是否有可能使用以下方法获取Alfresco文件夹的nodeRef:search.luceneSearch("PATH_OF_THE_FOLDER")
答案 0 :(得分:1)
没有理由为每个结果调用findNode,因为luceneSearch调用返回一个节点数组。
如果您知道查询返回单个节点,则可以执行以下操作:
<Grid Background="LightGray">
<Grid.RowDefinitions>
<RowDefinition Height="Auto"/>
<RowDefinition Height="Auto"/>
<RowDefinition Height="*"/>
</Grid.RowDefinitions>
<Grid Grid.Row="0">
<Grid.ColumnDefinitions>
<ColumnDefinition Width="3*"/>
<ColumnDefinition Width="150"/>
<ColumnDefinition Width="Auto"/>
<ColumnDefinition Width="Auto"/>
<ColumnDefinition Width="Auto"/>
</Grid.ColumnDefinitions>
<TextBlock Text="Items" Style="{StaticResource FormHeaderStyle}"/>
<xctk:WatermarkTextBox Grid.Column="1" Name="ItemNameSearchTxt" TextChanged="ItemNameSearchTxt_TextChanged"/>
<Button Grid.Column="2" Name="AddNewBtn" Content="Add New" Padding="20, 0" HorizontalAlignment="Right" Click="AddNewBtn_Click"/>
<Button Grid.Column="3" Name="DeleteBtn" Content="Delete" Padding="20, 0" HorizontalAlignment="Right" Click="DeleteBtn_Click"/>
<Button Grid.Column="4" Name="ExportBtn" Content="Export" Padding="20, 0" HorizontalAlignment="Right"/>
</Grid>
<Separator Grid.Row="1" Height="1" Width="Auto" VerticalAlignment="Bottom"/>
<DataGrid Grid.Row="2" Name="ItemsDataGrid" ItemsSource="{Binding ItemDS}" IsReadOnly="True" AutoGenerateColumns="False">
<DataGrid.Columns>
...
</DataGrid.Columns>
</DataGrid>
</Grid>
答案 1 :(得分:0)
我终于找到了解决方案。
nodeRef
返回一个对象列表,所以我循环遍历此列表并获取列表元素的var folder;
var nodes = search.luceneSearch("PATH:\"/app:company_home/app:dictionary/cm:StarXpert_x0020_Workflow\"");
for each(node in nodes){
folder=search.findNode(node.nodeRef);
}
,如下所示:
folder
现在在nodeRef
我有luceneSearch
返回的单个节点的import pandas as pd
filename1 = 'filename1.xlsx'
filename2 = 'filename2.xlsx'
df1 = pd.read_excel(filename1, index_col=0)
df2 = pd.read_excel(filename2, index_col=0)
。