I have a php script which randomly needs to show images based on a specific page title.
So I managed to find some code with an if statement, which works with a single image but when I want it to show random images I don't get it to work.
Example of what works:
<?php if(is_page('Page title')){ echo '<img src="/imageA.jpg" />'; } ?>
Example of what I am after:
<?php if(is_page('Another page title')){ echo '<img src="/image-<?php echo rand(B,C; ?>.jpg">'; }
What above needs to do is load randomly imageB.jpg or imageC.jpg
This doesn't happen now because in the url you can see that the code after image- is being rendered wrong, so it can't find the image.
Alternatively, you could map out an array of letters and take use of array_rand() to randomly select letters inside the mapped letters:
<?php
if(is_page('Another page title')) {
$letters = range('B', 'D'); // generates an array of letters B to D, just change it into your liking
echo '<img src="/image-' . $letters[array_rand($letters)] . '.jpg" />';
// take out that `<?php` inside the string, you don't need those
}
?>
Sidenote: That range() function is optional, if you do not want it, just explicitly set what values that you want randomized:
$letters = array('Z', 'X', 'T', 'banner', 'logo', 'header'); // if you don't want that function above
You doing it wrong way. You are trying to insert php code inside a string, which is not going to be evaluated, but rather just outputted as is.
What you have to do, is to select a random image, for example, that way, using array_rand function:
<?php
$images = array('A', 'B', 'C');
$random_image = array_rand($images);
if (is_page('Another page title')) {
echo '<img src="/image-' . $images[$random_image] . '.jpg">';
}
use the string contatination operator - '.' - in order to append additional output (such as a function call) to the string.
<?php if(is_page('Another page title')){ echo '<img src="/image-' . rand(...) . '.jpg">';}?>
Related
I'm using the echo command in PHP, but I want to enter PHP code like <?php echo $variable [id_login]?>, while echoing something else out, but this does not work.
Is this possible to do, and if so, how would I do it?
echo "<script>location='member.php?&id=<?php echo $taruh[id_login] ?></script>";
You cannot use echo or open/cloce php twice like you did, you might want to try something like the line below,
echo '<script>location=member.php?id=' . $taruh[id_login] . '</script>';
after echo you can write enything you'd like, even if it's a php variable, just use single quote and dot where you need it (like I do here), as you can see, echo is only used once..
For example:
<?php
$your_variable = 'some text';
$other_variable = 'some PHP code';
echo 'I wrote: ' . $your_variable . ' and ' . $other_variable . '!';
?>
Output will be:
I wrote: some text and some PHP code!
I hope this will bring you into the right direction..
EDIT
Also important: if you use query string in URLs, the first 1 can be a ? every other part after should be a & for example see the url below
http://www.example.com/index.php?id=12345&coder=yes&country=usa
before id I used a quest sign, for all others I didn't use the quest sign...
I am writing an application that will look at a single record, obtain values from about 12 flags (0 or 1), look up those flags against a status table (in MySQL) and return a variable called $status_message which is in that table.
In this table I need to have hyperlinks (working fine) but also echo some variables, i.e.
You have no bids for {{$row->_item_name}}
or
View this item now by clicking here
Now I need item name and the other example to be translated into <?php echo $row->_item_name; ?>
I have tried a preg_replace with the following:
<?php
$find = array('/{{/', '/}}/');
$replace = array('<?php echo ', ' ?>');
echo preg_replace($find, $replace, $status_message);
?>
but this is not working.
Can anyone advise how I can get the desired result and 'echo' the variable in the MySQL field?
Had a brainwave. Much simpler,
instead of $row->_item_name I just put {{itemname}} in the string. I then use the following code:
<?php
$message_buyer = str_replace('{{itemname}}', $row->_item_name , $message_buyer);
echo $message_buyer;
?>
so no need to have <?php calls in the string at all.
I'm trying to setup my WordPress website so that if a post / page has a featured image assigned, this image will be used as the page banner. If however the page doesn't have a featured image, it must onload select a random image out of six available options. I've tried to used this if statement below:
<div id="slider">
<div class="theslide">
<?php
if ( has_post_thumbnail() ) {
the_post_thumbnail();
}
else {
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . $random = rand(1,6); '.jpg">';
}
?>
</div>
</div>
It works, but the random number function is not closing preperly, so the code ends up looking like this:
<div id="slider">
<div class="theslide">
<img src="/wp-content/uploads/link-ship-chandlers-banner-6 </div>
</div>
</div>
Instead of like this:
<div id="slider">
<div class="theslide">
<img src="/wp-content/uploads/link-ship-chandlers-banner-6.jpg">
</div>
</div>
I'm assuming my syntax is wrong for using php inside of the echo, but everything I try either has the same problem or cause a php error.
Any help would be appreciated.
