arunpatal Posted November 8, 2012 Share Posted November 8, 2012 Hi, i am trying to add php variables inside the java but its not really working i saw some place that it works....... <TABLE border="0" cellspacing="0" cellpadding="0"> <TR> <TD> <form name="Oneslideform" > <DIV align="center"> <TABLE width="150" border="1" cellspacing="0" cellpadding="4" bordercolor="#330099"> <TR> <TD bgcolor="#330099"> <DIV align="center"><B><FONT color="#FFFFFF">Image Slideshow</FONT></B></DIV> </TD> </TR> <TR> <TD bgcolor="#FFFFFF"> <DIV align="center"><A href="javascript:Onetransport()"><IMG src="slideimage1.gif" width="90" height="90" name="Oneshow" border="0"></A></DIV> </TD> </TR> <TR> <TD bgcolor="#330099"> <DIV align="center"> <SELECT name="Oneslide" onchange="Onechange();"> <OPTION value="slideimage1.gif">Image 1</OPTION> <OPTION value="<?php echo $image2; ?>">Image 2</OPTION> <OPTION value="slideimage3.gif">Image 3</OPTION> </SELECT> </DIV> </TD> </TR> <TR> <TD bgcolor="#330099"> <DIV align="center"> <INPUT type=button onclick="Oneprevious();" value="<<" title="Previous"> <INPUT type=button name="Oneslidebutton" onclick="Oneap(this.value);" value="Start" title="AutoPlay"> <INPUT type=button onclick="Onenext();" value=">>" title="Next"> </DIV> </TD> </TR> </TABLE> </DIV> </form> </TD> </TR> </TABLE> <p>This free script provided by<br /> <a href="http://javascriptkit.com">Javascript Kit</a></p> Quote Link to comment Share on other sites More sharing options...
MDCode Posted November 8, 2012 Share Posted November 8, 2012 There is no java anywhere on the page. You are also not using php anywhere inside the javascript. There is no way to help unless you reveal the functions you're using Quote Link to comment Share on other sites More sharing options...
arunpatal Posted November 8, 2012 Author Share Posted November 8, 2012 There is no java anywhere on the page. You are also not using php anywhere inside the javascript. There is no way to help unless you reveal the functions you're using If you check there is a small php coding inside <OPTION value="<?php echo $image2; ?>">Image 2</OPTION> and i have coped this complete script from http://www.javascriptkit.com/script/script2/3slide.shtml Can you please check Quote Link to comment Share on other sites More sharing options...
Maq Posted November 8, 2012 Share Posted November 8, 2012 1) That's not Java, that's javascript. 2) What does "but its not really working" mean? What's supposed to happen? Errors? Please elaborate. Quote Link to comment Share on other sites More sharing options...
arunpatal Posted November 8, 2012 Author Share Posted November 8, 2012 (edited) 1) That's not Java, that's javascript. 2) What does "but its not really working" mean? What's supposed to happen? Errors? Please elaborate. I mean when i am giving complete URL of image then it show the image... But when i am give php variable then its not show the image The displayed fields look like this in browser image 1 ">image 2 (here is php coding) image 3 <SELECT name="Oneslide" onchange="Onechange();"> <OPTION value="http://localhost/images/dsadas.jpg">Image 1</OPTION> <option value="<?php echo $image2; ?>">Image 2</option> <OPTION value="slideimage3.gif">Image 3</OPTION> </SELECT> Edited November 8, 2012 by arunpatal Quote Link to comment Share on other sites More sharing options...
Barand Posted November 8, 2012 Share Posted November 8, 2012 Have a look at the page source and see what you have for the contents of $image in that option value Quote Link to comment Share on other sites More sharing options...
arunpatal Posted November 8, 2012 Author Share Posted November 8, 2012 Have a look at the page source and see what you have for the contents of $image in that option value It is Image......... <?php $image1= '<a href="inventory_images/' . $img1 . '.jpg"><img style="border:#666 1px solid;" src="inventory_images/' . $img1 . '.jpg" alt="' . $product_name . '" width="80" height="80" border="1" /></a>'; ?> But it dose not show that image <form name="Oneslideform" > <DIV align="center"><A href=""><IMG src="slideimage1.gif" width="90" height="90" name="Oneshow" border="0"></A></DIV> </TD> </TR> <TR> <TD bgcolor="#330099"> <DIV align="center"> <SELECT name="Oneslide" onchange="Onechange();"> <OPTION value="http://localhost/inventory_images/dsadas.jpg">Image 1</OPTION> <option value="<?php echo $image1; ?>"</option> <OPTION value="slideimage3.gif">Image 3</OPTION> </SELECT> Quote Link to comment Share on other sites More sharing options...
MDCode Posted November 8, 2012 Share Posted November 8, 2012 You're using a and img tags? I'm not too sure that would work...why not just enter the image name as they are doing in the example they provided Quote Link to comment Share on other sites More sharing options...
Jessica Posted November 8, 2012 Share Posted November 8, 2012 Look at the SOURCE in your browser. What is rendered. Also your code keeps changing from 1 to 2 - what is the actual code? When do you define $image2? Quote Link to comment Share on other sites More sharing options...
arunpatal Posted November 8, 2012 Author Share Posted November 8, 2012 You're using a and img tags? I'm not too sure that would work...why not just enter the image name as they are doing in the example they provided I have a variable which is $image1 <?php $image1= '<a href="inventory_images/' . $img1 . '.jpg"><img style="border:#666 1px solid;" src="inventory_images/' . $img1 . '.jpg" alt="' . $product_name . '" width="80" height="80" border="1" /></a>'; ?> Now i want to add this variable inside form. This is from look like <form name="Oneslideform" > <DIV align="center"><A href=""><IMG src="slideimage1.gif" width="90" height="90" name="Oneshow" border="0"></A></DIV> </TD> </TR> <TR> <TD bgcolor="#330099"> <DIV align="center"> <SELECT name="Oneslide" onchange="Onechange();"> <OPTION value="http://localhost/inventory_images/dsadas.jpg">Image 1</OPTION> <option value="<?php echo $image1; ?>">Image 1</option> <OPTION value="slideimage3.gif">Image 3</OPTION> </SELECT> </form> But this form dose not show the image from this variable ($image1) It do show the Image1 which is direct URL to image. How can i show this variable in this form.......... Quote Link to comment Share on other sites More sharing options...
