Cypher questions - return updates count, update 2 things in a single query - Neo4j

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Message from discussion Cypher questions - return updates count, update 2 things in a single query
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Date: Thu, 18 Oct 2012 20:12:50 +0200
Message-ID: <CAAuGTBgetnug+Uwb3DtDNQM1cdH8a68UAJ6vp8zir-3na7-...@mail.gmail.com>
Subject: Re: [Neo4j] Cypher questions - return updates count, update 2 things
 in a single query
From: Wujek Srujek <wujek.sru...@gmail.com>
To: neo4j@googlegroups.com
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I start the transaction myself before the first query, and finish it after
the second. As far as I know, cyhper will start own transactions for write
queries if there is no active transaction - so in this case the transaction
penalty should not exist, I think?
Will try your approach as soon as I can.
wujek

On Thu, Oct 18, 2012 at 7:41 PM, Wes Freeman <freeman....@gmail.com> wrote:

> It is better to do it with one query, if possible, because it creates a
> sort of transaction (and is obviously more efficient over REST especially,
> since it's one request instead of two).
>
> I didn't have time to look at this before, but here goes a pretty hacky
> attempt:
>
> start me=node(3), like=node(2)
> match me-[l:likes]->()
> // this where is optional, but this way there's less modified
> where not(me-[l]->like) and l.preferred! = true
> set l.preferred = false
> // hack to always return a row (we don't even use the count)
> with count(*)
> // we have to re-start the query, since the with wiped out our previous
> named vars
> start me=node(1), like=node(3)
> match me-[l:likes]->like
> set l.preferred=true
> return l,me, like;
>
> Wes
>
> On Thu, Oct 18, 2012 at 12:31 PM, wujek <wujek.sru...@gmail.com> wrote:
>
>> Hi. I still can't get it to work.
>> Your comment about getting rid of the r.preferred predicate as it is
>> always false is wrong - it is false only the first time, when no preference
>> is defined; after the initial preferred is set, setting the next one must
>> unset the previous one, which will then mean the predicate is will find
>> something. And this predicate proves to be the problematic one, as the
>> 'with' subquery doesn't work if the where is false. I am not sure why that
>> is, and the documentation on 'with' is very vague on this.
>>
>> The second option of inverting the query parts also doesn't work. I also
>> don't follow it logically - whereas the first part sets the new preferred
>> to true for a reference to the node I want, the second part will just reset
>> it to false, as it does that for every relationship of type LIKE, right?
>>
>> Maybe I wasn't specific enough about what I want to achieve. So: A person
>> has many things that she likes, and there can be either one or no preferred
>> thing. If there is none, setting a new preferred will just set it; if there
>> is one already, setting another thing as preferred must unset the
>> previously set one.
>>
>> I am able to do it with 2 queries - first unset the previous preferred
>> thingy, and then set the new one. It seems it is not possible to do it in
>> just one query - it's fine, I was just curious if this is possible.
>>
>> Thanks for your help, though.
>>
>> wujek
>>
>>
>> On Wednesday, October 17, 2012 10:07:37 PM UTC+2, Abdul Azeez Shaik wrote:
>>
>>> Also, you can do other way round,
>>>
>>> start me = node({me}), like=node{{like}} match me-[r:LIKES]->like set
>>> r.preferred = true
>>> WITH me
>>> match me-[r:LIKES]->t set r.preferred = false where r.preferred = true
>>>
>>> On Thu, Oct 18, 2012 at 1:33 AM, Abdul Azeez Shaik <azeezte...@gmail.com
>>> > wrote:
>>>
>>>>  AFIK, WHERE predicate can be removed, as anyways r.preferred is false.
>>>> But, i don't know what would be the performance effect on that. May be
>>>> Peter or Michael can answer this better.
>>>>
>>>> BTW, what is your data size?
>>>>
>>>> Thanks,
>>>> Abdul
>>>>
>>>>
>>>> On Thu, Oct 18, 2012 at 12:31 AM, wujek <wujek....@gmail.com> wrote:
>>>>
>>>>> Hi. This:
>>>>>
>>>>> start me = node({me}), like=node{{like}} match me-[r:LIKES]->t set
>>>>> r.preferred = false where r.preferred = true
>>>>> with me, like
>>>>> match me-[r:LIKES]->t set r.preferred = true
>>>>>
>>>>> doesn't work for me in the following scenario: there is no preferred
>>>>> yet (all relations have preferred = false), so the first part doesn't
>>>>> update anything, and then the second doesn't update anything as well.
>>>>> When I first set a preferred to some node, then the query works fine.
>>>>> Which means I cannot be used as is in my use case, as there is no preferred
>>>>> initially, and this seems to prevent setting one.
>>>>>
>>>>> Am I doing something wrong?
>>>>>
>>>>> wujek
>>>>>
>>>>>
>>>>> On Monday, October 15, 2012 12:57:03 PM UTC+2, Abdul Azeez Shaik wrote:
>>>>>>
>>>>>> Hi Wujek,
>>>>>>
>>>>>> You can use WITH clause to have two sub-queries in one query.
>>>>>> Your query would be,
>>>>>> start me = node({me}), like=node{{like}} match me-[r:LIKES]->t set
>>>>>> r.preferred = false where r.preferred = true
>>>>>> WITH me, like
>>>>>> match me-[r:LIKES]->t set r.preferred = true
>>>>>>
>>>>>>
>>>>>>  --
>>>>>
>>>>>
>>>>>
>>>>
>>>>
>>>  --
>>
>>
>>
>
>  --
>
>
>

