我在MongoDB中有一个文档,我试图在PHP中解开它。我想展开一个文档,其中包含一个包含anothersubdocument的子文档。如果文档只包含字符串和数字,我能够成功地执行此操作,但如果它包含另一个子文档,那么我无法使其工作。我收到这个错误:
exception: $unwind: value at end of field path must be an array
你能解除包含另一级子文档的子文档吗?如果没有,你会怎么做呢?
提前致谢!
这是查询:
$project = array(
'$project' => array(
'_id' => 1,
'items' => 1,
)
);
$unwind = array(
'$unwind' => '$items'
);
$query = $mongo->store->aggregate($project,$unwind_items);
这是结构:
{
"_id": {
"$oid": "526fdc1fd6b0a8182300009c"
},
"items": [
{
"quantity": "1",
"category_id": {
"$oid": "526fdc1fd6b0a81823000029"
},
"category": "test",
"images": [
{
"name": "9by9easy.PNG",
"path": "upload_files/nibh-vulputate-mauris-corporation/",
"file_path": "upload_files/nibh-vulputate-mauris-corporation/68e7c50bde1476e96ca2461dc553cce5528fb70e41b1f.PNG",
"size": 8761
},
{
"name": "9by9hard.PNG",
"path": "upload_files/nibh-vulputate-mauris-corporation/",
"file_path": "upload_files/nibh-vulputate-mauris-corporation/8cd2dcf4fcd476262db2eba3fdb2c39a528fb70e42757.PNG",
"size": 11506
}
],
"link": "fghfhfhfg"
}
],
"name": "Nibh Vulputate Mauris Corporation",
}
答案 0 :(得分:14)
我知道你可以用MongoDB做的事情。 aggregate()
命令可以根据需要使用尽可能多的参数。
在mongo shell中,有一个像这样的命令
db.collection.aggregate(
{ $project: {
_id: 1,
items: 1
} },
{ $unwind: '$items' },
{ $unwind: '$items.images' }
);
将展开items
子文档,然后展开images
子文档。
根据您问题中的代码,也许这会起作用
$project = array(
'$project' => array(
'_id' => 1,
'items' => 1,
)
);
$unwind_items = array(
'$unwind' => '$items'
);
$unwind_images = array(
'$unwind' => '$items.images'
);
$query = $mongo->store->aggregate($project,$unwind_items,$unwind_images);