我正在尝试在Test1类中的顶点和与它们相邻的顶点之间执行连接,但由于不允许使用关键字join
,因此我收到了解析错误。我试图沿关系Path
显示源节点和目标节点的ID,并获取两个顶点的名称
select @rid as P, name as N from Test1 join
(select @rid as R, name as M from (select expand(out('Path')) from Test1))
on t.P=u.R
更新
我尝试了以下解决方案,但所有元素都打包在一个集合中,而在我的情况下,我想执行一个笛卡尔积:
select @rid, out('Path').@rid as rid2, name, out('Path').name as name2 from Test1
答案 0 :(得分:1)
select in.name as name1, out.name as name2, in.@rid as id1, out.@rid as id2 from Path
答案 1 :(得分:1)
我有这个简单的数据集:
检索两个相邻节点的所有元素的简单方法可能是使用 $('#file_input').on('click', function(e){
e.preventDefault();
navigator.camera.getPicture(onSuccess, onFail, {
quality: 100,
destinationType: Camera.DestinationType.FILE_URI,
sourceType: 2, // 1 for camera, 2 for gallery
allowEdit: true, // Can edit the pictures and lets the user pick gallery app
mediaType: Camera.MediaType.PICTURE,
encodingType: Camera.EncodingType.JPEG,
popoverOptions: CameraPopoverOptions,
// targetWidth: 500,
// targetHeight: 500,
saveToPhotoAlbum: false,
correctOrientation: true // false: switches the height and width });
});
function onSuccess(imageData) {
// here I can upload imageData to the server
}
function onFail(message) {
alert('Failed because: ' + message);
}
});
函数。
<强>查询:强>
bothV()
<强>输出:强>
希望有所帮助