hi there i am new to php ,have been working in asp classic and access but now have switched to php and mysql,and for the practice sake i started to convert my asp and access based projects into php and mysql, i have a inventory system in which i have a stock entry Portion which has aloots of fields concerning to the codeno,partno, auto , product,quantit y, size, height and lots of other fields too but our main concern is with these two
"CODENO and PARTNO
sample values for these tow are here
codeno--------part no
1000--------CSV-3345
1000--------CSV-3346
1001--------CSV-3345
by looking at the values above you can figure outsome how that the values for the code no can be repeated more than once for eg you can enter part nos for the CODENO=1000 as many times as u want lets say partno=CSV-3345 but you cannot enter the existing partno for the codeno=1000 twice means the partno=csv-3345 can not be entered twice under codeno=1000 but the same partno=CSV-3345 can be entered for another code no say
"1001"in asp and access what i did for checking if the record exists this is what i did
this runs accurately perfect foor asp and access but in php the case is different what i have done in my php code is here
BUT THE PROB IS THAT IF I ENTER THE PART NO WHICH ALREADY EXISTS IT TELLS THAT THE VALUE EXISTS BUT IF I ENETER ANY OTHER PART NO THAT DOESNOT EXIST IT JUST HANKS
ANY SUGGESTIONS WOULD BE HIGHLY APPRECIATED
REGARDS,
OMER
"CODENO and PARTNO
sample values for these tow are here
codeno--------part no
1000--------CSV-3345
1000--------CSV-3346
1001--------CSV-3345
by looking at the values above you can figure outsome how that the values for the code no can be repeated more than once for eg you can enter part nos for the CODENO=1000 as many times as u want lets say partno=CSV-3345 but you cannot enter the existing partno for the codeno=1000 twice means the partno=csv-3345 can not be entered twice under codeno=1000 but the same partno=CSV-3345 can be entered for another code no say
"1001"in asp and access what i did for checking if the record exists this is what i did
Code:
sqlf="SELECT Stk_items.Prt_no from Stk_items where Stk_items.Code_no="&lngcodeno
rf.Open sqlf, connf, 3
foundit=False
do until rf.EOF OR foundit //runs until foundit becomes true
if (StrComp(rf.Fields("Prt_no"), partno, vbTextCompare) = 0) then foundIt=True //assign true to foundit
else
rf.MoveNext //else move to next record
End if
Loop
rf.Close
set rf = Nothing
set connf= Nothing
Code:
$hostname="localhost";
$username="root";
$password="66456";
$database="db1";
$conn_stk = mysql_pconnect($hostname, $username, $password) or
trigger_error(mysql_error(),E_USER_ERROR);
mysql_select_db($database, $conn_stk);
$codeno=$_POST["code_no"];
$prtno=$_POST["prt_no"];
$prd=$_POST["prd"];
$brand=$_POST["brand"];
$auto=$_POST["auto"];
$uprc=$_POST["unit_price"];
$qtty=$_POST["qtty"];
$sroom=$_POST["show_room"];
$st1=$_POST["store_1"];
$st2=$_POST["store_2"];
$st3=$_POST["store_3"];
$size=$_POST["size"];
$height=$_POST["height"];
$od1=$_POST["out_d1"];
$od2=$_POST["out_d2"];
$id1=$_POST["in_d1"];
$id2=$_POST["in_d2"];
$trep=$_POST["t_rep"];
$rp1=$_POST["rp_no1"];
$rp2=$_POST["rp_no2"];
$rp3=$_POST["rp_no3"];
$rp4=$_POST["rp_no4"];
$rp5=$_POST["rp_no5"];
$rp6=$_POST["rp_no6"];
$sqlf="select Prt_no from stk_items where Code_no='$codeno'";
$rs=mysql_query($sqlf) or die(mysql_error().'<br />'.$sqlf);
$foundit=false;
if(mysql_num_rows($rs))
{
while($f=mysql_fetch_array($rs));
{
$ret=strcmp($f['Prt_no'],$prtno);
if($ret==0)
{
$foundit=true;
}
else
{}
}
}
ANY SUGGESTIONS WOULD BE HIGHLY APPRECIATED
REGARDS,
OMER
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