Jump to content

totajamal

New Members
  • Posts

    2
  • Joined

  • Last visited

totajamal's Achievements

Newbie

Newbie (1/5)

0

Reputation

  1. ok here is my code: I have two tables in database recipes and cookie. i am trying to copy id from recipes and add to cookie for each row. <?php //connect to datatbase $connect = mysql_connect("localhost","kosterb_jamaltag","jamaltag") or die (mysql_error()); mysql_select_db("kosterb_jamaltag") or die ("Could not find db") or die (mysql_error()); $uname = $_POST['membername']; $cname = $_POST['cookiename']; $ingDis = str_replace("%&%","'",$_POST['inge']); $dirDis = str_replace("%&%","'",$_POST['dire']); //New order if(!isset($_SESSION['id'])) { $query = "SELECT MAX(id) FROM Recipes"; $result = mysql_query( $query ); if($result==0) { echo "<b>Error ".mysql_errno().": ".mysql_error()."</b>"; } else { $row = mysql_fetch_array ($result); $id= $row[0]+1; $_SESSION['id'] = $id; $_SESSION['ingrediants'] = $ingDis; $query="SELECT From Recipes WHERE id='$id'"; $result=mysql_query($query); if($result==0) { echo "<b>Error ".mysql_errno().": ".mysql_error()."</b>"; } } } $id= $_SESSION['id']; //mysql_query("SELECT FROM Recipes WHERE id='$id'"); //mysql_query("INSERT INTO Cookie VALUES( '','$id','$dirDis')"); $query="insert into Cookie SET " ." id='$id',membername='$uname', cookiename='$cname'"; if ($bug) { echo "=$query"; } $result = mysql_query($query); if($result==0) { echo "<b>Error ".mysql_errno().": ".mysql_error()."</b>"; } // Get user/cookie name from Cookie table mysql_query("SELECT * FROM Cookie WHERE membername='$uname' AND cookiename='$cname'" ); echo "<h1 style=text-align:\"center\"><font color=\"darkbrown\">$cname by $uname<br /></h1></font><br /><br />" ; //grap file $file = $_FILES['image']['tmp_name']; if(!isset($file)) echo "please select an image"; else { $image=addslashes(file_get_contents($_FILES['image']['tmp_name'])); $image_name = addslashes($_FILES['image']['name']); $image_size = getimagesize($_FILES['image']['tmp_name']); if($image_size==FALSE) echo "That is not an image"; else { if(!$insert = mysql_query("INSERT INTO Cookie VALUES ('','','$image_name','$image','','')")) echo "There is a problem uploading the image."; else { $lastid = mysql_insert_id(); echo "<img src=get.php?id=$lastid width=\"300\" \> <br>"; } } } // Insert Ingeriadiants to database mysql_query("INSERT INTO Recipes VALUES( '','$ingDis','$dirDis')"); //Get discriptions from Ingrediants table mysql_query("SELECT * FROM Recipes WHERE ingrediants='$inge' AND directions='$dire'"); echo nl2br("<p><h1><font color= \"green\">Ingrediants</font></h1> $ingDis<br /> </> <p><h1><font color= \"green\">Directions</font></h1> $dirDis<br /></p>") ; ?>
  2. Hello,I am new with php. I have proplems with id and my picture, the id in recipe table will be the same 72 or 0 evry time I insert information to the table. and the picure did not upload. this is my code: displayshare.php
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.