将PHP输出为JSON的变量

问题描述：

file1.php
<?php
$listMenu=array('Menu #1','Menu #2','Menu #3');
?>

<div class="wjNavButton"><a><?php echo($listMenu[0]); ?></a></div>
<div class="wjNavButton"><a><?php echo($listMenu[1]); ?></a></div>
<div class="wjNavButton"><a><?php echo($listMenu[2]); ?></a></div>


file2.php
$buff=include('file1.php');
$rest='[{sectionId:"LT", sectionType:"menu", sectionData="'.$buff.'"}]';
echo($rest);

result:
[{sectionId:"LT", sectionType:"menu", sectionData="1"}]

question:
- is this possible to put result of php page in variable?
- how i can result as output of file1.php and not sectionData="1"?

  file1.php 
＆lt;？php 
 $ listMenu = array（'Menu＃1'，  '菜单＃2'，'菜单＃3'）; 
？＆gt; 
 
＆lt; div class =“wjNavButton”＆gt;＆lt; a＆gt;＆lt;？php echo（$ listMenu [0]）;  ？＆gt;＆lt; / a＆gt;＆lt; / div＆gt; 
＆lt; div class =“wjNavButton”＆gt;＆lt; a＆gt;＆lt;？php echo（$ listMenu [1]）;  ？＆gt;＆lt; / a＆gt;＆lt; / div＆gt; 
＆lt; div class =“wjNavButton”＆gt;＆lt; a＆gt;＆lt;？php echo（$ listMenu [2]）;  ？＆gt;＆lt; / a＆gt;＆lt; / div＆gt; 
 
 
 
file2.php 
 $ buff = include（'file1.php'）; 
 $ rest ='[{sectionId：“LT”，sectionType  ：“menu”，sectionData =“'。$ buff。'”}]'; 
echo（$ rest）; 
 
结果：
 [{sectionId：“LT”，sectionType：“menu”，sectionData =“  1“}] 
 
问题：
-这可以将php页面的结果放在变量中吗？
-我怎么能得到file1.php的输出而不是sectionData =”1“？
  code  >  pre> 
  div>

答

ob_start();
include 'file1.php';
$buff = ob_get_clean();

$data = array(array('sectionId' => 'LT', 'sectionType' => 'menu', 'sectionData' => $buff));
echo json_encode($data);

You can't get the contents of a page using $buff = include. Imagine include as copy and pasting the contents of one file into another. It doesn't return anything. (Unless you structure your include files differently so it does, read the documentation.)
You can capture and get the content using output buffering though.
Never write your own JSON, use json_encode to make sure your syntax is correct and values are escaped properly.

答

file2.php

$buff=include('file1.php');
$rest='[{sectionId:"LT", sectionType:"menu", sectionData="'.$buff.'"}]';
echo($rest);

Looks like you needed a closing quote on that string.

将PHP输出为JSON的变量

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