[RESOLVED] HELP with Short Script

fotofx

Working on a Zen Cart..

I am trying to set up a real simple script that allows us to enter a model number in and automatically add 1 to the stock quantity.

We are going to use this with a bar code scanner and as the items come in we will scan each box and want each scan to add 1.

I thought I had this right but it keeps giving me an error..

Can someone please take a look?

Thanks

<?php 
  require('includes/application_top.php'); 
?> 

<HTML> 
<HEAD> 
<STYLE TYPE="text/css"> 

</STYLE> 
</HEAD> 

<BODY> 
<FORM METHOD="POST" ACTION="<?php echo $_SERVER['PHP_SELF'];?>"> 
<INPUT TYPE="text" SIZE=25 NAME="model" /> 
<BR><BR /> 
<INPUT TYPE="submit" NAME="submit" VALUE="Submit"> 
</FORM> 

<?php 
  mysql_query("UPDATE products SET products_quantity = "products_quantity + 1"" 
WHERE products_model = $_POST['model']); 

?> 
</BODY> 
</HTML>

HalfaBee

change it to this
mysql_query("UPDATE products SET products_quantity = products_quantity + 1 WHERE products_model = '{$_POST['model']}' ");

fotofx

Yup... That worked great..

Now I need to pretty it up a bit.. I want it to accept each item and go back to get another. Back in the old days that was done with a label and a GOTO command..

How would I do it?

Thanks

HalfaBee

It should reload itself as is.

fotofx

The cursor is not returning to the box to accept the next entry..

fotofx

Also need to add a little code to echo something if not found..

HalfaBee

Maybe something like this.

<?php
  require('includes/application_top.php');


if( isset( $_POST['submit'] ) ) {
    $result = @mysql_query("UPDATE products SET products_quantity = products_quantity + 1 WHERE products_model = '{$_POST['model']}' ");

if( mysql_num_rows( $result ) != 1 ) {
    $message = "{$_POST['model']} is not in the database.";
}
}
?>

<HTML>
<HEAD>
<STYLE TYPE="text/css">

</STYLE>
</HEAD>

<BODY>
<? echo $message ?>

<FORM METHOD="POST" ACTION="<?php echo $_SERVER['PHP_SELF'];?>">
<INPUT TYPE="text" SIZE=25 NAME="model" />
<BR><BR />
<INPUT TYPE="submit" NAME="submit" VALUE="Submit">
</FORM>

</BODY>
</HTML>

fotofx

That is sending the not in database message every time.. Even when it updates the record.

HalfaBee

To make it jump to model, you need to use javascript

eg
<body onload="document.getElementByName('model').focus()" >

and put id='model' in the input tag for model.

fotofx

Also the cursor is still not going back to the box

HalfaBee

heh, I always get that mixed up

    if( mysql_affected_rows( ) != 1 ) {
        $message = "{$_POST['model']} is not in the database.";
    }