?
Solved

problem with odbc_num_rows

Posted on 2003-11-04
4
Medium Priority
?
537 Views
Last Modified: 2013-12-12
hi everybody,

I have a problem with the count of the rows  after a query.
I work with a DB MS Access and I know that odbc_num_rows doesn't work.

I thied this:


$sql1 = "SELECT COUNT(*) FROM contribuente where con_numero_controllo in (".$sql.") as righe";
     $sql_result = odbc_prepare($cn,$sql1) or die("ERROR");
     odbc_execute($sql_result) or die("ERROR");
     $rc = odbc_fetch_into($sql_result, $my_array);
     echo ("Total rows: " . $my_array[0]);
     odbc_free_result($sql_result);


the subselect  $sql is:

$sql="select con_numero_controllo as N_controllo,con_numero_registro as N_registro, con_cognome as Cognome, con_nome as Nome,con_data_nascita as Data_nascita
from contribuente
where (1=1) and (con_cognome like'ca%')";

con_cognome is a choice that the uses makes. for example i write "ca" and it gives me back the all name that gegin with ca.

there is something wrong in the the subselect bacause it can't perform it.

Can anyone help me please..

thanks in advance



0
Comment
Question by:campanal
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 2
  • 2
4 Comments
 
LVL 5

Expert Comment

by:Karitz
ID: 9677013
If you want to get the number of rows you use something like

$sql1 = "SELECT COUNT(*) as righe FROM contribuente where con_numero_controllo in (".$sql.")";

but if you want to get all the results and put them in an array you use

$sql1 = "SELECT * FROM contribuente where con_numero_controllo where (1=1) and (con_cognome like'ca%')";


this way you are sure you are getting all the rows (of data) in your table
0
 

Author Comment

by:campanal
ID: 9677042
But I only want the number of rows because if the result is > 2 then I show a messagge..
0
 
LVL 5

Accepted Solution

by:
Karitz earned 150 total points
ID: 9677640
I that case use

$sql1 = "SELECT COUNT(*) as righe FROM contribuente where where (1=1) and (con_cognome like'ca%')";
 $sql_result = odbc_prepare($cn,$sql1) or die("ERROR");
 odbc_execute($sql_result) or die("ERROR");
     $rc = odbc_fetch_into($sql_result, $my_array);
     echo ("Total rows: " . $my_array[0]);
     odbc_free_result($sql_result);

this should give you the nyumber of rows

alternatively


$sql1 = "SELECT * FROM contribuente where con_numero_controllo where (1=1) and (con_cognome like'ca%')";
 $sql_result = odbc_prepare($cn,$sql1) or die("ERROR");
 odbc_execute($sql_result) or die("ERROR");

     $total=odbc_num_rows( $sql_result )
     echo ("Total rows: " . $total);
     odbc_free_result($sql_result);
0
 

Author Comment

by:campanal
ID: 9678245
thanks for the help
0

Featured Post

Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

This article discusses four methods for overlaying images in a container on a web page
3 proven steps to speed up Magento powered sites. The article focus is on optimizing time to first byte (TTFB), full page caching and configuring server for optimal performance.
The viewer will learn how to count occurrences of each item in an array.
This tutorial will teach you the core code needed to finalize the addition of a watermark to your image. The viewer will use a small PHP class to learn and create a watermark.
Suggested Courses

718 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question