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Closed 7 years ago.
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I am trying to display an image using a variable "$images" that contains the URL parsed form an API.
This is my code:
echo "<td>""<img src='",$image,"'>""</td>\n";
I assume there is a typo I cannot detect because I get a blank screen when running this.
echo "<td>"."<img src='".$image."'>"."</td>";
Or
echo "<td><img src='".$image."'></td>";
You were missing the dots/commas after/before the td tags
use . not ,
<img src='".$image."'>
PHP requires that you chain your strings together using a .
E.g.
echo 'Test' . ' ' . 'Hello'; // Test Hello
Or simply :
echo "<td><img src='$image'></td>";
Check documentation.
Related
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Closed 4 years ago.
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I have a simple link that needs to pass a variable through $_GET and set it to a variable in the page that is being opened. It is failing every time however and I am not sure why.
<a class='section_header' href='category/index.php?`id='" . $catid . "'><b>" . $row[0] ."</b></a>
Code in category/index.php
$id = $_GET['id'];
echo $id;
<a class='section_header' href="category/index.php?id=<?php echo $catid?>"><b><?php echo $row[0];?></b></a>
Try
<a class='section_header' href='category/index.php?id='" . $catid . "'><b>" . $row[0] ."</b></a>
You had a random ` in there. Once it prints out you can use inspect element to make sure the url looks correct and id is populated.
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Closed 8 years ago.
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It's a basic question, but I've googled all sorts of variations of it & nothing that quite answers me. I want to display one of two images. One if the criteria is met & one if it is not met. This code works. If criteria is met, it does display the image. But there is no alternative
<? if(stripslashes($getfeedbackQryRow['CompanyID'])=='344'){?><img src='img1.png' alt='Todays Bite'><? }?>
Then I put this together, but it's still wrong
<?PHP
if ($getfeedbackQryRow['CompanyID']) == '344') {
print ("<IMG SRC =/img1.png>");
}
else {
print ("<IMG SRC =img2.png>");
}
?>
This one looks like it should be right, but it throws an error... It's probably a syntax issue. Does anyone know what I'm doing wrong?
You have an extra ")"
if ($getfeedbackQryRow['CompanyID']) == '344') {
Remove the ")" here ['CompanyID'])
if ($getfeedbackQryRow['CompanyID']) == '344') {
// ^------- this is causing your error
You should do something simple like this:
$image_name = $getfeedbackQryRow['CompanyId'] == '344' ? 'img1.png' : 'img2.png';
echo '<img src="' . $image_name . '">';
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Closed 8 years ago.
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im getting this error but cannot figure out why
while($info = mysql_fetch_array( $data ))
{
echo itg_fetch_image('.$info['meta_value'].'); //the line giving error
Print "<tr>";
Print "<th>ID:</th> <td>".$info['meta_id'] . "</td> ";
Print "<th>VALUE:</th> <td>".$info['meta_value']. "</td> ";
Print "<th>DONE:</th> <td>YES</td> ";
}
if i comment out this line it shows fine in the value td, ive tried everthing. taking the dots out the taking the comments out then adding "". itg_fetch_image is http://www.intechgrity.com/automatically-copy-images-png-jpeg-gif-from-remote-server-http-to-your-local-server-using-php/#
the code is ment to be echo itg_fetch_image('url') and all meta_values return a url string
The quotes on the echo line are causing the problem.
It should look like this:
echo itg_fetch_image($info['meta_value']);
it seems that itg_fetch_image() is a function
try this:
echo itg_fetch_image($info['meta_value']);
you can just pass variable . do not need to use quotes.
You must write it as
echo itg_fetch_image($info['meta_value']);
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Closed 9 years ago.
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how can I print out a variable name in text in php?
I want to accomplish something as simple as having this:
echo "|" . "$myvariable" . "|";
print out this:
|$myvariable|
instead of this:
||
how can I accomplish this?
Use single quotes so that the variables don't get interpolated:
echo "|" . '$myvariable' . "|";
As others pointed out, use single quotes to show the variable name "as is".
echo "|" . '$myvariable' . "|";
Be sure to read this part of the manual.
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Closed 9 years ago.
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I have in a foreach loop:
echo "<span style=\"" . myCss($value) . "\">lol</span>";
Which turns into (in source):
<span style="">lol</span>color: #999999;background-color: transparent;font-weight:normal;text-decoration: none;<span style="">...
Why? how to prevent the browser? Same for Chrome and Firefox. Note there is a reason for it being in-line, an I want to avoid doing it via javascript.
Try this
echo "<span style='" . myCss($value) . "'>lol</span>";
How about a little separation of PHP and HTML:
<span style="<?php echo myCss($value); ?>">lol</span>
Notice I encapsulate the PHP within the quotes, rather than echo the entire line. In a foreach loop it would look something like:
<?php
foreach($array as $key => $value){
?>
<span style="<?php echo myCss($value); ?>">lol</span>
<?php
}
?>
This separation of PHP and HTML has been the standard practice everywhere that I have worked, and I personally find it to be much more transparent.
Without seeing your function and the values of the variables, I an only assume that there are characters in the echoed result that mess up the html. You should always use htmlspecialschars() when you output to html:
echo "<span style=\"" . htmlspecialschars(myCss($value)) . "\">lol</span>";
Although you would probably use it in your function.