Thanks in advance
Willem
If you don't need to know which header was chosen, just do this:
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . rand(1,6) . '.jpg">';
Using more than one ; on a line is a no-go (you are essentially telling the php-interpreter that '.jpg">'; is an independant command - which will do nothing)
Assigning a variable (ie. $whatever = 'something) inside the echo statement won't be a problem in this case - though it really doesn't do any good either. What it does is create a new variable called $random that you could use afterwards to find out which header was used - but results will be unpredictable if used in the echo statement (ie. in your case the variable would contain [random number].jpg, not just the random number), instead assign the random number to $random first like this:
$random = rand(1,6);
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . $random . '.jpg">';
echo "We are using header {$random}, which was chosen at random.";
Note that the above example also shows an alternate way to include the variable in a string - using double-quotes you can simply write the variable directly inside the string itself. When doing so I recommend always using {} to wrap the variable though not needed in this case - this allows referencing more complex variables (such as array elements or object properties), and it makes the whole thing more readable too.
Other (IMO less readable, possibly more error-prone) solutions include:
$random = rand(1,6);
echo "<img src=\"/wp-content/uploads/link-ship-chandlers-banner-$random.jpg\">";
echo sprintf('<img src="/wp-content/uploads/link-ship-chandlers-banner-%d.jpg">', $random);
printf('<img src="/wp-content/uploads/link-ship-chandlers-banner-%d.jpg">', $random);
Actually this can work, but you end the line with ;.
So removing the ; and adding a .
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . $random = rand(1,6) . '.jpg">';
do this alternatively you don't need to set this variable
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . rand(1,6) . '.jpg">';
I am trying to echo a dynamic image path in PHP. I have this code that works
<?php
//this code works
$image = '<img src="img/newlogo.jpg">';
echo($image);
?>
//gives me an image
And this code that doesn't
<?php
//this code doesn't
$lineofstring='newlogo.jpg';
$image = '<img src="img/$lineofstring">';
echo($image);
?>
$lineofsting is actually going to be an image path which is stored in a mysql database where one row is filled with, for example: pictureabc.jpg, second row is picturexyz.jpg etc.
I am trying to pass on the searched imagepath name onto $lineofstring, which is then echo'd but no picture comes out. Where am i going wrong?
To interpolate variables in PHP, you need to use double quote marks ", e.g.
$image = "<img src=\"img/$lineofstring\">";
In this case, you need to escape the inner ".
You can also concatenate strings with .:
$image = '<img src="img/' . $lineofstring . '">';
if it's easier.
Your $lineofstring variable is in a string using single quotes. Single quotes tell PHP to use the string literally so your variable is not being recognized as a variable. Change it to double quotes or use concatenation to accomplish your goals.
$image = "<img src=\"img/$lineofstring\">";
Or:
$image = '"<img src="img'.$lineofstring.'">';
$profilepicture="images/profiles/".$myid.".gif";
$noprofilepicture="images/profiles/no_avatar.gif";
echo "<IMG SRC='";
if (file_exists($profilepicture)) {
echo "".$profilepicture."";
} else {
echo "".$noprofilepicture."";
}
echo "'>";
what I want to do is want to save php variable in database (suppose {$baseUrl} ) and I am getting the data on php page with echo command and on the same page I have defined $baseUrl='/public'. I want to get the value for the base url but I'm getting simply {$baseUrl} not '/public'
in db I have <img src="{$baseUrl}/img.jpg" />
on page I have
$baseUrl = "/public";
echo $content
it is giving <img src="{$baseUrl}/img.jpg /">
how can I get <img src="/public/img.jpg" />
Of course. Strings are strings, not PHP code. You'll have to replace it yourself, e.g.:
<?php
$baseUrl = '/public';
$string = '<img src="{$baseUrl}/img.jpg" />';
$replacements = array(
'{$baseUrl}' => $baseUrl,
);
echo strtr($string, $replacements);
I think what you want to do is a string replace. The variable pointer ($baseUrl) is a string that comes from the database. If you echo it, it is still just a string. What you need to do is something like this:
<?php
echo str_replace('$baseUrl', $baseUrl, $varFromDB);
?>
Or do I understand your question wrong?
With
str_replace($content, '{$baseUrl}', $baseUrl)
.
I am guessing that you are using the wrong quotes:
echo '<img src="{$baseURL}/img.jpg" />';
rather than
echo "<img src='{$baseURL}/img.jpg'/>";
You want the $baseUrl variable to be evaluated? You have to put the variable outside of the string definition:
echo '<img src="' . $baseUrl . '/img.jpg" />';
or put it between double quotes (strings enclosed in double quotes are parsed by PHP):
echo "<img src=\"{$baseUrl}/img.jpg\" />";
Based on your comments, you need this solution:
$content = str_replace("{$baseUrl}", $baseUrl, $content);
You may want to use the eval() function...
This is not a good idea, but it would achieve what you want.