Php 显示重复json输出的html选项

所以，我在这里面临一个奇怪的情况：

我有一个表单，这个表单有几个类别，它们通过Json加载

我引用的表单部分的代码如下所示：

<div>
  <select>
    <?php include_once("php_loader/getAllCategoriesOptions.php"); ?>
  </select>
</div>

到目前为止还不错，我有一个php文件：

<?php
$con = $con = mysqli_connect("localhost","XXX","XXX","XXX");
mysqli_query($con,"SET NAMES UTF8");  /
if (!$con)
    {
    die('Could not connect: ' . mysqli_error());
    }
$result = mysqli_query($con,"SELECT * FROM categories WHERE IdCategory > 1") or die;
while($row = mysqli_fetch_assoc($result))
  {
    $output[]=$row;
  }
  print json_encode($output);
foreach($output as $json){
    echo'<option value="'.$json['IdCategory'].'">'.$json['Category'];
}
mysqli_close($con);
?>

Wich返回以下输出：

<div>
<select>
[{"IdCategory":"0","Category":"All Categories"},{"IdCategory":"1","Category":"Technology"},{"IdCategory":"2","Category":"Hobbies"},{"IdCategory":"3","Category":"Tattoos"},{"IdCategory":"4","Category":"Trips"},{"IdCategory":"5","Category":"Esoteric"},{"IdCategory":"6","Category":"Books"},{"IdCategory":"2","Category":"Hobbies"},{"IdCategory":"3","Category":"Tattoos"},{"IdCategory":"4","Category":"Trips"},{"IdCategory":"5","Category":"Esoteric"},{"IdCategory":"6","Category":"Books"}]
<option value="0">All Categories</option>
<option value="1">Technology</option>
<option value="2">Hobbies</option>
<option value="3">Tattoos</option>
<option value="4">Trips</option>
<option value="5">Esoteric</option>
<option value="6">Books</option>
<option value="2">Hobbies</option>
<option value="3">Tattoos</option>
<option value="4">Trips</option>
<option value="5">Esoteric</option>
<option value="6">Books </option>
</select>
</div>

[{IdCategory:2，Category:嗜好}，{IdCategory:3，Category:纹身}，{IdCategory:4，Category:Trips}，{IdCategory:5，Category:Esteric}，{IdCategory:6，Category:Books}]

爱好 塔托斯 旅行 深奥的 书

到目前为止还不错。但当我在html页面中包含此内容时，它会提供以下输出：

<div>
<select>
[{"IdCategory":"0","Category":"All Categories"},{"IdCategory":"1","Category":"Technology"},{"IdCategory":"2","Category":"Hobbies"},{"IdCategory":"3","Category":"Tattoos"},{"IdCategory":"4","Category":"Trips"},{"IdCategory":"5","Category":"Esoteric"},{"IdCategory":"6","Category":"Books"},{"IdCategory":"2","Category":"Hobbies"},{"IdCategory":"3","Category":"Tattoos"},{"IdCategory":"4","Category":"Trips"},{"IdCategory":"5","Category":"Esoteric"},{"IdCategory":"6","Category":"Books"}]
<option value="0">All Categories</option>
<option value="1">Technology</option>
<option value="2">Hobbies</option>
<option value="3">Tattoos</option>
<option value="4">Trips</option>
<option value="5">Esoteric</option>
<option value="6">Books</option>
<option value="2">Hobbies</option>
<option value="3">Tattoos</option>
<option value="4">Trips</option>
<option value="5">Esoteric</option>
<option value="6">Books </option>
</select>
</div>

我从mozzila firebug中获取了这个值——在网站中，这个值只出现了两次

这里可能发生什么事？？我无法理解哪里错了

可能需要注意的是，我之前运行过一个Similar php文件，php文件会加载该表的每个条目-我怀疑这可能与此有关，但我无法得到解决方案，我在同一html页面中查询同一个表两次，以获得不同的输出，一个给了我几个，每个类别一个，另一个给我几个，如上所述

提前谢谢

PS：如所问：这是我之前在同一代码中使用的另一个php文件：

<?php
$con = $con = mysqli_connect("localhost","ZZZ","ZZZ","sametable");
mysqli_query($con,"SET NAMES UTF8");  //é preciso meter isto em utf-8 senão não manda nada
if (!$con)
    {
    die('Could not connect: ' . mysqli_error());
    }
$result = mysqli_query($con,"SELECT * FROM categories") or die;
while($row = mysqli_fetch_assoc($result))
  {
    $output[]=$row;
  }

foreach($output as $json){
   echo 
   '<li>
   <a class="item">'.$json['Category'].'</a>
   </li>';   
}
mysqli_close($con);
?>

---

所以，我似乎发现了错误，这是一个非常愚蠢的错误

我认为在同一代码中运行两个php文件没有问题，但我不知道我不能以同样的方式命名这两个文件变量，因此，第一个php的$output出现在第二个php文件中，结果是$output$output

---

将第二个文件的名称更改为$output2解决了这个问题。

我们可以在html页面上看到调用它的php吗？那真的很有帮助。