I have an IF/ELSE statement and I would like to print out some images that I am getting from my Drupal site. I can't figure out how to print those IMG tags without getting errors.
This is what I have so far:
<?php
$field = field_get_items('node', $node, 'field_visitor_image');
if($field){
<img src="<?php print image_style_url('lead_teaser', $node->field_visitor_image['und'][0]['uri']); ?>">
}
else
{
<img src="<?php print image_style_url('lead_teaser', $node->field_banner_image['und'][0]['uri']); ?>">
}
?>
You have to break out of PHP mode when you start outputting HTML.
if($field){
?>
<img src="<?php print image_style_url('lead_teaser', $node->field_visitor_image['und'][0]['uri']); ?>">
<?php
}
Use echo and string concatenation:
if ($field) {
echo '<img src="' . image_style_url('lead_teaser', $node->field_visitor_image['und'][0]['uri']) . '">';
}
You cannot nest
<?php > inside another <?php >.
One option for you could be to concatenate using ".".
Related
Is there anyway I can get in-line styling within a PHP echo statement. I want to set the height and the width of an image, as when I try to apply the height and `width externally it makes no difference as the image loads before the style sets in.
I have tried this but doesn't seem to be making any difference what so ever...
<p><b>Profile Picture: </b>
<?php
$picture = $row['imagePath'];
if (empty($picture)){
echo "<img src='profiles/no-image.png' 'width=500' 'height=600' >";
} else {
echo "<img src='".$row['imagePath']."' 'width=500' 'height=600' >";
};
?></p>
That doesn't work because you are not setting the quotes right.
This should do the trick:
echo "<img src='profiles/no-image.png' width='500' height='600' >";
You can apply styles the same way:
echo "<img src='profiles/no-image.png' style='width:500px;height:600px;'>";
Here is an alternative. Use PHP like a template engine. This approach does not use echo statements to output HTML. Instead, dynamic elements are introduced as needed.
<p><b>Profile Picture: </b>
<?php
$picture = $row['imagePath'];
if (empty($picture)) :
?>
<img src="profiles/no-image.png" width="500" height="600" >
<?php else : ?>
<img src="<?= $row['imagePath'] ?>" width="500" height="600" >
<?php endif ?>
</p>
I apologize for any misuse of terminology...I'm a noob...
I have a dynamically created page that contains a dynamic link. I added an IF/ELSE statement to display a different word based on the number of items in the variable $rowsphoto.
The different words display correctly, but the URL that is generated contains all of the PHP instead of generating the correct URL.
This is the original code, which works fine:
<?php if($portfolioid != 0) { ?>
<div class="extrafield">Additional works in Portfolio:</div>
This is the code I have after adding the IF/ELSE statement:
<?php if($portfolioid != 0) { ?>
<div class="extrafield">
<?php
if ($rowsphoto <= 4){
echo "Additional works in <a href='index.php?option=com_jartists&view=portfolio&aid=<?php echo $artistid;?>&pid=<?php echo $portfolioid; ?>&album=<?php echo $albumid;?>&id=<?php echo $photoidd; ?>&Itemid=105' class='portfoliocol'>Edition:</a>";
} else {
echo "Additional works in <a href='index.php?option=com_jartists&view=portfolio&aid=<?php echo $artistid;?>&pid=<?php echo $portfolioid; ?>&album=<?php echo $albumid;?>&id=<?php echo $photoidd; ?>&Itemid=105' class='portfoliocol'>Portfolio:</a>";
}
?>
</div>
I ran the code through a couple syntax checks and they all came back with no errors. What am I doing wrong? Is this even possible?
You'd be better off using it correctly like this:
<?php if($portfolioid != 0): ?>
<div class="extrafield">
<?php if($rowsphoto <= 4): ?>
Additional works in <a href='index.php?option=com_jartists&view=portfolio&aid=<?php echo $artistid; ?>&pid=<?php echo $portfolioid; ?>&album=<?php echo $albumid; ?>&id=<?php echo $photoidd; ?>&Itemid=105' class='portfoliocol'>Edition:</a>
<?php else: ?>
Additional works in <a href='index.php?option=com_jartists&view=portfolio&aid=<?php echo $artistid; ?>&pid=<?php echo $portfolioid; ?>&album=<?php echo $albumid; ?>&id=<?php echo $photoidd; ?>&Itemid=105' class='portfoliocol'>Portfolio:</a>
<?php endif; ?>
</div>
<?php endif; ?>
You're mixing html and php very badly. You should be separating them to keep your code clean and concise.
Your issue with it displaying the php instead of the correct variables is because (as #scrowler said):
You can't use PHP tags inside PHP, you just need to escape the string
bounds and use the . concatenation operator instead of trying to open
new PHP tags, e.g. echo "String here" . $varname; as opposed to echo
"String here"
While Darren's answer is correct, alternatively you can just stay inside of php
<?php
if($portfolioid != 0) {
echo '<div class="extrafield">';
if ($rowsphoto <= 4){
echo "Additional works in <a href='index.php?option=com_jartists&view=portfolio&aid=$artistid&pid=$portfolioid&album=$albumid&id=$photoidd&Itemid=105' class='portfoliocol'>Edition:</a>";
} else {
echo "Additional works in <a href='index.php?option=com_jartists&view=portfolio&aid=$artistid&pid=$portfolioid&album=$albumid&id=$photoidd&Itemid=105' class='portfoliocol'>Portfolio:</a>";
}
echo '</div>';
}
?>
I'm trying to show an image (or rather a link to an image) stored in a database and I'd like to get the image to show only if the link is set in the database.
