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dave204

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Everything posted by dave204

  1. Looking like I'll have to work this one out myself, because no one on here can give me the correct php code to produce a link to the http inside the Android app. echo row['image'] just won't cut it inside an Android app..
  2. Is it not http://www.padihamca...splay.php/image ?? I need it in the above format, to use inside the Android app..
  3. OK - I've pulled together 2 scripts - and come up with the following, but how do I then add the resulting $row['image']; into the URL I had previously, to use inside an Android app?? Previous URL: http://www.padihamcars.com/file_display.php/?id=1 Quite obviously, id=1 no longer applies. So what do I add to this URL from echo $row['image']; ?? Many thanks in advance!! <?php include "file_constants.php"; $link = mysql_connect("$host", "$user", "$pass") or die("Could not connect: " . mysql_error()); $sql = "SELECT image FROM test_image ORDER BY id DESC LIMIT 1"; $result = mysql_query("$sql"); $row = mysql_fetch_assoc($result); mysql_close($link); header("Content-type: image/jpeg"); echo $row['image']; ?>
  4. Hi there! I have a script (which works), for obtaining the first (or any id number submitted) image from a database. But recently, I asked on here how to obtain the LAST id from this database ie most recent image. I was given a very good solution, but I'm not 100% sure how to correctly add this to my current script, and then using the result in a URL I am using within an Android app I've developed. Can anyone give me the full, altered script from my original, and solution below, and also what I should then add to the end of the URL (in place of ?id=1.. Many thanks in advance!! ORIGINAL PHP SCRIPT: <?php include "file_constants.php"; // just so we know it is broken error_reporting(E_ALL); // some basic sanity checks if(isset($_GET['id']) && is_numeric($_GET['id'])) { //connect to the db $link = mysql_connect("$host", "$user", "$pass") or die("Could not connect: " . mysql_error()); // select our database mysql_select_db("$db") or die(mysql_error()); // get the image from the db $sql = "SELECT image FROM test_image WHERE id=" .$_GET['id'] . ";"; // the result of the query $result = mysql_query("$sql") or die("Invalid query: " . mysql_error()); // set the header for the image header("Content-type: image/jpeg"); echo mysql_result($result, 0); // close the db link mysql_close($link); } else { echo 'Please use a real id number'; } ?> SOLUTION: <?php $sql = "SELECT image FROM test_image ORDER BY id DESC LIMIT 1"; $result = mysqli_query($conn, $sql); $row = mysqli_fetch_row($result); ?> CURRENT URL: "http://padihamcars.com/file_display.php?id=1"
  5. Just to add to above, submitting id=1,id=2, etc as a query from the URL line, fits the 'GET $id' inside the php file. So I need to be able to submit a query to the SELECT line, based on id number..
  6. Thanks for that last post - I'm very grateful - but I need an 'id' number as the result, which I can then use in the URL I illustrated. Although I appreciate the clarity of your answer, it doesn't fit my current code (which gives me dynamic access to the id's). To explain, in this line: $sql = "SELECT image FROM test_image WHERE id=" .$_GET['id'] . ";"; which then gives me access to the id at the end of this URL: "http://padihamcars.c...splay.php?id=1" (where the query for id can be id=1, id=2, id=3, etc. How can I achieve what you're proposing around my "Select image.. as above??? Many thanks once again!
  7. At this stage, I'm looking to get the last image - my current Android code just covers that (although replication, or a For/While loop will easily add id=2, id=3 in the http statement. However, if I understand correctly, $GET(id) would retrieve the 12 id's - then my http statement would display id=1 (or in this case the inverse, id=12, or whatever is last). Have I understood it right?
  8. Thanks. I added this line to file_display.php: $sql = "SELECT image FROM test_image ORDER BY id DESC LIMIT 12 WHERE id=" .$_GET['id'] . ";"; However, it failed to download any image to App???
  9. Hi there, I have a PHP script, which is used to extract id's of uploaded images in a database. I then use id=1 etc tp display the relevent image using Picasso library in an Android App. However my customer will be adding images to the database, which will be auto-incremented by id number, but I want to display the LATEST image ie the highest id number. How do I achieve this in this URL? Here is my code for the file_display.php I want to refresh last 12 images if possible, ie each new entry should be added from id=1 to id=12, but I don't know how to achieve this. That's why I want to obtain LAST ID, and use this in URL below.. THANKs!! <?php include "file_constants.php"; // just so we know it is broken error_reporting(E_ALL); // some basic sanity checks if(isset($_GET['id']) && is_numeric($_GET['id'])) { //connect to the db $link = mysql_connect("$host", "$user", "$pass") or die("Could not connect: " . mysql_error()); // select our database mysql_select_db("$db") or die(mysql_error()); // get the image from the db $sql = "SELECT image FROM test_image WHERE id=" .$_GET['id'] . ";"; // the result of the query $result = mysql_query("$sql") or die("Invalid query: " . mysql_error()); // set the header for the image header("Content-type: image/jpeg"); echo mysql_result($result, 0); // close the db link mysql_close($link); } else { echo 'Please use a real id number'; } ?> AND URL inside Android app: "http://padihamcars.com/file_display.php?id=1"
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