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I start the transaction myself before the first query, and finish it after =
the second. As far as I know, cyhper will start own transactions for write =
queries if there is no active transaction - so in this case the transaction=
 penalty should not exist, I think?<div>
Will try your approach as soon as I can.</div><div>wujek<br><br><div class=
=3D"gmail_quote">On Thu, Oct 18, 2012 at 7:41 PM, Wes Freeman <span dir=3D"=
ltr">&lt;<a href=3D"mailto:freeman....@gmail.com" target=3D"_blank">freeman=
....@gmail.com</a>&gt;</span> wrote:<br>
<blockquote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1p=
x #ccc solid;padding-left:1ex">It is better to do it with one query, if pos=
sible, because it creates a sort of transaction (and is obviously more effi=
cient over REST especially, since it&#39;s one request instead of two).<div=
>
<br></div><div>I didn&#39;t have time to look at this before, but here goes=
 a pretty hacky attempt:<br>

<div><br></div><div><div>start me=3Dnode(3), like=3Dnode(2)=A0</div><div>ma=
tch me-[l:likes]-&gt;()=A0</div><div>// this where is optional, but this wa=
y there&#39;s less modified</div><div>where not(me-[l]-&gt;like) and l.pref=
erred! =3D true=A0</div>


<div>set l.preferred =3D false=A0</div><div>// hack to always return a row =
(we don&#39;t even use the count)</div><div>with count(*)=A0</div><div>// w=
e have to re-start the query, since the with wiped out our previous named v=
ars</div>


<div>start me=3Dnode(1), like=3Dnode(3)=A0</div><div>match me-[l:likes]-&gt=
;like=A0</div><div>set l.preferred=3Dtrue=A0</div><div>return l,me, like;=
=A0</div><span class=3D"HOEnZb"><font color=3D"#888888"><div><br></div><div=
>Wes</div></font></span><div>
<div class=3D"h5"><br><div class=3D"gmail_quote">On Thu, Oct 18, 2012 at 12=
:31 PM, wujek <span dir=3D"ltr">&lt;<a href=3D"mailto:wujek.sru...@gmail.co=
m" target=3D"_blank">wujek.sru...@gmail.com</a>&gt;</span> wrote:<br>

<blockquote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1p=
x #ccc solid;padding-left:1ex">Hi. I still can&#39;t get it to work.<div>Yo=
ur comment about getting rid of the r.preferred predicate as it is always f=
alse is wrong - it is false only the first time, when no preference is defi=
ned; after the initial preferred is set, setting the next one must unset th=
e previous one, which will then mean the predicate is will find something. =
And this predicate proves to be the problematic one, as the &#39;with&#39; =
subquery doesn&#39;t work if the where is false. I am not sure why that is,=
 and the documentation on &#39;with&#39; is very vague on this.</div>


<div><br></div><div>The second option of inverting the query parts also doe=
sn&#39;t work. I also don&#39;t follow it logically - whereas the first par=
t sets the new preferred to true for a reference to the node I want, the se=
cond part will just reset it to false, as it does that for every relationsh=
ip of type LIKE, right?</div>


