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What is reason for following sql code couldn't insert data?

pstmtJ.executeUpdate(sql); → You're sending the original SQL string, since that is calling Statement.executeUpdate(String).

Use pstmtJ.executeUpdate(); instead (PreparedStatement.executeUpdate()) to execute the prepared statement.

Categories : Java

Related to : What is reason for following sql code couldn't insert data?
Couldn't start riemann.bin?
The error seems to state that port 5555 is already in use. Perhaps a previously (aborted) start left a process that still has a handle on the port ? Do a netstat -ao (Windows) or netstat -ap (Linux) to see if there's already a listening port 5555, and which process has a handle on it, then kill that process.

Categories : Clojure
two .so files couldn't load from two different package in an application why?
Read, ldd(1) and ldconfig(8). You may want to set your LD_LIBRARY_PATH environment variable, or configure /etc/ then run ldconfig In both cases, you want the two directories (Linux -including Android- has directories, not "folders") containing your .so ELF shared object files to appear. Read also Program Library HowTo & Drepper's paper: How To Write A Shared Library

Categories : Android
Couldn't find Product without an ID build association
Change this and try: def create @cart = Cart.find(params[:line_item][:cart_id]) @product = Product.find(params[:line_item][:product_id]) @line_item = @product) respond_to do |format| if format.html { redirect_to @line_item.cart, notice: 'Line item was successfully created.' } format.json { render action: 'show', status: :creat

Categories : Ruby On Rails
I couldn't find my error; empty returns from Ajax request
Following this line data: { m: 't', f: 'fu' },. It will do the else statement because m == 't' else if($_POST['m'] === 't') In this else statement it needs to be t == 'fu', but this isn't true because $_POST['t'] doensn't exists. I think this need to be f instead of t. So change this line: if($_POST['t'] === 'fu') { To if($_POST['f'] === 'fu') {

Categories : PHP
What is the reason behind the result of this code?
The constructor of a sub-class (Bb in your case) must call a constructor of the super-class (Abc in your case) before executing its own constructor. This code : public Bb() { System.out.println("constructor B"); } Is equivalent to : public Bb() { super (); // calls the super-class constructor System.out.println("constructor B"); }

Categories : Java
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