staci Posted June 7, 2011 Share Posted June 7, 2011 Ok, so here's my problem. I am trying to create a website where I resell various items for people. One way I will be doing this is by posting their items on my website. Since I will be adding new items all the time I thought I would use some php to make it easier. Problem is, I'm very much new to php, though I do have some training with visual basic, kinda similar. ANYWAY, so here's what I have going: I made a form where I can enter various pieces of information about an item. The form then stores each field into it's own variable and then sends them to a mySQL database. After it updates the database it will generate an html page based off of the same variables. My problem is: It creates everything just fine, but none of the variables in the html page show up. I know html cannot read variables, but well, here's my code... <?php $table=$_POST['table']; $host="myhost"; // I changed the name for the forum. $username="myusername"; // Mysql username $password="mypassword"; // Mysql password $db_name="mydatabase"; // Database name $tbl_name="$table"; // Table name // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect server "); mysql_select_db("$db_name")or die("cannot select DB"); //These are the variables I get from my form. $bra=$_POST['brand']; $mod=$_POST['model']; $con=$_POST['condition']; $pri=$_POST['price']; $ims=$_POST['images']; $but=$_POST['button']; $des=$_POST['description']; $url=$_POST['url']; $inc=$_POST['includes']; $qua=$_POST['quantity']; $img=$_POST['image']; //check if query successful $sql="INSERT INTO `electronics` (`id`, `brand`, `model`, `condition`, `price`, `image1`, `button`, `description`, `url`, `includes`, `quantity`, `image2`) VALUES ('', '$bra', '$mod', '$pri', '$con', '$ims', '$but', '$des', '$url', '$inc', '$qua', '$img')"; $result=mysql_query($sql); //If it was successful in updating the database I then want it to generate the html code for the page it will generate. if($result){ $html =' <head> <title>Online Resale</title> <meta http-equiv="content-type" content="text/html; charset=iso-8859-1" /> <link href="style.css" rel="stylesheet" type="text/css" /> </head> <body> <div id="header"> <ul id="menu"> </ul> </div> <div id="gutter"></div> <div id="col1"> </div> <div id="col2"> <div align="center"><font size="5"><? echo $bra; ?> </font><br><font size="3">Model: <? echo $mod; ?> </div><p> <table> <tr> <td> <div align="center"> <img src="<?php echo $ims; ?> " width="330"></div></td> <td> <div align="center"> <img src="<?php echo $img; ?> " width="330"></div></tr></td> <tr><td> Quantity: <?php echo "$qua"; ?> <p> Condition: <?php echo $con; ?> <p> Description: <?php echo $des; ?> <p> Included Items: <?php echo $inc; ?> <p> </td> <td><div align="center"><font size="2"><img src="/paypal.jpg" height="150"><br>Cash and Money Orders also available<br> when pick-up is selected.</font><br> <?php echo $but; ?> </div> </td> </tr> </table> <p> <hr>All of our items come with a <a href="../return.html">Return Policy</a>. We guarantee all information is accurate and described to the best of our abilities. If you have any questions please <a href="../contact.html">Contact Us</a>.<p> All items will be shipped next business day. Please expect 3-7 days for delivery.<p> <div align="center"><FORM><INPUT TYPE="BUTTON" VALUE="Go Back" ONCLICK="history.go(-1)"></FORM> </div></div> <div id="col3"><div align="center"> <?php include (\'../adverts.php\'); ?> </div> </div> <div id="footer"><?php include (\'../footer.php\'); ?> </div> </div> </body> '; //This is where it creates the url I specified in my form. $fh = fopen("$url", "w+"); if($fh==false) die("unable to create file"); //This is where it writes the html code and saves the page. $fp=fopen("$url","w+"); fwrite($fp,$html); fclose($fp); mysql_close(); } else echo "error"; ?> I must again say that it generates everything just fine... my ONLY problem is that the variables I want to show up in the created html page are blank. If I view source for the page it actually shows the <?php $variable; ?> as though it thinks it is an html tag. (which of course doesn't display anything). So, I'm just wondering what I'm doing wrong here... Maybe there is a way to store the variables, maybe change the page to .php? Any help would be greatly appreciated! Link to comment https://forums.phpfreaks.com/topic/238735-form-to-sql-to-generated-html-page/ Share on other sites More sharing options...
joel24 Posted June 8, 2011 Share Posted June 8, 2011 you need to break the quotation if you echo this way, or use double quotes. i.e. $string = 'this is a string'; $single = '<p>the sting is: '.$string.'</p>'; $double = "<p>the string is: $string</p>"; echo $single . $double; You only need <?php echo $whatever; ?> if the code is not located within PHP <?php ?> i.e. <?php $string = 'this is a string'; ?> <html> <head> </head> <body> My String: <?php echo $string; ?> </body> </html> Link to comment https://forums.phpfreaks.com/topic/238735-form-to-sql-to-generated-html-page/#findComment-1226776 Share on other sites More sharing options...
staci Posted June 8, 2011 Author Share Posted June 8, 2011 It worked flawlessly. You are my hero. Link to comment https://forums.phpfreaks.com/topic/238735-form-to-sql-to-generated-html-page/#findComment-1226795 Share on other sites More sharing options...
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