Subquery returns more than 1 row - Mysql with Visual Studio 2010 -


i have next code in mysql:

("select id_viaje,         (select nombre            unidades,                 viaje           id_unidad = id_unidades),         (select nombre            empleados,                 viaje           id_empleado = id_conductor),         (select nombre            empleados,                 viaje           id_empleado = id_guarda),         (select nombre            ciudad,                 viaje           id_ciudad = id_salida),         (select nombre            ciudad,                 viaje           id_ciudad = id_llegada),         fecha_salida,         fecha_llegada    viaje; ")

i have tried limit 1 @ end of each one, have replaced = in , have no idea why showing me error:

22:30:30    select id_viaje, (select nombre unidades, viaje id_unidad in (id_unidades)) , (select nombre empleados, viaje id_empleado in (id_conductor)) ,(select nombre empleados, viaje id_empleado in (id_guarda)) , (select nombre ciudad, viaje id_ciudad in (id_salida)) , (select nombre ciudad, viaje id_ciudad in (id_llegada)) , fecha_salida, fecha_llegada viaje limit 1      error code: 1242 subquery returns more 1 row

please need quickly!

this error returned when subquery returns multiple rows when supposed return single row.

your subquery logic wrong. when put subquery inside select statement, must sure subquery return 0 or 1 row.

what can putting left join , use few aliases in query, this:

select id_viaje,         u.nombre unidade_nombre,        e_con.nombre conductor_nombre,         e_grd.nombre guarda_nombre,         c_salida.nombre salida_nombre,        c_llegada.nombre llegada_nombre,         fecha_salida,         fecha_llegada    viaje v left join unidades u on v.id_unidad = u.id_unidades left join empleados e_con on e.id_empleado = v.id_conductor left join empleados e_grd on e.id_empleado = v.id_guarda left join ciudad c_salida on e.id_salida = c_salida.id_ciudad left join ciudad c_llegada on e.id_llegada = c_llegada.id_ciudad ;

try running in mysql workbench first , if runs successfylly there should put application code.

also note code not supposed run in first attempt. please make applicable corrections before running code.

read more mysql joins , mysql subqery


Comments

Post a Comment

Popular posts from this blog

c# - Send Image in Json : 400 Bad request -

from android trying send image class of data, iis webservice. (c#) the problem 400 bad request . the image being encoded base64 . , placed json rest of class elements. my guess base64 not valid inside json. server not understand it. if set string "" , post accepted fine. so question is, how make base64 valid within json array? ( tried url.encode no success). or how should send image android webservice? gson gson = new gson(); string json = gson.tojson(record); // record has param { string base64photo } how big image? i'm pretty sure surpassing iis json size limit (default 4 mb). check http://geekswithblogs.net/frankw/archive/2008/08/05/how-to-configure-maxjsonlength-in-asp.net-ajax-applications.aspx or http://www.webtrenches.com/post.cfm/iis7-file-upload-size-limits good luck!
Read more

javascript - addthis share facebook and google+ url -

i attempting use addthis create simple url shares whichever current page user on when click share button. of right now, have twitter working. when user clicks, able see url populated in tweet box. when user clicks facebook or google+, url doesn't seem show in share box. <ul class="addthis_toolbox addthis_default_style "> <li><a class="addthis_button_facebook"><img src="/img/facebook-social.png" width="20" height="20" border="0" alt="share" /></a></li> <li><a class="addthis_button_twitter"><img src="/img/twitter-social.png" width="20" height="20" border="0" alt="share" /></a></li> <li><a class="addthis_button_google_plusone_share"><img src="/img/google-social.png" width="20" height="20" border="0" alt="s...
Read more

ios - Show keyboard with UITextField in the input accessory view -

i have buttona , when buttona clicked, want keyboard show uitextfield in inputaccessoryview. there way have keyboard show manually , set inputaccessoryview without having uitextfield initially? thanks. you cannot summon keyboard without object can become first responder. there 2 ways work around: subclass uiview , implement uikeyinput protocol in it. example: in .h file: @interface inputobject : uiview<uikeyinput> @property (nonatomic, copy) nsstring *text; @property (nonatomic, strong) uiview *inputaccessoryview; // must override inputaccessoryview , since it's readonly default @end in .m file, implement protocol: - (bool)canbecomefirstresponder { return yes; } - (bool)hastext { return [self.text length]; } - (void)inserttext:(nsstring *)text { nsstring *selftext = self.text ? self.text : @""; self.text = [selftext stringbyappendingstring:text]; [[nsnotificationcenter defaultcenter] postnotificationname:kinputobjec...
Read more