Currently, if the link is not set (value is null), it shows a broken link.
Is there a way to for example use an if-statement and echo a HTML-code?
Something like this:
(The value have been fecthed to array $current in this example:)
<?php
if(isset($current['image']) {
echo "<img src='<?php echo $current['image'];
?>' class='left' style='max-height:20em; max-width:15em; margin-right:1em; margin-top:0;'})">
You used <?php twice, you have problem with quotes, brackets, etc.
<?php
if (!empty($current['image'])) {
echo "<img src='" . $current['image'] . "' class='left' style='max-height:20em; max-width:15em; margin-right:1em; margin-top:0;'>";
} else {
// here you can write for example default no-image image or whatever yo want, if you want
}
Nevermind, got it.
-Solution:
<?php if(isset($current['image'])): ?><img src="<?php echo $current['image']; ?>" class="left" style="max-height:20em; max-width:15em; margin-right:1em; margin-top:0;})">
<?php endif; ?>
<?php
if(isset($current['image'])) {
?>
<img src='<?php echo $current['image'];?>' class='left' style='max-height:20em; max-width:15em;
margin-right:1em; margin-top:0;'>
<?php
}
?>
Hello I have this src path in codeigniter
src="<?php echo base_url();?>photos/<?php echo $data1; echo '"/>';?>
The problem is that when i'm going to continue my code it seems that all the below code is inside an echo ""; it looks like the echo is still open.. For example i want to continue my code with a html divand thediv` is still on an echo area.. Thanks
I think you are inside of quotes of src-property.
src="<?php echo base_url();?>photos/<?php echo $data1;?>" />
try this
src="<?php echo base_url();?>photos/<?php echo $data1;?>" />
try this
<?php echo 'src="'.base_url().'photos/'.$data.'" />'; ?>
or
<?php echo '<img src="'.base_url().'photos/'.$data.'" />'; ?>
it seems like you haven't closed the last echo . tyr this
src="<?php echo base_url();?>photos/<?php echo $data1;?>" />
there is more better way of doing this in codeigniter. pass the url as the arguement to the base_url function
src="<?php echo base_url('photos/'.$data1);?>" />
try it. for more details: click
this is my first post for help on here, and man do I really need it. This is the first time I've developed a client's site using multisite, and I'm having trouble applying the appropriate header image to it's site. There are six sites in all, and I'm using the same template for all six sites' front pages. Also, the front page is static and doesn't have a specific page selected.
The conditional below is my attempt at specifying specific images depending on which sub-site I'm on. It keeps throwing a syntax error, (sublime calls it a parse error). I would be so grateful for any help!
<?php
if( get_bloginfo('All in with Laila Ali')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-LailaAli.jpg" />
} elseif{
if( get_bloginfo('Jaimies 15 Minute Meals')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-JamieOliver.jpg" />
}
} elseif{
if( get_bloginfo('Lucky Dog')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-LuckyDog.jpg" />
}
} elseif{
if( get_bloginfo('Game Changers with Kevin Frazier')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-GameChangers.jpg" />
}
} elseif{
if( get_bloginfo('Recipe Rehab')) {
<img src="<?php bloginfo('template_directory');?>/images/Banner-RecipeRehab.jpg" />
}
} else {
<img src="<?php bloginfo('template_directory');?>/images/Banner-PetVet.jpg" />
}
?>
You are getting this error because you have consecutive <?php tags with HTML in between them. Things like <img src=" aren't valid php, but you are inside <?php tags, so PHP gives you an error.
One of the ways you can fix this is by ending the tags when you need to switch back to HTML. Like this:
<?php
if( get_bloginfo('All in with Laila Ali')) {
?>
<img src="<?php bloginfo('template_directory');?>/images/Banner-LailaAli.jpg" />
<?php
} elseif{
if( get_bloginfo('Jaimies 15 Minute Meals')) {
?>
<img src="<?php bloginfo('template_directory');?>/images/Banner-JamieOliver.jpg" />
<?php
}
}
?>
Another way, is to have your PHP echo the HTML you want. Which would look something like this:
<?php
if( get_bloginfo('All in with Laila Ali')) {
echo '<img src="' . bloginfo('template_directory') . '/images/Banner-LailaAli.jpg" />';
} elseif{
if( get_bloginfo('Jaimies 15 Minute Meals')) {
echo '<img src="'. bloginfo('template_directory') . '/images/Banner-JamieOliver.jpg" />';
}
}
?>
Basically, you have to remember that once you open a <?php tag, you can only put valid PHP until you close it with ?>.