['commitment'])){ echo $characters[$nb]['categories']['commitment'];} ?></td> <td> <?php if(isset($characters[$nb]['descriptionPlain'])){echo $characters[$nb]['descriptionPlain']; }?> </td> <td> <?php if(isset($characters[$nb]['applyUrl'])){echo $characters[$nb]['applyUrl'];} ?> </td> </tr> <?php } ?> </tbody>

将嵌套的JSON回送到PHP

2 个答案: