SQL check XML node exists via a variable, add & delete -


I'm exploring a way to check whether NODE exists in an XML data type using any MODQL 2012 I want to see the varchar in a stored procedure as node, I have tried something like the bottom, but it does not appear to work, what I want to do is "null" Var2 (SP) As passed in varchar), and if it exists, So if it does not exist, then insert it into the root of all the nodes found in all the examples are hardcodes, which works.

Code as a variable with node: 

  declared @node varchar (255) declared @ x xml set @ annode = 'var2' set @ x = ' & Lt; Root & gt; & Lt; Var1 & gt; 0 & lt; / Var1 & gt; & Lt; Var2 & gt; 123 & lt; / Var2 & gt; & Lt; var3 & gt; 1000 & lt; / Var3 & gt; & Lt; var4 & gt; 989 & lt; / var4 & gt; & Lt; / Root & gt; 'Code with hardcod node: @ x.exist (' //. [Node (=) (select "variable: variable (" node ")] '' - always gives false returns   < P>  
  select @ x.exist ('// var2') - returns true   
 If the node is present, then back into the root element node  
 Edit:  
 I have done the work I can do the following:  
  select @ x.exist ('// * [local-name () = sql: variable ("node") ') - 1 when node exists   
 If the node does not exist, then I add a new node this way Can be set to:  
  set @add = '& lt; newNode & gt; test & lt; / newNode & gt;' @ X.modify ('enter sql: variable ("@ add" ) I can remove the existing nodes in this way:  
  set @del = 'newNode' (/ root) [1] ')   
 Can not be removed to work on @x.modify ('delete / root / * [local-name () = sql: variable ("@ del")])   
 .   
 
 
 I have done my answer to my question:  
  - Check to see if NODE exists @ node varchar (255) set @node = "existingnode" @x.exist ('// * [local-name] = sql: variable ("@ node")]') - Add a new node: Announce @add as xml set @add = '& lt; Abc & gt; Testing & lt; / Abc & gt; '@ X.modify (') in sql: variable ("@ add") (/ root) [1] ') - To delete a node: @ DL, declare it as varchar (255) @ del = 'abc' @x .modify ('delete / root / * [local-name () = sql: variable ("@ del")]')   
 The reason for my question and answer It is that I wanted to work on node names, not attribute values.

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