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I am trying to add a <div> to my php, but everything I've found on google hasn't worked for me. I don't know what I'm doing wrong. Any ideas on how to add a <div> in a php function for styling purposes only?
<?php
//get_additional is a custom function
ob_start();
echo get_additional('Cosmos','Trades');
echo nl2br("\n");
//I want the div to start here
$buffered = ob_get_clean();
echo $buffered;
$times = $buffered;
$imgurl = 'Images/cat.png';
for($i=0;$i<$times;$i++){
echo '<img src="'.$imgurl.'" />';
}
?>
You need to echo it.
Like you've already used once in your code, echo '<img src="'.$imgurl.'" />';
Now coming back to your code:
<?php
//get_additional is a custom function
ob_start();
echo get_additional('Cosmos','Trades');
echo nl2br("\n");
//I want the div to start here
echo '<div class = "blah">
// whatever you want to do here
</div>';
$buffered = ob_get_clean();
echo $buffered;
$imgurl = 'Images/cat.png';
for($i=0;$i<$times;$i++){
echo '<img src="'.$imgurl.'" />';
}
?>
Related
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I am working on a website in which I want to do inline styling inside php tags.
<?php
if($data['item']->hello_world!=null)
{
echo "policy:";
echo strtolower($data['item']->hello_world->name);
echo "<br>";
echo strtolower($data['item']->hello_world->description);
}
?><?php
if($data['item']->hello_world!=null)
{
echo "policy:";
echo strtolower($data['item']->hello_world->name);
echo "<br>";
echo "<br>";
echo strtolower($data['item']->hello_world->description);
}
?>
The data which I want to style coming from the php code is:
echo strtolower($data['item']->hello_world->name);
Problem Statement:
I am wondering what changes I should make in the php code so that I am able to do styling for the above mentioned line.
To inline style the element:
echo '<span style="color: red">' . strtolower($data['item']->hello_world->name) . '</span>';
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I'm trying to call variable with in variable and below attempts are working fine
Attempt 01:
<?php
$var1 = "Hello";
echo $var1;
?>
Attempt 02:
<?php
$view = '$var1 = "Hello";
echo $var1;';
echo $view;
?>
But I'm trying to call variable without using "echo" or "print" command, as mentioned below
<?php
$view = '$var1 = "Hello";
echo $var1;';
$view;
?>
Is there any other way to achieve this? Please advise.
<?php
$view = '$var1 = "Hello";
echo $var1;';
eval($view);
?>
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echo <a href = 'test.php'> "CategoryID: " . $row["CategoryID"]. " - Category Name: ".$row["CategoryName"]. </a> "<br>";
This is what i have an is not working properly.
This:
echo "<a href = 'test.php'>CategoryID: {$row['CategoryID']} - Category Name: {$row['CategoryName']}</a><br />";
I am using the { and } as they allow you to include an array in a string and ignore the concatenation which I find harder to read.
I find it funny that you can loop through a MySQL array but can't echo a simple string :P
Some links (teach a man to fish...):
W3Schools
PHP documentation
Codecademy
Tutorials Point
Try this:
<?php
$link = "";
$link = sprintf("<a href = 'test.php'>CategoryID: %d - Category Name: %s </a><br />", $row['CategoryID'], $row['CategoryName']);
echo $link;
?>
Assuming that $row['CategoryID'] is an integer and $row['CategoryName'] is a string.
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I'm programming in PHP, but I don't think its the right way.
I'm programming something like this:
<?php
$apple = 10
if(apple >= 4){
?>
<img src="Bladieblabla">
Biep
<?php
}else{
print "Awhh we don't have 4 apples";
}
?>
You can use echo, or print like this
if($apple > 5) {
echo '<img src="Bladieblabla">';
echo 'Biep';
} else {
print "Awhh we don't have 4 apples";
}
Update: Using echo only one time:
echo '<img src="Bladieblabla">
Biep';
<?php
$apple = 10;
if(apple > 5){
echo '<img src="Bladieblabla">';
echo 'Biep';
}else{
print "Awhh we don't have 4 apples";
}
?>
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<?php
$response = file_get_contents("http://api.trakt.tv/shows/trending.json/5d7588c188eeea0074b8d2664d12fffc");
$result = json_decode($response, true);
echo $result['title'][0];
echo "<br>";
echo $result['network'][0];
echo "<br>";
echo $result['air_day'][0];
echo "<br><img style='width:200px;' src='";
echo $result['images'][0]['poster'];
echo "'>";
?>
Ain't working. I don't know why.
I use the the trakt.tv shows API.
Write
echo $result[0]['title'];
instead of
echo $result['title'][0];
Besides, PHP's echo function will print integers and strings, but will fail with array-alike structures. You could use var_dump or var_export instead. Thanks to them, you could scan the structure and you wouldn't ask this question ;)