使用 PHP 转换为 json 后解析 XML 文件

Question

我有一个 XML url，我需要在 jquery ajax 请求上解析和分页。我在 PHP 方面没有太多经验，我想弄清楚如何获取每个项目的值？另外，我怎样才能在结果中搜索匹配项？

我的代码：

$xmlFile = simplexml_load_file($xmlURL);
$json = json_encode($xmlFile);
$array = json_decode($json,TRUE);



echo "<pre>";



foreach ($array['item'] as $item) {
    $title = $item['title'];
    $date = $item['date'];






   echo "<div id='event-item'>";
   echo "<span id='title'>" . $title . "</span>" . "<br />";
   echo "<span id='date'>" . $date . "</span><br/>";
   echo "</div>";
}

输出（试图遍历和搜索）：

Array
(
    [item] => Array
        (
            [0] => Array
                (
                    [title] => English Matters Book Group for English Language Learners
                    [date] => Wednesday, June 15, 2016
                    [time] => 10:00 AM 


ib=0
                )

            [1] => Array
                (
                    [title] => Teen Advisory Board (TAB)
                    [date] => Wednesday, June 15, 2016
                    [time] => 5:30 PM
                    [enddate] => Array
                        (
                        )

Answer 1

一个标准的 foreach 循环就足够了：

<?php
foreach ($array['item'] as $item) {
    echo $item['title'] . '<br />';
}