<div><br></div><div>Maybe I wasn&#39;t specific enough about what I want to=
 achieve. So: A person has many things that she likes, and there can be eit=
her one or no preferred thing. If there is none, setting a new preferred wi=
ll just set it; if there is one already, setting another thing as preferred=
 must unset the previously set one.</div>


<div><br></div><div>I am able to do it with 2 queries - first unset the pre=
vious preferred thingy, and then set the new one. It seems it is not possib=
le to do it in just one query - it&#39;s fine, I was just curious if this i=
s possible.</div>


<div><br></div><div>Thanks for your help, though.</div><div><br></div><div>=
wujek<div><br><br>On Wednesday, October 17, 2012 10:07:37 PM UTC+2, Abdul A=
zeez Shaik wrote:</div><blockquote class=3D"gmail_quote" style=3D"margin:0;=
margin-left:0.8ex;border-left:1px #ccc solid;padding-left:1ex">


<div>Also, you can do other way round,<br><br><div>start me =3D node({me}),=
 like=3Dnode{{like}}=A0match me-[r:LIKES]-&gt;like set r.preferred =3D true=
</div><div>WITH me</div><div>match me-[r:LIKES]-&gt;t set r.preferred =3D f=
alse where r.preferred =3D true</div>




<div><br></div></div><div class=3D"gmail_quote"><div>On Thu, Oct 18, 2012 a=
t 1:33 AM, Abdul Azeez Shaik <span dir=3D"ltr">&lt;<a>azeezte...@gmail.com<=
/a>&gt;</span> wrote:<br></div><blockquote class=3D"gmail_quote" style=3D"m=
argin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">


<div>

AFIK, WHERE predicate can be removed, as anyways r.preferred is false.<div>=
But, i don&#39;t know what would be the performance effect on that. May be =
Peter or Michael can answer this better.</div><div><br></div><div>BTW, what=
 is your data size?</div>





<div><br></div><div>Thanks,</div></div><div>Abdul<div><div><div><br><div><b=
r><div class=3D"gmail_quote">On Thu, Oct 18, 2012 at 12:31 AM, wujek <span =
dir=3D"ltr">&lt;<a>wujek....@gmail.com</a>&gt;</span> wrote:<br>




<blockquote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1p=
x #ccc solid;padding-left:1ex">Hi. This:<div><div><div><br></div><div>start=
 me =3D node({me}), like=3Dnode{{like}} match me-[r:LIKES]-&gt;t set r.pref=
erred =3D false where r.preferred =3D true</div>





</div><div>with me, like</div><div><div>match me-[r:LIKES]-&gt;t set r.pref=
erred =3D true</div></div></div><div><br></div><div>doesn&#39;t work for me=
 in the following scenario: there is no preferred yet (all relations have p=
referred =3D false), so the first part doesn&#39;t update anything, and the=
n the second doesn&#39;t update anything as well.<div>





When I first set a preferred to some node, then the query works fine. Which=
 means I cannot be used as is in my use case, as there is no preferred init=
ially, and this seems to prevent setting one.</div><div><br></div><div>





Am I doing something wrong?</div><span><font color=3D"#888888"><div><br></d=
iv></font></span><div><span><font color=3D"#888888">wujek</font></span><div=
><br><br>On Monday, October 15, 2012 12:57:03 PM UTC+2, Abdul Azeez Shaik w=
rote:<blockquote class=3D"gmail_quote" style=3D"margin:0;margin-left:0.8ex;=
border-left:1px #ccc solid;padding-left:1ex">





Hi Wujek,<div><br></div><div>You can use WITH clause to have two sub-querie=
s in one query.</div><div>Your query would be,</div><div>start me =3D node(=
{me}), like=3Dnode{{like}} match me-[r:LIKES]-&gt;t set r.preferred =3D fal=
se where r.preferred =3D true</div>







<div>WITH me, like</div><div>match me-[r:LIKES]-&gt;t set r.preferred =3D t=
rue</div><div><br></div></blockquote><blockquote class=3D"gmail_quote" styl=
e=3D"margin:0;margin-left:0.8ex;border-left:1px #ccc solid;padding-left:1ex=
">





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<p></p>

-- <br>
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<p></p>

-- <br>
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