diddyd Posted May 16, 2011 Share Posted May 16, 2011 Hello There, I have been trying to get my page going for some time. I have managed to setup my database on mysql. I have managed to create myself a few pages that look how I want and do what I want. I have managed to get the data I want into my database. Where I am struggling big time is to find some way to get the images that I have gotten into my database back out again with a php page. I have managed to do a dynamic drop down, from one of the tables, but when I bring back the image files I get the binary, or a readout of the file. (I have a .gpx file as well as a load of .jpg files). The code I have managed so far is below, does anyone feel inclined to help me get a little further with this? I am new to HTML and to PHP, although databases are perhaps a little more familiar to me. I would like if possible to show the images in the lower half of the same page that the drop down is on, although if this is not possible then please let me know. <?php require_once 'login.php'; $db_server = mysql_connect($db_hostname, $db_username, $db_password); if (!$db_server) die("unable to connect to MySQL: " . mysql_error()); //connecting to the database. mysql_select_db($db_database) or die("Unable to select database: " . mysql_error()); // If submitted, check the value of "select". If its not blank value, get the value and put it into $select. if(isset($select)&&$select!=""){ $select=$_GET['select']; } ?> <body> <form method="POST" action="predroptest.php"> Hills : <select name="selecthill"> <option value="">--- Select Hill---</option> <? // Get records from database (table "Hills"). $list=mysql_query("select ID, Hill from Hills order by Hill"); // Show records by while loop. while($row_list=mysql_fetch_assoc($list)){ ?> <option value="<? echo $row_list['ID']; ?>" <? if($row_list['ID']==$select){ echo "selected"; } ?>><? echo $row_list['Hill']; ?></option> <? // End while loop. } ?> </select> <input type="submit" name="Submit" value="Select" /> </form> <hr> <p> <? // Get records from database (table "Hills"). $select1=$_POST['selecthill']; $query="SELECT * FROM Objects Where Data_ID='$select1'"; $result=mysql_query($query); $num=mysql_numrows($result); echo "<b><center>Hill Info</center></b><br><br>"; $i=0; while ($i < $num) { $Data_ID=mysql_result($result,$i,"Data_ID"); $Object=mysql_result($result,$i,"Object"); $Type=mysql_result($result,$i,"Type"); $Description=mysql_result($result,$i,"Description"); header('Content-type: image/jpg'); echo "<b>$Data_ID<br>$Object<br>$Type<br>$Description<hr><br>"; $i++; } mysql_close(); ?> </p> </body> Thanks very much for the help Diana Tracy www.up-a-hill.me.uk Quote Link to comment Share on other sites More sharing options...
dragon_sa Posted May 16, 2011 Share Posted May 16, 2011 A better way is just to store the name of the image in the database and possibly the path to the stored image, and use your script to upload the image to a directory on your web server where it will be displayed from. there is a sample script on how to do that in this post which also gives you resize options http://www.phpfreaks.com/forums/index.php?topic=333194.0 Quote Link to comment Share on other sites More sharing options...
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