noob help please

[superblunt]

Hi I got one Mysql INSERT and one Mysql UPDATE that I want to do

 

$sql1="INSERT INTO purchase (date, time, supplier, grade, mainitem, itemcode, itemmass, itemamount, gender, price)

VALUES('$date','$time','$companyname','$grade','$mainitem','$itemcode','$itemmass','$itemamount','$gender','$itemprice')";

 

$sql2="UPDATE mainitem SET totalstock = 36 WHERE mainitem = '$mainitem'";

 

if($companyname=='' or $grade=='' or $mainitem=='' or $itemcode=='' or $itemmass=='' or $itemamount=='' or $gender=='' or $itemprice=='')

{

echo "<b>Please add all fields</b>";

exit;

 

}

 

 

if (!mysql_query($sql1))

  {

  die('Error: ' . mysql_error());

  }

echo "1 record added";

 

if (!mysql_query($sql2))

  {

  die('Error: ' . mysql_error());

  }

echo "1 record Updated";

 

when it is done it only adds the INSERT and not the update even though it shows bought echo's

[scottybwoy]

You can't have single quotes round a $var

 

Change $sql2 to :

 

<?php

$sql2="UPDATE mainitem SET totalstock = 36 WHERE mainitem = '" . $mainitem . "'";

?>

[bluejay002]

though directly implying a  variable inside a string is quite handy but i dont use it...personnal preference i guess. i usually use concatenation of values.

[superblunt]

Thx allot

 

but it still seems to only insert the INSERT and not the update

[superblunt]

The variable is = to cow if i insert cow it works

[superblunt]

o and the variable comes from a $_POST

[superblunt]

I changed it input value to the item ID that seems to work

 

Thx for you help