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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 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 5 years ago.
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i want to concat two variable and assign it to another variable
$name = Drew;
$counter = $counter + 1;
$num_padded = sprintf("%05d", $counter);
$new_padded_value = $name . $num_padded;
echo $new_padded_value;
Im expecting it to echo Drew.00001
its not working. how can I achieve this.
You're missing a string dot between the vars. The dot that you used is just the string concatenation operator; it doesn't actually add a dot to the string result. Change:
$new_padded_value = $name . $num_padded;
To:
$new_padded_value = "$name.$num_padded";
Also change:
$name = Drew;
To:
$name = "Drew";
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Closed 6 years ago.
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Can anyone help me for this why doesn't my code work?
$h.= ' 'mysql_result($res,$z,"name")'';
It gives me an error because of the <a href
$h .= ''.mysql_result($res,$z,"name").'';
Should work now
. before mysql_result and after.
Also take a look here it will help you to understand String Operators and etc.
You need to concatenate.
$h.= ' ' . mysql_result($res,$z,"name") . '';
Anyway, I would suggest you to put mysql_result($res,$z,"name") into a variable, to increase readability and avoid executing the function every time you call it.
I mean:
$name = mysql_result($res,$z,"name");
$h.= ' ' . $name . '';
Try this
$h.= ''.mysql_result($res,$z,"name").'';
You are missing . the concatenation operator
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Closed 7 years ago.
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I am trying to display an image using a variable "$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='$image'></td>";
Check documentation.
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Closed 8 years ago.
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im getting this error but cannot figure out why
while($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']);