如何删除img src标签中的变量？

Question

I have a sample code:

$filename = 'http://thebox.vn/Uploaded/catmy/2013_04_23/couple_t2.jpg?maxwidth=480';

And I using this code to remove variable (maxwidth)

echo preg_replace('/(\?)$/', '', $filename)

=> How to remove variable (maxwidth), how to fix it ?

Answer 1

if you want to get rid of query , just do that:

$filename = 'http://thebox.vn/Uploaded/catmy/2013_04_23/couple_t2.jpg?maxwidth=480';
$parts = explode("?",$filename);
$filename = $parts[0];

Answer 2

you could do:

$filename = "http://thebox.vn/Uploaded/catmy/2013_04_23/couple_t2.jpg?maxwidth=480";
$filename = array_shift(explode('?', $filename));
echo $filename;

Answer 3

You may try this

$filename = 'http://thebox.vn/Uploaded/catmy/2013_04_23/couple_t2.jpg?maxwidth=480';
echo preg_replace("/\?[a-z]+=\d+/", '', $filename);

DEMO.

Answer 4

Your current regular expression says: replace the last character of $filename with the empty string if that last character is the question mark character.

Here is a fixed regular expression that works for your particular example: /\\?maxwidth=.*$/

There are many other expressions that could do the job for various circumstances. However, perhaps it would be better to use PHP's parse_url() function to split the URL into its various parts and then just discard the parts that you do not care about and merge back into a string. For example:

$filename = 'http://thebox.vn/Uploaded/catmy/2013_04_23/couple_t2.jpg?maxwidth=480';

// Parse the filename into parts.
$filename_parsed = parse_url( $filename );

// Merge the parsed filename back into a string,
// discarding any irrelevant parts.
$filename_merged = $filename_parsed[ 'scheme' ] . '://' . $filename_parsed[ 'host' ] . $filename_parsed[ 'path' ];

// Prints: http://thebox.vn/Uploaded/catmy/2013_04_23/couple_t2.jpg
echo $filename_merged;