melting_dog
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Posts posted by melting_dog
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Hmm it kind of is -
Say if I had an SQL query that returned a set of results that I would use across my site over and over again. I imagine that it would be better practice to have that query on its own file and to 'call' that file (via include?) whenever you need it.
I just want to confirm that this is the best practice to do this
Thanks!
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OK so sorry if my understandings a bit off, but if a I want to introduce a class into my doc is the best way to fo this with the include() function?
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Its ok - I've figured it out. I realised I had to set permissions on the upload directory to read/write on my computer
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OK I get a new meesage now:
Warning: move_uploaded_file(uploads/ferriswheel.jpg) [function.move-uploaded-file]: failed to open stream: Permission denied in /Applications/XAMPP/xamppfiles/htdocs/xampp/whatmycitywears/submitphotoprocess.php on line 35
So I think I need to somehow set the permissions in XAMPP to allow upload?
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Hi guys,
I need a little help. I am running a site in XAMPP whilst I develop it locally. I am trying to upload an image file. Heres my code:
$photo=$_POST['photo'];
$target_path = "uploads/";
$target_path = $target_path . basename( $_FILES['photo']['name']);
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target_path)) {
echo "The file has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
But I get these errors:
Warning: move_uploaded_file(upload/ferriswheel.jpg) [function.move-uploaded-file]: failed to open stream: No such file or directory
Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move '/Applications/XAMPP/xamppfiles/temp/phpNwxq6I' to 'upload/ferriswheel.jpg' in /Applications/XAMPP/xamppfiles/htdocs/xampp/whatmycitywears/submitphotoprocess.php on line 53
Am I right in thinking this is some sort of issue with the temp directory? It seems to be putting it somewhere is XAMPSS folders rather then in my directory
Any help would be great!
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Thanks guys,
Was able to do it just with CSS and display:inline.
Thanks for all the help!
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Thanks,
I'm actually getting the image and all other information (caption, alt text, etc) fine. Im just having a good deal of difficulty with the styling - I want to create rows of three images next to each other (so if I had 12 images I'd get 4 rows of three) . I was wondering if there was a better way to go about the gallery that I was missing. List? Tables? Or maybe arrays?
Thanks again!
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Hi guys,
Just after a bit of advice really. I want to create an image gallery. Currently, my images (which are sourced from a DB) are brought about in a loop and set in their own div, side by side. I was just thinking though: would it be better to do this with an array? Its a bit difficult coding for this just using the basic loop. Anyone have any suggestions to make this better?
PS, here is my current code:
$sql = "SELECT * FROM imagegal";
if ($result = mysql_query($sql)) {
if (mysql_num_rows($result)) {
while($row = mysql_fetch_array($result)){
echo '<span class="imagegal">';
echo '<span><img src="' . $row['image'] . '"</span>';
echo '</span>';
}}}
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Sweet thanks!
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Hi guys,
Sorry - I know this is a basic question but I've been out of the game for a year now and need to jog my memory.
I am trying to count the number of rows per group. e.g:
$sql = "SELECT *, COUNT(id) FROM image GROUP BY city";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
echo $row['COUNT(id)'];
So if there are, for example, 5 New York's in the table, 3 London's and 2 Sydney's I want the echo to display 5, 3, 2
Should I use a loop for this?
Cheers
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Well I found and used this:
if (filter_var($useremail, FILTER_VALIDATE_EMAIL)) {
and it seems to work. Is it really this simple though?
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Hi guys,
Thanks for your help and sorry for the delayed reply. Yes, as it was the loop was working fine. The extra </div> was the spanner in the works hiding my results completely. Sorry - should have properly checked my code before posting.
Thanks guys!
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Hi guys,
Trying to ge this to work:
function checkEmail($useremail) {
if (!preg_match("/^( [a-zA-Z0-9] )+( [a-zA-Z0-9\._-] )*@( [a-zA-Z0-9_-] )+( [a-zA-Z0-9\._-] +)+$/" , $useremail)) {
return false;
$error = '<div id="blackText"><p>Sorry! Please check your email address and try again!<p><p><a href="forumsSignUp.php">Back</a></p></div>';
}
return true;
//PASSWORD AND OTHER VALIDATION STUFF, blah blah blah
}
But its just not. Everything works fine if I remove this though. Can anyone suggest an alternative to email address validation? Thanks!
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Hi guys
Probably a simple but trying to figure out why I cant get this while loop to work (at least I think its the loop)
Query works as it returns 1 row, but not all of them.
Any help would be great!
$sql = "SELECT * FROM requests ORDER BY leave_after DESC";
if ($result = mysql_query($sql)) {
if (mysql_num_rows($result)) {
while ($row = mysql_fetch_assoc($result))
{
echo '<div id="resultholder">';
echo "<p>Request ID:" . $row['id'] . ", User ID: " . $row['user_id'] . ", Departing From: " . $row['departing'] . ", Going To: " . $row['destination'] . " , Leaving After: " . $row['leave_after'] . " , but Arriving Before: " . $row['arrive_before'] ."</p></div>";
echo '</div>';
}
}
}
Thanks!
