Question

人们，下午好！我认为这可能是一个愚蠢的问题..

我在Hibernate中有以下代码：

xmnValue = pixelToLonFxn( ext.rr@xmin, coeff ) 
xmxValue = pixelToLonFxn( ext.rr@xmax, coeff ) 
ymnValue = pixelToLatFxn( ext.rr@ymax, coeff )
ymxValue = pixelToLatFxn( ext.rr@ymin, coeff )
rr2 = raster(nrows=nrow(imageData ), ncols=ncol(imageData ), xmn=xmnValue, xmx=xmxValue, ymn=ymnValue, ymx=ymxValue )
values( rr2 ) = (values( imageData ) - 7 ) * 5
rr2[rr2<minDbzValue]  <- NA
pol < - rasterToPolygons(rr2, fun=function(x){ x >= minDbzValue } )
pol2 = spTransform( pol, "+proj=utm +zone=14 +datum=WGS84 +units=km" )
ptsUTM = SpatialPoints( pol2 )
coords = ptsUTM@coords
n = length( pol2@data$layer )
wxData = data.frame( x = numeric( n ), y = numeric( n ), z = numeric( n ),  c = character( n ), stringsAsFactors = FALSE )
wxData$x = coords[,1]
wxData$y = coords[,2]
wxData$z = pol2@data$layer
wxData$c = colors[ match( wxData$z, colorRanges ) ]
xmnValue = pixelToLonFxn( ext.rr@xmin, coeff ) 
xmxValue = pixelToLonFxn( ext.rr@xmax, coeff ) 
ymnValue = pixelToLatFxn( ext.rr@ymax, coeff )
ymxValue = pixelToLatFxn( ext.rr@ymin, coeff )
rr2 = raster(nrows=nrow(imageData ), ncols=ncol(imageData ), xmn=xmnValue, xmx=xmxValue, ymn=ymnValue, ymx=ymxValue )
values( rr2 ) = (values( imageData ) - 7 ) * 5
rr2[rr2<minDbzValue]  <- NA 
pol <- rasterToPolygons(rr2, fun=function(x){ x >= minDbzValue } )
pol2 = spTransform( pol, "+proj=utm +zone=14 +datum=WGS84 +units=km" )
ptsUTM = SpatialPoints( pol2 )
coords = ptsUTM@coords
n = length( pol2@data$layer )
wxData = data.frame( x = numeric( n ), y = numeric( n ), z = numeric( n ), c = character( n ), stringsAsFactors = FALSE )
wxData$x = coords[,1]
wxData$y = coords[,2]
wxData$z = pol2@data$layer
wxData$c = colors[ match( wxData$z, colorRanges ) ]

有谁知道如何在此查询中获取字段的值？

示例：我如何获得此人（员工）的姓名。

注意：我想使用一个很好的Hibernate练习......我相信重复是不好的：

query = session.createQuery ("select F from Employee F where F.email =" + email);

你能帮帮我吗？：）

谢谢。

Answer 1

在Hibernate 5.2之前：

String sql = "SELECT e.person FROM Employee e WHERE e.email = :email";
Query query = session.createQuery( sql )     
query.setParameter( "email", emailAddress );
List<Person> people = (List<Person>) query.getResultList();

在Hibernate 5.2及更高版本中：

String sql = "SELECT e.person FROM Employee e WhERE e.email = :email";
Query query = session.createQuery( sql, Person.class );
query.setParameter( "email", emailAddress );
List<Person> people = query.getResultList();

通过将Hibernate EntityManager合并到Core作为5.2.x +的一部分，您现在可以获得更好的类型安全查询，以避免以后再进行转换：）。

使用良好的Hibernate实践获取值

1 个答案: