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.
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>';
}
}
$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 . '/' . $photo2->name . '" />';
}
echo '</div>';
}
Just beginning PHP to bear with me.
Results I'm trying to achiever:
I have a table of YouTube URL's and MetaData.
Trying to build this:
<div class="slide">
<iframe></iframe>
<iframe></iframe>
</div>
<div class="slide">
<iframe></iframe>
<iframe></iframe>
</div>
Two videos per slide, then I'm going to paginate through results using Deck.js.
I suspect I'm going about this completely the wrong way, not that experienced at programmin g logic;
while($data = mysql_fetch_array($result)) {
for ($counter = 1; $counter<=2; $counter++) {
echo "<div class=\"slide\">";
echo "<h3>" . $data['VIDEO_TITLE'] . "</h3>";
echo "<iframe width=\"560\" height=\"315\" src=\"" . $data['VIDEO_URL'] . "\" frameborder=\"0\" allowfullscreen></iframe>";
/* If Video 1, increment counter for 2nd video */
if ($counter == 1) {
$counter++;
}
/* If Video 2, close div and reset counter */
else if ($counter == 2) {
echo "</div>";
$counter = 1;
}
/* If error break out */
else {
echo "</div>";
break;
}
}
}
Basically trying to nest loops to keep track of how many videos per div and start a new one when a div has two.
I've tried a few different ways, this being the latest. Results in:
<div class="slide">
<iframe></iframe>
<div class="slide>
<iframe></iframe>
Hit the blank wall now, not sure what to try next. Willing to use/learn any method to accomplish the results, just not sure where to go at this point.
Cheers.
You could remove the second loop all together using the % operator (modulus). The idea is that a % b === 0 then the number a was evenly divisible by b. Using this, you can easily check for even or odd or every Nth row.
$k = 1;
echo "<div class=\"slide\">";
while($data = mysql_fetch_array($result)) { // you should really change to mysqli or PDO
if($k % 3 === 0){
echo "</div>";
echo "<div class=\"slide\">";
}
echo "<h3>" . $data['VIDEO_TITLE'] . "</h3>";
echo "<iframe width=\"560\" height=\"315\" src=\"" . $data['VIDEO_URL'] . "\" frameborder=\"0\" allowfullscreen></iframe>";
$k++;
}
echo "</div>";
Put the echo <div> before the for loop (still inside the while loop) and the </div> after the for loop
In your while loop you're retrieving just one row, but then you're iterating over it twice with a nested loop. Do away with the inner loop and just use a flip-flop variable to track left and right. I think this will do what you want:
$left=true; // track whether we're emitting HTML for left or right video
while($data = mysql_fetch_array($result)) {
if ($left) {
echo "<div class=\"slide\">";
echo "<h3>" . $data['VIDEO_TITLE'] . "</h3>";
}
echo "<iframe width=\"560\" height=\"315\" src=\"" . $data['VIDEO_URL'] . "\" frameborder=\"0\" allowfullscreen></iframe>";
if (!$left) {
echo "</div>";
}
$left = !$left; // invert $left to indicate we're emitting the right iFrame
}
// end of loop. If we had an odd number of
// videos, tidy up the HTML
if (!$left) {
echo "</div>";
}
PHPFiddle
<?php
$x = 10;
$counter = 0;
while($x > 0)
{
if($counter != 0 && $counter % 2 == 0)
{
echo "ENDOFSLIDE</br>";
}
if($counter == 0 || $counter % 2 == 0)
{
echo "SLIDE</br>";
}
echo "iframe => $x </br>";
$x--;
$counter++;
}
echo "ENDOFSLIDE";
?>
It won't work because the for loop is inside the fetch loop for the SQL data. The second iteration of the for loop does not have a new SQL row. A better solution would be to capture the common column that identifies the two videos (the title) and generate a new div whenever that value changes. Try something like this, which will work for any number of SQL rows with the same title. This will also give proper results if the SQL query returns no rows and will handle the potential of a title with only one URL - which could get ugly if you merely flip-flop and end up with URLs on the wrong title. Of course, as in your current solution, your SQL query must ORDER BY VIDEO_TITLE so the rows are adjacent. I didn't run it, but should be close.
$lastTitle = "";
$n = 0; //count SQL rows processed
while($data = mysql_fetch_array($result)) {
// See if the title changed from last
if( $data['VIDEO_TITLE'] != $lastTitle ) {
// if we processed any rows, must end the current div before starting a new one
if( $n > 0 )
echo "</div>";
echo "<div class=\"slide\">";
echo "<h3>" . $data['VIDEO_TITLE'] . "</h3>";
//save the title as last title
$lastTitle = $data['VIDEO_TITLE'];
}
$n++; //count the SQL row
echo "<iframe width=\"560\" height=\"315\" src=\"" . $data['VIDEO_URL'] . "\" frameborder=\"0\" allowfullscreen></iframe>";
}
if( $n > 0 )
echo "</div>";
i am trying to get my data from the database and wrap it with html tags. here is the working code so far:
function homethumb(){ $this->count; $i = 0;
while($row = mysqli_fetch_object($this->result))
{
$this->count++; $i++;
if($i == 1){echo '<div class="gal1">';}
echo ' <div class="gal"><img src="img/' . $row->thumb2 . '.jpg"></div>';
if($i == 2){
echo '</div> <!-- gal1 -->';
$i=0;
}
}
}
Here I am getting everything from the database (Select * from portfolio), but in the portfolio I have, websites, demos and graphics; so I wanted to get only the data where category = "web" from the above code, so I tried this:
function homethumb(){ $this->count; $i = 0;
while($row = mysqli_fetch_object($this->result))
{
if($row->category = "web"){
$this->count++; $i++;
if($i == 1){echo '<div class="gal1">';}
echo ' <div class="gal"><img src="img/' . $row->thumb2 . '.jpg"></div>';
if($i == 2){
echo '</div> <!-- gal1 -->';
$i=0;
}
}
}
}
now the nested if statements do not generate the divs I need, how can I get this working
thanks for your help
I can't see your SQL based on your question, but you could just modify your SELECT query to include WHERE category="web"
This way, you're only selecting the rows you need, instead of looping over every row in that table.
