$pgroup = $Sys->db->query("SELECT dj_photo_group.photo_group, dj_photo.name FROM dj_photo_group, dj_photo WHERE dj_photo_group.photo_group = dj_photo.img_group ORDER BY dj_photo.img_group DESC");
while($photo = $pgroup->fetch_object()) {
echo '<div class="photo_head">' . ucfirst($photo->photo_group) . '</div>';
echo '<div class="photo_sub">';
while($photo->name) {
echo '<img src="photogallery/' . $photo->photo_group . '/' . $photo->name . '" />';
}
echo '</div>';
}
This is showing the same 1 image over and over again. I have 3 groups: Wedding, 5points and Misc that have 2 images each. It only shows Weddings with 1 image infinite number of times.
What am I doing wrong and how can I fix it?
$photo->name is always true. So if you use it in a while loop, you're essentially saying:
while(true){
// do this
}
Remember, while loops will keep going as long as the condition specified is true.
You should really be using an if statement instead.
if($photo->name){
// do this
}
while($photo->name) will repeat until $photo->name is false. Because in the loop there is nothing that will change it, it will always be true, and so the loop continues forever.
Like Nishant said, you want to replace it with an if statement.
while($photo->name) is entering into infinite loop as the condition is always true. You should put an end condition to this while loop or otherwise you can go for if condition also
if($photo->name)
Try this.. this solution will provide you your desired layout....
$pgroup1 = $Sys->db->query("SELECT dj_photo_group FROM photo_group GROUP BY dj_photo_group");
while ($photo1 = $pgroup1->fetch_object())
{
echo '<div class="photo_head">' . ucfirst($photo1->dj_photo_group) . '</div>';
echo '<div class="photo_sub">';
$pgroup2 = $Sys->db->query("SELECT name FROM dj_photo WHERE img_group = '$photo1->dj_photo_group'");
while ($photo2 = $pgroup2->fetch_object())
{
echo '<img src="photogallery/' . $photo1->dj_photo_group . '/' . $photo2->name . '" />';
}
echo '</div>';
}
Related
Hey everyone I am having an issue with echoing an image or an iframe, the idea is there is an image link or an iframe in the same row in the database & i want to only echo what the string content is, if it echos the iframe there is a blank space of an image & underneath is the youtube video, if it echos the image the iframe does not show, they both currently echo the same row id labled url, I only want one to echo based on the string content instead of both echo's.
Here is the code I use to fetch the stuff, its only the echo area you want to pay attention to, I am not sure what kind of, if or else statement i could put in here.
< ?php
require_once("config.php");
$limit = (intval($_GET['limit']) != 0 ) ? $_GET['limit'] : 4;
$offset = (intval($_GET['offset']) != 0 ) ? $_GET['offset'] : 0;
$sql = "SELECT * FROM bookmarks WHERE 1 ORDER BY id DESC LIMIT $limit OFFSET
$offset";
try {
$stmt = $DB->prepare($sql);
$stmt->execute();
$results = $stmt->fetchAll();
} catch (Exception $ex) {
echo $ex->getMessage();
}
if (count($results) > 0) {
foreach ($results as $row) {
echo '<li><kef>' . $row['title'] . '</kef></br>';
echo '<img src=' . $row['url'] . '></br>'; /*here */
echo '<kef1>' . $row['url'] . '</kef1></br>'; /*and here*/
echo '<kef2>' . $row['desc'] . '</kef2></br>';
echo '<kef3>' . $row['tag'] . '</kef3></br> </br> </li>';
} }
?>
This script is a get results php
The best solution would be to save the data in different columns in the database. But if you want to keep the current structure would be able to check if the string contains HTML, then it is an iframe. Otherwise it is a link to the image.
<?php
if($row['url'] == strip_tags($row['url'])) {
//No HTML in string = link
echo '<img src="' . $row['url'] . '" alt="" />';
} else {
//Contains HTML = iframe
echo $row['url'];
}
?>
This assumes that you in your database have saved the values as either http://www.exempel.com or <iframe src =" http: // www ... ">. If the iframe-value also consists only of a URL, you must use another solution.
If you know whether the url is an image or video when you save them in the database then you could add another field for the media type. You could then use this in an if statement in your code, for example:
if (count($results) > 0) {
foreach ($results as $row) {
echo '<li><kef>' . $row['title'] . '</kef></br>';
if ($row['mediaType'] === 'image') {
echo '<img src=' . $row['url'] . '></br>'; /*here */
} if ($row['mediaType'] === 'video') {
echo '<kef1>' . $row['url'] . '</kef1></br>'; /*and here*/
}
echo '<kef2>' . $row['desc'] . '</kef2></br>';
echo '<kef3>' . $row['tag'] . '</kef3></br> </br> </li>';
}
}
I have a page and I would really like some help / advice to create a better way / function to call in information from another table.
