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javahibernatehql
Camille R
Camille R

Reputation: 1433

Hibernate multiple parameters setString generated Java

I'm using hibernate, and trying to do a LIKE on certain fields.

I'm splitting a string, and then generating the HQL, with 

table.entry LIKE :argsearch_0 OR table.entry LIKE :argsearch_0 OR 
table.entry LIKE :argsearch_1 OR table.entry LIKE :argsearch_1

(0 and 1 is in fact incremented with a counter).

But i get :

Not all named parameters have been set: [argsearch_0]

First question :

Second question :

Upvotes: 0

Views: 2129

Answers (2)

Vinay Johnwesly
Vinay Johnwesly

Reputation: 34

  You can Try this  



    **Step 1--:** Add How many parameter you need just add in hashmap
   -------------------------------------------------------------------
    
    HashMap param_List=new HashMap();
    param_List.put("contactNo",22);

    
    **Step 2--:** Just You pass your Query
   -------------------------------------------------------------------
    
    Query query1 = session.createQuery("select * from emailTemplate  where c.contactNo =:contactNo");


  **Step 3--:** What ever Data type is no matters but get an output.
  -------------------------------------------------------------------

    for(Object paramKey : param_List.keySet())
    {
    query1.setParameter(paramkey.toString(), param_List.get(paramKey);
    }


 **Step 4--:**
-------------------------------------------------------------------
    String finalResult=query1.getSingleResult().toString();

Upvotes: 0

Steve Ebersole
Steve Ebersole

Reputation: 9443

Depends what you mean when you ask "Can I used 2 named parameter and do only 1 setParameter".

In your original query you have 2 named parameters ('argsearch_0' and 'argsearch_1') and each has 2 usages in the query. So you have to call set for both 'argsearch_0' and 'argsearch_1'. But you only call set once for each (actually you can call set multiple times for each parameter if you really want, but only the last once is used.

As for your second question, as someone already pointed out, its because you have a bug in your code. You are not setting the value for the 'argsearch_0' parameter.

Upvotes: 1

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