arunpatal Posted November 8, 2012 Author Share Posted November 8, 2012 This thing dose not work for me.......... PHP variable to Form variable: <form name="myform4"> <input type="hidden" name="formvar" value="<?php $phpvar=10; echo $phpvar; ?>"> // PHP code inside HTML!! </form> Source: http://www.skytopia.com/project/articles/compsci/form.html Quote Link to comment Share on other sites More sharing options...
arunpatal Posted November 8, 2012 Author Share Posted November 8, 2012 (edited) OK I try to explain.......... This is the code for complete page. Here there are 4 image variables $img1 $img2 $img3 $img4 I want to show these Images like in this example http://www.javascrip...t2/3slide.shtml Please help me with this......... <?php // Script Error Reporting error_reporting(E_ALL); ini_set('display_errors', '1'); ?> <?php // Check to see the URL variable is set and that it exists in the database if (isset($_GET['id'])) { // Connect to the MySQL database include "storescripts/connect_to_mysql.php"; $id = preg_replace('#[^0-9]#i', '', $_GET['id']); // Use this var to check to see if this ID exists, if yes then get the product // details, if no then exit this script and give message why $sql = mysql_query("SELECT * FROM products WHERE id='$id' LIMIT 1"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { // get all the product details while($row = mysql_fetch_array($sql)){ $product_name = $row["product_name"]; $price = $row["price"]; $img1 = $row["img1"]; $img2 = $row["img2"]; $img3 = $row["img3"]; $img4 = $row["img4"]; $details = $row["details"]; $category = $row["category"]; $subcategory = $row["subcategory"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); } } else { echo "That item does not exist."; exit(); } } else { echo "Data to render this page is missing."; exit(); } mysql_close(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title><?php echo $product_name; ?></title> <link rel="stylesheet" href="style/style.css" type="text/css" media="screen" /> </head> <body> <div align="center" id="mainWrapper"> <?php include_once("template_header.php");?> <div id="pageContent"> <table width="100%" border="0" cellspacing="0" cellpadding="15"> <tr> <td width="12%" valign="top"><div style='float:left;'><img src="inventory_images/<?php echo $id; ?>.jpg" width="81" height="100" alt="<?php echo $product_name; ?>" /><br /> <a href="inventory_images/<?php echo $id; ?>.jpg">Full Size</a><br /> <p> <img src="inventory_images/<?php echo $img1; ?>.jpg" width="81" height="100" alt="<?php echo $product_name; ?>" /><br /> <a href="inventory_images/<?php echo $img1; ?>.jpg">Full Size</a> </div></td> <td width="12%" valign="top"><div style='float:left;'><img src="inventory_images/<?php echo $img2; ?>.jpg" width="81" height="100" alt="<?php echo $product_name; ?>" /><br /> <a href="inventory_images/<?php echo $img2; ?>.jpg">Full Size</a><br /> <p> <img src="inventory_images/<?php echo $img3; ?>.jpg" width="81" height="100" alt="<?php echo $product_name; ?>" /><br /> <a href="inventory_images/<?php echo $img3; ?>.jpg">Full Size</a> </div></td> <td width="76%" valign="top"><h3><?php echo $product_name; ?></h3> <p><?php echo "$".$price; ?><br /> <br /> <?php echo "$subcategory $category"; ?> <br /> <br /> <?php echo $details; ?> <br /> </p> <form id="form1" name="form1" method="post" action="cart.php"> <input type="hidden" name="pid" id="pid" value="<?php echo $id; ?>" /> <input type="submit" name="button" id="button" value="Add to Shopping Cart" /> </form> </td> </tr> </table> </div> <?php include_once("template_footer.php");?> </div> </body> </html> Edited November 8, 2012 by arunpatal Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted November 8, 2012 Share Posted November 8, 2012 Your file must end in .php (or you must specifically configure your server to parse .html files as php code.) Quote Link to comment Share on other sites More sharing options...
arunpatal Posted November 8, 2012 Author Share Posted November 8, 2012 Your file must end in .php (or you must specifically configure your server to parse .html files as php code.) Its a .php file Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted November 8, 2012 Share Posted November 8, 2012 Are you browsing to the file using a URL in your browser? What exactly does the address bar of your browser show for this page? Quote Link to comment Share on other sites More sharing options...
arunpatal Posted November 8, 2012 Author Share Posted November 8, 2012 Are you browsing to the file using a URL in your browser? What exactly does the address bar of your browser show for this page? URL look like this http://localhost/product.php?id=4 I first have to select item from index.php page and it takes me to this product.php page Quote Link to comment Share on other sites More sharing options...
MMDE Posted November 8, 2012 Share Posted November 8, 2012 (edited) URL look like this http://localhost/product.php?id=4 I first have to select item from index.php page and it takes me to this product.php page PHP must be parsed server-side. It is not parsed by client. Why are you using PHP? ^ Should explain how it works. You use Javascript or HTML5 or something like that for client-side scripting. Things that the server decide and is dynamic etc can be done by for example PHP, or actually Java. http://en.wikipedia.org/wiki/Server-side_scripting Edited November 8, 2012 by MMDE Quote Link to comment Share on other sites More sharing options...
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