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haha Thanks Zurev...surprised I didnt think of that. Works like a charm
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Hi guys,
I want to create a piece of script that takes my users to a website based of there input in a form. Basiaclly what they input into a text field will become part of the URL used in action at the top of the form.
So it would look like this:
http://www."their input from form".mydomain.com
does anyone have any suggestions on how to accomplish this?
Thanks
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Ok this may seem like a noob question but could someone please explain to me what loader classes actually do?
Cheers
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Hi,
I want to have a require() and a header() on the same page. This is just impossible right?
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Turning on errors, error_reporting = E_ALL, will give us a real good clue as to what's wrong
Hmm Thanks but it came up with nothing.... any other suggestions?
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Hi all,
Yeah just a fresh pair of eyes needed. It just wont insert into the table and the alert returns a blank. Only thing I can think is thats its posting variables from two different queries...but that shouldnt make a difference...
Cheers
FORM PAGE************************* //*****************************LOGGED CHECKER********************* $sql = "SELECT * FROM users WHERE firstName= '$logged'"; if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { if ($row = mysql_fetch_assoc($result)) { //**************QUESTION FORM***************************** $sql2 = "SELECT * FROM questions WHERE questionID= '$orderCounter'"; if ($result2 = mysql_query($sql2)) { if (mysql_num_rows($result2)) { if ($row2 = mysql_fetch_assoc($result2)) { echo "<p>" . $row2['questionText'] . "</p>"; echo '<form name="form1" method="post" action="questionProcess.php">'; echo '<input type="radio" name="q1" value="' . $row2['answer1'] . '" />' . $row2['answer1'] . '<br />'; echo '<input type="radio" name="q1" value="' . $row2['answer2'] . '" />' . $row2['answer2'] . '<br />'; echo '<input type="radio" name="q1" value="' . $row2['answer3'] . '" />' . $row2['answer3'] . '<br />'; echo '<input type="radio" name="q1" value="' . $row2['answer4'] . '" />' . $row2['answer4'] . '<br />'; echo '<input type="hidden" name="userid" value="' . $row['userid'] . '" />'; echo '<input type="hidden" name="orderCounter" value="' . $row['orderCounter'] . '" />' echo '<input type="submit" name="Submit" id="Submit" value="Next" />'; echo '</label>'; echo '</form>'; } } } } } } ?> PROCESSING PAGE**************************************** $q1=$_POST['q1']; $userid=$_POST['userid']; $orderCounter=$_POST['orderCounter']; // To protect MySQL injection $q1 = stripslashes($q1); $userid = stripslashes($userid); $orderCounter = stripslashes($orderCounter); $q1 = mysql_real_escape_string($q1); $userid = mysql_real_escape_string($userid); $orderCounter = mysql_real_escape_string($orderCounter); //ALERT CHECKER function confirm($msg) { echo "<script langauge=\"javascript\">alert(\"".$msg."\");</script>"; } $msg = $userid; confirm($msg); //ALERT CHECKER if ($q1 == "" ) { $wrong = '<div id="blackText"><p>Please Answer the Question<p></div>'; } else { $sql="INSERT INTO $tbl_name (userid, value, orderCounter) VALUES ('$userid', '$q1', '$orderCounter')"; $result=mysql_query($sql); } ob_end_flush(); ?> -
Yeah kinda glad to hear that actually. Just sussing out options. I suppose the best option would be to integrate pay pal
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So sorry if this doesnt belong in this section but i was wondering if anyone knew of a way to encrypt sensitive information (i.e credit card details) passed from a form to a database/email?
I am building a shopping cart and apart from protecting against SQL injection was wondering if there are any other ways to protect information.
Cheers
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The correct format for putting a GET parameter on the end of a URL is -
your_page.php?name=value&name2=value2&name3=value3
In your case you are only using one name/value pair, so you would need to produce -
your_page.php?name=value
If the name you want is comDiscussNo and the value is in the variable $comDiscussNo, you would need to use -
your_page.php?comDiscussNo=$comDiscussNo
Thanks heaps...thats done the trick!
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Hey again all,
This is a follow on from another post but now i think the issues a bit more simpler.
I am having issues with getting a variable from a header:location()
The first page has:
header("location:forumsDiscussion.php?$comDiscussNo");
The second page has:
$discussIDtrans = (int)$_GET['comDiscussNo'];
function confirm($msg)
{
echo "<script langauge=\"javascript\">alert(\"".$msg."\");</script>";
}
$msg = $discussIDtrans;
confirm($msg);
But its comeing up as nothing (0) even though i can see the correct number in the URL.
Cheers, any help is appreciated
Query outputs the word 'Array'?
in PHP Coding Help
Posted
Hi guys,
I have a simple query here:
$sql="SELECT user_id FROM $tbl_name WHERE email='$loginuseremail' AND password='$loginpassword'";
$result=mysql_query($sql);
$row = mysql_fetch_assoc($result);
//START SESSION
session_start();
$_SESSION['username']=$loginuseremail;
$_SESSION['user_id']=$row;
echo $_SESSION['user_id'];
It should obviously output an ID number but instead I just get the word 'Array'. If I remove the mysql_fetch_assoc() I just get 'Resource ID #4'.
Can anyone point me in the right direction?
Thanks!