Asked  7 Months ago    Answers:  5   Viewed   36 times

I have a string that looks like this:

$str = "bla_string_bla_bla_bla";

How can I remove the first bla_; but only if it's found at the beginning of the string?

With str_replace(), it removes all bla_'s.



Plain form, without regex:

$prefix = 'bla_';
$str = 'bla_string_bla_bla_bla';

if (substr($str, 0, strlen($prefix)) == $prefix) {
    $str = substr($str, strlen($prefix));

Takes: 0.0369 ms (0.000,036,954 seconds)

And with:

$prefix = 'bla_';
$str = 'bla_string_bla_bla_bla';
$str = preg_replace('/^' . preg_quote($prefix, '/') . '/', '', $str);

Takes: 0.1749 ms (0.000,174,999 seconds) the 1st run (compiling), and 0.0510 ms (0.000,051,021 seconds) after.

Profiled on my server, obviously.

Wednesday, March 31, 2021
answered 7 Months ago

A regex would be simplest:

$input = 'foo_left.jpg';
if(!preg_match('/_(left|right|center)/', $input, $matches)) {
    // no match

$pos = $matches[0]; // "_left", "_right" or "_center"

See it in action.


For a more defensive-minded approach (if there might be multiple instances of "_left" and friends in the filename), you can consider adding to the regex.

This will match only if the l/r/c is followed by a dot:

preg_match('/(_(left|right|center))./', $input, $matches);

This will match only if the l/r/c is followed by the last dot in the filename (which practically means that the base name ends with the l/r/c specification):

preg_match('/(_(left|right|center))\.[^\.]*$/', $input, $matches);

And so on.

If using these regexes, you will find the result in $matches[1] instead of $matches[0].

Wednesday, March 31, 2021
answered 7 Months ago

strip doesn't mean "remove this substring". x.strip(y) treats y as a set of characters and strips any characters in that set from both ends of x.

On Python 3.9 and newer you can use the removeprefix and removesuffix methods to remove an entire substring from either side of the string:

url = ''
url.removesuffix('.com')    # Returns 'abcdc'
url.removeprefix('abcdc.')  # Returns 'com'

The relevant Python Enhancement Proposal is PEP-616.

On Python 3.8 and older you can use endswith and slicing:

url = ''
if url.endswith('.com'):
    url = url[:-4]

Or a regular expression:

import re
url = ''
url = re.sub('.com$', '', url)
Tuesday, June 1, 2021
answered 5 Months ago

If it is acceptable that newline (and other whitespace) characters are removed from both ends of the string then you can use

let string = "nnBLAnblub"
let trimmed = string.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
// In Swift 1.2 (Xcode 6.3):
let trimmed = (string as NSString).stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())

To remove leading newline/whitespace characters only you can (for example) use a regular expression search and replace:

let trimmed = string.stringByReplacingOccurrencesOfString("^\s*",
    withString: "", options: .RegularExpressionSearch)

"^\s*" matches all whitespace at the beginning of the string. Use "^\n*" to match newline characters only.

Update for Swift 3 (Xcode 8):

let trimmed = string.replacingOccurrences(of: "^\s*", with: "", options: .regularExpression)
Thursday, July 1, 2021
answered 4 Months ago

This can't work properly. Stored with Unicode there are many more Characters than with ANSI. So if you "convert" to ANSI, you will loose lots of charackters.

You can use Unicode (UTF-8) charset with htmlentities:

string htmlentities ( string $string [, int $flags = ENT_COMPAT [, string $charset [, bool $double_encode = true ]]] )

htmlentities($myString, ENT_COMPAT, "UTF-8"); should work.

Thursday, August 5, 2021
answered 3 Months ago
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