At the moment, my code looks like:
[I do know this is deprecated SQL and would really like to do some nice SQLi for this.]
<?
$menuid = "100";
$imageid = "50";
// ** Talk to 'imagedirectory' table
mysql_select_db($database_db, $BASEdb);
$query_displayimage = "SELECT * FROM imagedirectory WHERE menuid = ".$menuid." AND imageid = ".$imageid."";
$displayimage = mysql_query($query_displayimage, $BASEdb) or die(mysql_error());
$row_displayimage= mysql_fetch_assoc($displayimage);
?>
<img src="/images/assets/<?php echo $menuid; ?>-<?php echo $imageid; ?>-<?php echo $row_displayimage['urlslug']; ?>.jpg" alt="<?php echo $row_displayimage['alttext']; ?>" />
I figure There really has to be a better way because if there is 10 images on a page, this is pretty intense way of doing it.
Since you seem to know that mysql_* is deprecated, I am assuming you have read up on, and are using mysqli_* instead.
You needn't query the database every time. mysqli_query() returns a mysqli_result, which you can iterate over, and read using functions like mysqli_fetch_assoc(). Here is one way of doing it:
<?php
// store your query in a variable.
$query_displayimage = "SELECT * FROM imagedirectory";
// query the database.
$displayimage = mysqli_query($query_displayimage, $BASEdb);
// check for errors.
$dbErrors = mysqli_error($BASEdb);
if (count($dbErrors))
{
print_r($dbErrors);
die();
}
// iterate over the returned resource.
while ($row_displayimage = mysql_fetch_assoc($displayimage))
{
echo '<img src="/images/assets/' . $menuid . '-' . $imageid . '-' . $row_displayimage['urlslug'] . '.jpg" alt="' . $row_displayimage['alttext'] . '" />';
}
?>
Hope that helped.
EDIT:
You can use this code in a function, too. For example:
<?php
function printImage($menuid, $imageid)
{
$query_displayimage = "SELECT * FROM imagedirectory";
$displayimage = mysqli_query($query_displayimage, $BASEdb);
$dbErrors = mysqli_error($BASEdb);
if (count($dbErrors))
{
print_r($dbErrors);
die();
}
if ($row_displayimage = mysql_fetch_assoc($displayimage))
{
echo '<img src="/images/assets/' . $menuid . '-' . $imageid . '-' . $row_displayimage['urlslug'] . '.jpg" alt="' . $row_displayimage['alttext'] . '" />';
}
else // if there is a problem getting the image
{
echo 'Error getting image.';
}
}
?>
and elsewhere in your HTML, you would do something like:
<div>
And here is an image!
<?php printImage(20, 50); ?>
</div>
What I have is a reply script, and I need the page to only use the username echo'd in for a variable, like if Meap sends me a PM and I want to reply to it, when I hit the reply link I need it to use his name, and not another name from the page. Currently it is using the first name on the page instead of the one that the reply link is for
Here's the code;
while ($row = mysqli_fetch_array($data)) {
session_start();
$reply = $row['from'];
echo '<div class="viewpost">';
echo '<div class="vpside">';
if(!empty($row['picture'])) {
echo '<img class="pictest" src="' . MM_UPLOADPATH . $row['picture'] . '" alt="' . MM_UPLOADPATH . 'nopic.png' . '" />';
}
if(!empty($row['from'])) {
echo '<p>From:<br />' . $row['from'] . '</p>';
$_SESSION['reply'] = $row['from'];
echo 'Reply';
}
if(!empty($row['rank'])) {
echo '<p>Rank:<br />' . $row['rank'] . '</p>';
}
if(!empty($row['gender'])){
echo '<p>Gender:<br /> ' . $row['gender'] . '</p>';
}
echo '</div>';
if(!empty($row['title'])) {
echo'<h4><u>' .$row['title']. '</u></h4>';
}
if(!empty($row['msg'])) {
echo '<p class="">' . $row['msg'] . '</p>';
}
echo '<div class="sig">';
if(!empty($row['bio'])) {
echo '<p>' . $row['bio'] . '</p>';
}
echo '</div>';
echo '</div><br />';
}
Here's a picture of what it looks like when you get a Private Message;
Evidently you have to be specific down to the last detail on here, the question - or rather the problem - is that it is not using the correct recipient for the reply. Say I click reply under Meaps name, it will use Lilpunish instead. Basically it will use the first one loaded, and I need it to use the correct username. How would I do this.?
