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v4.2
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中文EN
v4.2

    Path Template

    A path template n().e()...n() can apply filters on each node and edge of a path independently. It is a very flexible way of path query.

    Syntax:

    • Statement alias supported (PATH type)
    • Optional parameters:
    Parameter Type Specification Description Structure Type of Custom Alias
    no_circle() / / To dismiss the paths with circles; see Basic Concept - Terminologies for the definition of circle Not supported
    limit() Int -1 or >=0 Number of results to return in single execution, -1 means to return all results Not supported

    Path template query supports the prefix optional. With this prefix, a pseudo path "0-0-0" will be returned if there is no result found, and the value of the alias reference of any node or edge in the pseudo path is 0.

    1 step

    Example: as the path structure shown below, find transactions from a customer's card to another customer's card, return 10 paths and their properties

    n({@customer}).re({@has}).n({@card})
      .re({@transfer}).n({@card})
      .le({@has}).n({@customer}) as p
    limit 10 
    return p{*}
    

    N steps

    Example: as the path structure shown below, find 3-step transaction paths from Customer CU001's card, return 10 paths and their properties

    n({_id == "CU001"}).re({@has}).n({@card})
      .re({@transfer})[3].n({@card}) as p
    limit 10 
    return p{*}
    

    1~N steps

    Example: find 1 to 3 steps' transaction paths from Customer CU001's card, return 10 paths and their properties

    n({_id == "CU001"}).re({@has}).n({@card})
      .re({@transfer})[:3].n({@card})
      .le({@has}).n({@customer}) as p
    limit 10 
    return p{*}
    

    M~N steps

    Example: find 2 to 3 steps' transaction paths from Customer CU001's card, return 10 paths and their properties

    n({_id == "CU001"}).re({@has}).n({@card})
      .re({@transfer})[2:3].n({@card})
      .le({@has}).n({@customer}) as p
    limit 10 
    return p{*}
    

    Non-weighted Shortest Path

    Example: as the path structure shown below, find the shortest transaction paths from Customer CU001's card to Customer CU002's card within 3 steps, return 10 paths and their properties.

    n({_id == "CU001"}).re({@has}).n({@card} as n1)
    n({_id == "CU002"}).re({@has}).n({@card} as n2)
    with n1, n2
    n(n1).re({@transfer})[*:3].n(n2) as p 
    limit 10
    return p{*}
    

    Analysis: in this example, two path templates are first applied to find Customer CU001 and Customer CU002's cards respectively, then one card is used as scr and another as dest in a third transaction path template to find the shortest path between them.

    Circle

    Example: as the path structure shown below, find the circular single-direction transaction paths from Customer CU001's card within 4 steps, return 10 paths and their properties

    n({_id == "CU001"}).re({@has}).n({@card} as n)
      .re({@transfer})[:4].n(n) as p
    limit 10
    return p{*}
    

    no_circle()

    Example: find the non-circular single-direction transaction paths from Customer CU001's card within 4 steps, return 10 paths and their properties

    n({_id == "CU001"}).re({@has}).n({@card})
      .re({@transfer})[:4].n({@card})
      .no_circle() as p
    limit 10
    return p{*}
    

    limit

    Example: find the transaction paths from Card CA001, CA002, and CA005 to Card CA016 within 3 steps, return 2 paths for each pair and their properties

    find().nodes({_id in ["CA001", "CA002", "CA005"]}) as n
    n(n).e()[:3].n({_id == "CA016"})
      .limit(2) as p
    return p{*}
    

    OPTIONAL

    Example: find the transaction paths from Card CA001, CA002, and CA005 to Card CA016 within 3 steps, return 2 paths for each pair and their properties; if no paths returned, return optional paths

    find().nodes({_id in ["CA001", "CA002", "CA005"]}) as n
    optional n(n).e()[:3].n({_id == "CA016"})
      .limit(2) as p
    return p{*}
    

    0~N steps

    Example: find customers who viewed or purchased Product P001, then find other customers 0 to 2 steps away from those customers, return all secondary customers' properties.

    n({_id == "P001"}).e().n({@customer})
      .e()[0:2].n({@customer}) as p
    return p{*}
    

    Inter-step Filter

    Example: find the outbound @transfer paths from Card CA001 to Card CA002 within 4 steps, with each @transfer time greater than the previous one within the same path, return 10 paths and their properties.

    n({_id == "CA001"}).re({@transfer.time > prev_e.time})[:4].n({_id == "CA002"}) as p
    limit 10
    return p{*}
    
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