Additionally, it appears that you're using assignment = instead of comparison == for your if statement.
Do you just need to have == instead of =?
if($row->category == "web"){
But it would be best to restrict the query to the results you need at the database level, unless you need the other rows for some reason.
1)You missed an equal sign:
if($row->category = "web") => if($row->category == "web")
Or better yet
if($row->category === "web")
2)If you want to only get fields with a specific category field, you can simply change your query:
[rest of your query] WHERE category="web"
OK, it should go like this, assuming the fields are sorted as follows
ID, category, website, thumb2, demo, graphics
function homethumb(){ $this->count; $i = 0;
while($row = mysqli_fetch_object($this->result))
{
if($row[1] == "web"){
$this->count++; $i++;
if($i == 1){echo '<div class="gal1">';}
echo ' <div class="gal"><img src="img/' . $row[3] . '.jpg"></div>';
if($i == 2){
echo '</div> <!-- gal1 -->';
$i=0;
}
}
}
}
and there is no need for the nested if, you can just use it in one line as follows:
if($row[1] = "web")
{
echo '<div class="gal1">';
echo ' <div class="gal"><img src="img/' . $row[3] . '.jpg"></div>';
echo '</div> <!-- gal1 -->';
}
1) Change your query to contain a WHERE category="web" clause
2) You have an assignment operator in your if clause (=), when you need an equality operator (==)
Sorry for the huge ignorance on the topic, but I really have no idea where to look other than this website when I come into trouble with my PHP.
What I'm trying to do here is use pre-designated IDs to call particular movies from a database. But all I get is an 'Invalid argument supplied for foreach()' message on the second and third foreach's below.
Here's my code in the head:
//Custom lists of movies to bring in
//New Releases list
$films_new_releases = array(40805, 46705, 41630, 44564, 39451, 20352, 43933, 49009, 49797, 42194);
//Most Popular list
$films_most_popular = array(27205, 16290, 10138, 41733, 37799, 18785, 19995, 17654, 10140, 12162);
//Get information from address bar
$list = $_GET['l'];
if ($list == 'new releases') {
$list_chosen = $films_new_releases;
}
elseif ($list == 'most popular') {
$list_chosen = $films_most_popular;
}
else {
$list_chosen = $films_new_releases;
}
And in amongst the body:
// Loop through each film returned
foreach ($list_chosen as $list_chosen_film) {
$films_result = $tmdb->getMovie($list_chosen_film);
$film = json_decode($films_result);
// 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 == 'poster') {
$backdrop_url = $backdrop->image->url;
}
}
echo '<div class="view-films-film">
<img src="' . $backdrop_url . '" alt="' . $film->name . '" />
<div class="view-films-film-snippet">
<h2>' . $film->name . '</h2>';
if ($film->certification != null) {
echo '<img src="images/bbfc-' . strtolower($film->certification) . '.png" alt="" />';
}
echo ' <h3>Starring</h3>
<p>';
$num_actors = 0;
foreach ($film->cast as $cast) {
if ($cast->job == 'Actor') {
echo '' . $cast->name . ' ';
$num_actors++;
if ($num_actors == 5)
break;
}
echo ' </p>
<h3>Director</h3>
<p>';
foreach ($film->cast as $cast) {
if ($cast->job == 'Director') {
echo '' . $cast->name . ' ';
}
}
echo ' </p>
</div>
</div>';
}
// End films
}
The little testing I've done is checking what $list_chosen, $list_chosen_film, $films_result and $film actually contain by printing them at the bottom of the page.
$list_chosen shows - Array, $list_chosen_film shows - 42194, $films_result shows the entire JSON string, $film shows - Array.
Try adding:
print_r($film->backdrop);
before the second foreach() loop. Before the error message it won't be an array or it will contain zero elements (not allowed). If you also add:
echo $films_result;
you will be able to debug it and fully understand what is wrong. If not, post the whole output in your question.
This happens, because - as error displayed by PHP informed you - you have provided wrong parameter to foreach loop, probably null or some other value. Make sure you are providing array to foreach.
Also, every time you use foreach, do it like that:
if (count($some_list) > 0) {
foreach ($some_list as $list_item) {
// code for each item on the list
}
} else {
// code when there is nothing on the list
}
This will ensure you will not see errors just because there is nothing on the list.
EDIT:
On the documentation page you can find some tip how to avoid such errors if the collection you are trying to iterate through is empty. Just cast the collection to array type:
foreach ((array) $some_list as $list_item) {
// code for each item on the list
}
Can you provide a dump of $film? The error is telling you that you are pointing to an object that cannot be iterated through (most likely null).