Here's your problem:
$_SESSION['reply'] = $row['from'];
echo 'Reply';
Consider what is happening when this code is run multiple times. The first time, it sets the session variable reply to Meap. The next time, when the second message is printed, it sets it to another value.
What you actually want is to pass who to reply as a part of the link, e.g.
echo 'Reply';
And then not retrieve the reply-to from $_SESSION. Session variables are not really supposed to be used like this.
I have the following code with the if...else statement within a while loop.
$colour = array("50305A", "E33600", "DBDD02", "73BE58");
$nextcolr = next($colour);
if ($nextcolr == FALSE)
{
reset($colour);
echo current($colour);
}
else
{
echo next($colour);
}
I can't work out why what ever is in the else statement isn't being executed, even if I switch the two statements and reverse the operator. Could anyone help me?
The entire while loop:
while($row = mysql_fetch_array($result))
{
echo "by <a href='/neuro/profile.php?userid=$row[MemberID]'>" . $row['FirstName'] . " " . $row['LastName'] . "</a> on " . $row['Timestamp'] . " | " . $row['NumberOfComments'] . " comments.";
echo "<div id='blog' style='background-color:#";
if ($nextcolr == FALSE)
{
reset($colour);
echo current($colour);
}
else
{
echo next($colour);
}
echo "'><a href='blog.php?threadid=" . $row['tID'] . "'>" . $row['Title'] . "</a></div>";
}
$colour = array("50305A", "E33600", "DBDD02", "73BE58");
while ... {
$nextcolr = next($colour);
if ($nextcolr === FALSE)
{
reset($colour);
}
echo current($colour);
}
is how your while loop should look like. If I am right, you are also defining $colour in the while loop, which might cause problems.
If all this is in the while loop, then you are re-declaring the array on each iteration, thus returning the array internal pointer to the beginning with each iteration.
If you want to iterate this array multiple times, you could do it this way:
$colour = array("50305A", "E33600", "DBDD02", "73BE58");
$i = 0;
while ... {
...
echo $colour[$i++ % count($colour)];
...
}
So you don't need this if-else block.
The problem with your original while loop is that you never change the value of $nextcolr.
Thus, it always remains FALSE and the else part never gets executed.
Please forgive the extreme-beginner style of coding. I'm working with PHP and JSON strings from an API for the first time.
What I'm trying to do here, which obviously doesn't work if you look through it, is print out multiple movie results that the user is searching for. This is a search results page - with a movie poster and some key details next to each item.
One of the things I want next to each result is a list of the first 5 actors listed in the movie, as "Starring" and then any directors listed in the movie as "Director".
The problem is no doubt the fact that I've nested foreach statements. But I don't know any other way of doing this.
Please please the simplest answer would be the best for me. I don't mind if it's bad practice, just whatever solves it quickest would be perfect!
Here's the code:
// Check if there were any results
if ($films_result == '["Nothing found."]' || $films_result == null) {
}
else {
// Loop through each film returned
foreach ($films as $film) {
// Set default poster image to use if film doesn't have one
$backdrop_url = 'images/placeholder-film.gif';
// Loop through each poster for current film
foreach($film->backdrops as $backdrop) {
if ($backdrop->image->size == 'thumb') {
$backdrop_url = $backdrop->image->url;
}
}
echo '<div class="view-films-film">
<img src="' . $backdrop_url . '" alt="' . strtolower($film->name) . '" />
<div class="view-films-film-snippet">
<h2>' . strtolower($film->name) . '</h2>
<img src="images/bbfc-' . strtolower($film->certification) . '.png" alt="" />
<h3>Starring</h3>
<p>';
$num_actors = 0;
foreach ($films[0]->cast as $cast) {
if ($cast->job == 'Actor') {
echo '' . $cast->name . ' ';
$num_actors++;
if ($num_actors == 5)
break;
}
}
echo ' </p>
<h3>Director</h3>
<p>';
foreach ($films[0]->cast as $cast) {
if ($cast->job == 'Director') {
echo '' . $cast->name . ' ';
}
}
echo ' </p>
</div>
</div>';
// End films
}
}
An example of the result would be this:
with line 120 being foreach ($films[0]->cast as $cast) { and line 131 being foreach ($films[0]->cast as $cast) {
Why are you using $films[0] when you're in a foreach ($films as $film)? You should be using $film->cast right? And you should dump $film at the start of your loop with var_dump and inspect it - make sure $cast is an array and is populated. If it's not, work back from there.