deansaddigh Posted June 22, 2009 Share Posted June 22, 2009 This is really simple i believe, i just need to know how to echo out an image that i have in my db.(image location) I have the following code <?php echo "Welcome back:" . " " ; echo $_SESSION['name']; /*get the session email*/ $email =$_SESSION['email']; //build sql statement to get user avatar $query = "SELECT User_id FROM user WHERE Email = '$email'"; $result = mysql_query($query,$conn) or die ("Unable to perform query"); //Retrieve the data that is stored in the $query array $row= mysql_fetch_array($result); $userid = $row['User_id']; //use the user id in the avatar table to bring back the correct avatar path $query = "SELECT Avatar_path FROM avatar WHERE User_id ='$userid'"; $result = mysql_query($query, $conn) or die ("Unable to perform query"); //retrieve user idea from array $row= mysql_fetch_array($result); $avatar_path =$row['Avatar_path']; echo $avatar_path; ?> im assuming the echo $avatar_path needs to be within an image tag or something?? Link to comment https://forums.phpfreaks.com/topic/163218-echo-an-image/ Share on other sites More sharing options...
Dathremar Posted June 22, 2009 Share Posted June 22, 2009 echo "<img src='".$avatar_path."'>"; Link to comment https://forums.phpfreaks.com/topic/163218-echo-an-image/#findComment-861164 Share on other sites More sharing options...
deansaddigh Posted June 22, 2009 Author Share Posted June 22, 2009 Thank you for the reply. Nothing seems to be echoing out. i think i know the problem i have attached the db screen shot, do i need to echo out the Avatar_name to get the image to display? [attachment deleted by admin] Link to comment https://forums.phpfreaks.com/topic/163218-echo-an-image/#findComment-861168 Share on other sites More sharing options...
Dathremar Posted June 22, 2009 Share Posted June 22, 2009 echo "<img src='".$avatar_path. $avatar_name ."'>"; in the src You need to give the exact location of the picture that needs to be displayed. Link to comment https://forums.phpfreaks.com/topic/163218-echo-an-image/#findComment-861171 Share on other sites More sharing options...
deansaddigh Posted June 22, 2009 Author Share Posted June 22, 2009 thank you very much Link to comment https://forums.phpfreaks.com/topic/163218-echo-an-image/#findComment-861177 Share on other sites More sharing options...
deansaddigh Posted June 22, 2009 Author Share Posted June 22, 2009 Im now getting undifined index errors, but i cant find the problem, it seems like my code is fine, or i cant find the issue. <?php session_start(); if (isset($_SESSION['email']) == false) { header("Location:../index.php"); exit(); } include '../db/connection.php'; ?> <!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> <link rel="icon" type="image/gif" href="site icon/animated_favicon1.gif"/> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Lifey/Home</title> <!-- Css style sheet--> <link href="css/user.css" rel="stylesheet" type="text/css" /> </head> <body> <!--top links --> <?php include "includes/homepage_top_nav.php";?> <div id="left_contents"> <div id="topimage"></div> <div id="BodyColumns"> <?php echo "Welcome back:" . " " ; echo $_SESSION['name'] . "<br />"; /*get the session email*/ $email =$_SESSION['email']; //build sql statement to get user avatar $query = "SELECT User_id FROM user WHERE Email = '$email'"; $result = mysql_query($query,$conn) or die ("Unable to perform query"); //Retrieve the data that is stored in the $query array $row= mysql_fetch_array($result); $userid = $row['User_id']; //Get the avatar path $query = "SELECT Avatar_path, Avatar_name FROM avatar WHERE User_id ='$userid'"; $result = mysql_query($query, $conn) or die ("Unable to perform query"); //retrieve user idea from array $row= mysql_fetch_array($result); $avatar_path =$row['Avatar_path']; $avatar_name =$row['Avatar_name']; echo "<img src='".$avatar_path. $avatar_name ."'>"; ?> </div> <div id="bottomimage"></div> </div> <!-- Container div--> <div id="container"> <div id="contents"> <div id="logo"></div> <?php //If student fail to login display error on the homepage if (isset($_GET['message'])) { echo $_GET['message']; } ?> <p> Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. It has survived not only five centuries, but also the leap into electronic typesetting, remaining essentially unchanged. It was popularised in the 1960s with the release of Letraset sheets containing Lorem Ipsum passages, and more recently with desktop publishing software like Aldus PageMaker including versions of Lorem Ipsum.</p> <?php include"includes/homepage_footer_links.php";?> </div> <!-- End Container div--> </div> </body> </html> Any ideas would be gratefully appreciated Link to comment https://forums.phpfreaks.com/topic/163218-echo-an-image/#findComment-861207 Share on other sites More sharing options...
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