mikebyrne Posted April 4, 2008 Share Posted April 4, 2008 At present I have a searchfield that dispalys the result in searchresults.php. Is it possible to have the "ProductName" linked so that it would return the results in a page called product.php? Something like product.php?id= (id= being the ProductNo) login <form name="ex2" method="post" action="searchresults.php"> <select name="cat" onChange="setAction(this.options[this.selectedIndex].value);"> <option value="CD" selected="selected" >CDs</option> <option value="DVD" >DVDs</option> <option value="Game" >Games</option> </select> <input type="text" name="searchfield" size="22" maxlength="40" id="srchdrop" /> <input type="submit" name="searchme" value="Search Now! »" id="gosrch" /> </td></tr> </form> </table> searchresults.php <?PHP if (isset($_POST['searchme'])) { include("adminconnect.php"); $tbl_name = "product"; $cat = mysql_real_escape_string($_POST['cat']); $input =mysql_real_escape_string($_POST['searchfield']); $query = "SELECT * FROM product WHERE Producttype = '$cat' AND Productname LIKE '%$input%'"; $result = mysql_query($query) or die ("error in the query" . mysql_error()); if (mysql_num_rows($result) > 0) { while($row = mysql_fetch_array($result)) { echo "Title ".$row['ProductName']."\n<br> <img src=\"".$row['Image']."\"></img>\n<br> Amount in stock ".$row['Stockamount']."\n<br> Price ".$row['Price']."\n<br> Description: ".$row['Description']."\n<br>"; } } else { echo "No search results found"; } } else { echo "No search terms entered."; } ?> Link to comment https://forums.phpfreaks.com/topic/99538-solved-display-results-and-link-to-productphpid/ Share on other sites More sharing options...
craygo Posted April 4, 2008 Share Posted April 4, 2008 Yes <?php if (isset($_POST['searchme'])) { include("adminconnect.php"); $tbl_name = "product"; $cat = mysql_real_escape_string($_POST['cat']); $input =mysql_real_escape_string($_POST['searchfield']); $query = "SELECT * FROM product WHERE Producttype = '$cat' AND Productname LIKE '%$input%'"; $result = mysql_query($query) or die ("error in the query" . mysql_error()); if (mysql_num_rows($result) > 0) { while($row = mysql_fetch_array($result)) { echo "Title <a href=\"product.php?id=".$row['productid']."\">".$row['ProductName']."</a>\n<br> <img src=\"".$row['Image']."\"></img>\n<br> Amount in stock ".$row['Stockamount']."\n<br> Price ".$row['Price']."\n<br> Description: ".$row['Description']."\n<br>"; } } else { echo "No search results found"; } } else { echo "No search terms entered."; } ?> I am assuming the product id is in a field called productid if not then change it to the field name. Ray Link to comment https://forums.phpfreaks.com/topic/99538-solved-display-results-and-link-to-productphpid/#findComment-509212 Share on other sites More sharing options...
mikebyrne Posted April 4, 2008 Author Share Posted April 4, 2008 So what would i have to do to dispaly the information in product.php? Link to comment https://forums.phpfreaks.com/topic/99538-solved-display-results-and-link-to-productphpid/#findComment-509318 Share on other sites More sharing options...
craygo Posted April 4, 2008 Share Posted April 4, 2008 you would just run a query based on the id <?php $id = $_GET['id']; $sql = "SELECT * FROM product WHERE productid = '$id'"; $res = mysql_query($sql) or die(mysql_error()); // Since you are getting only one row no neew to loop $row = mysql_fetch_assoc($res); // now just echo all your fields below and format it the way you like ?> Ray Link to comment https://forums.phpfreaks.com/topic/99538-solved-display-results-and-link-to-productphpid/#findComment-509347 Share on other sites More sharing options...
mikebyrne Posted April 4, 2008 Author Share Posted April 4, 2008 Thats geat, thanks for your help!! Link to comment https://forums.phpfreaks.com/topic/99538-solved-display-results-and-link-to-productphpid/#findComment-509399 Share on other sites More sharing options...
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