$sql="select * from mytable "

[rabmerab]

Hi every one,

 

May be this is a just a basic but I am stuck.

 

I am using the sql query  $sql="select * from mytable " to try to display 4 rows but I getting  only one.

when I am using mysql command line with the same query I get all the 4 rows.

the problem only in php.

The other things like connecting to mysql and selecting the database been done.

Any help please.

Thanks in advance.

 

Rabmerab

 

Screen shot:

 

#connect to my sql

$conn=@mysql_connect("localhost","me","mypass") or die ("Couldn't connect");

 

 

# select database

$rs=@mysql_select_db("mydatabase",$conn)or die("could not  select database");

 

 

#Create the sql query

 

$sql="select * from mytable ";

       

# Execute query

 

$rs= mysql_query($sql,$conn);

 

 

#get the number of rows

    $num=mysql_numrows($rs);

   

    while($row= Mysql_fetch_array($rs) )

    {

    $list="<br>".$row["imei"];

  echo($list);

    }

 

?>

 

 

 

 

[fenway]

You have list =, not list .= -- you're replacing it, and only getting the last row's value.

[rabmerab]

Hi there,

 

Thanks for replaying.

Is not that the problem;  in the program above, using $num=mysql_numrows($rs) to count number of rows should deliver 4, even if I add more rows always  echo($num) only show 1.

Any more ideas.

 

rabmerab

 

 

[fenway]

That seems impossible... wait, it's it mysql_num_rows()?

[rhodesa]

Yeah, it should be erroring out at mysql_numrows...do you have error reporting turned on?

[rabmerab]

Hi again,

 

I don't know how come it didn't display the error even i got display_errors in phpinfo shows it is on.

It did shows me other errors.

Any way even I change it to $num=mysql_num_rows($rs), now it shows 2 but in reality, the table has 6 rows.

I don't know what's going on.

 

thanks for any more help.

 

rabmerab

[gizmola]

rahmerab:

 

Paste your real actual code here please. Everyone is guessing, because you're not providing the information we need.

[fenway]

I don't know what's going on.

Neither do any of us... turn on error reporting, and as gizmola said, let's make sure we're talking about the same code here.