📁
SKYSHELL MANAGER
PHP v8.2.30
Create
Create
Path:
root
/
home
/
qooetu
/
costes.qooetu.com
/
Name
Size
Perm
Actions
📁
.well-known
-
0755
🗑️
🏷️
🔒
📁
2e19d9
-
0755
🗑️
🏷️
🔒
📁
6b114
-
0755
🗑️
🏷️
🔒
📁
Modules
-
0755
🗑️
🏷️
🔒
📁
app
-
0755
🗑️
🏷️
🔒
📁
assets
-
0755
🗑️
🏷️
🔒
📁
bootstrap
-
0755
🗑️
🏷️
🔒
📁
cgi-bin
-
0755
🗑️
🏷️
🔒
📁
config
-
0755
🗑️
🏷️
🔒
📁
css
-
0755
🗑️
🏷️
🔒
📁
database
-
0755
🗑️
🏷️
🔒
📁
images
-
0755
🗑️
🏷️
🔒
📁
js
-
0755
🗑️
🏷️
🔒
📁
nbproject
-
0755
🗑️
🏷️
🔒
📁
public
-
0755
🗑️
🏷️
🔒
📁
resources
-
0755
🗑️
🏷️
🔒
📁
routes
-
0755
🗑️
🏷️
🔒
📁
storage
-
0755
🗑️
🏷️
🔒
📁
tests
-
0755
🗑️
🏷️
🔒
📁
uploads
-
0755
🗑️
🏷️
🔒
📁
vendor
-
0755
🗑️
🏷️
🔒
📁
wp-admin
-
0755
🗑️
🏷️
🔒
📁
wp-content
-
0755
🗑️
🏷️
🔒
📁
wp-includes
-
0755
🗑️
🏷️
🔒
📄
.htaccess
0.23 KB
0444
🗑️
🏷️
⬇️
✏️
🔒
📄
COOKIE.txt
0.2 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
X7ROOT.txt
0.27 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
defaults.php
1.29 KB
0444
🗑️
🏷️
⬇️
✏️
🔒
📄
engine.php
0 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
error_log
813.08 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
features.php
11.28 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
googlecfb82e09419fc0f6.html
0.05 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
index.php
12.43 KB
0555
🗑️
🏷️
⬇️
✏️
🔒
📄
index.php0
1.56 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
inputs.php
0.12 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
kurd.html
1.07 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
library.php
0 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
min.php
6.83 KB
0444
🗑️
🏷️
⬇️
✏️
🔒
📄
p.php
2.75 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
php.ini
0.04 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
product.php
1.78 KB
0444
🗑️
🏷️
⬇️
✏️
🔒
📄
qpmwztts.php
0.74 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
robots.txt
0.32 KB
0444
🗑️
🏷️
⬇️
✏️
🔒
📄
tovmbkwh.php
0.74 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
tyyffovi.php
0.74 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
📄
veoxv.html
1.23 KB
0644
🗑️
🏷️
⬇️
✏️
🔒
Edit: string.js
export function longestCommonPrefix(str1, str2) { let i; for (i = 0; i < str1.length && i < str2.length; i++) { if (str1[i] != str2[i]) { return str1.slice(0, i); } } return str1.slice(0, i); } export function longestCommonSuffix(str1, str2) { let i; // Unlike longestCommonPrefix, we need a special case to handle all scenarios // where we return the empty string since str1.slice(-0) will return the // entire string. if (!str1 || !str2 || str1[str1.length - 1] != str2[str2.length - 1]) { return ''; } for (i = 0; i < str1.length && i < str2.length; i++) { if (str1[str1.length - (i + 1)] != str2[str2.length - (i + 1)]) { return str1.slice(-i); } } return str1.slice(-i); } export function replacePrefix(string, oldPrefix, newPrefix) { if (string.slice(0, oldPrefix.length) != oldPrefix) { throw Error(`string ${JSON.stringify(string)} doesn't start with prefix ${JSON.stringify(oldPrefix)}; this is a bug`); } return newPrefix + string.slice(oldPrefix.length); } export function replaceSuffix(string, oldSuffix, newSuffix) { if (!oldSuffix) { return string + newSuffix; } if (string.slice(-oldSuffix.length) != oldSuffix) { throw Error(`string ${JSON.stringify(string)} doesn't end with suffix ${JSON.stringify(oldSuffix)}; this is a bug`); } return string.slice(0, -oldSuffix.length) + newSuffix; } export function removePrefix(string, oldPrefix) { return replacePrefix(string, oldPrefix, ''); } export function removeSuffix(string, oldSuffix) { return replaceSuffix(string, oldSuffix, ''); } export function maximumOverlap(string1, string2) { return string2.slice(0, overlapCount(string1, string2)); } // Nicked from https://stackoverflow.com/a/60422853/1709587 function overlapCount(a, b) { // Deal with cases where the strings differ in length let startA = 0; if (a.length > b.length) { startA = a.length - b.length; } let endB = b.length; if (a.length < b.length) { endB = a.length; } // Create a back-reference for each index // that should be followed in case of a mismatch. // We only need B to make these references: const map = Array(endB); let k = 0; // Index that lags behind j map[0] = 0; for (let j = 1; j < endB; j++) { if (b[j] == b[k]) { map[j] = map[k]; // skip over the same character (optional optimisation) } else { map[j] = k; } while (k > 0 && b[j] != b[k]) { k = map[k]; } if (b[j] == b[k]) { k++; } } // Phase 2: use these references while iterating over A k = 0; for (let i = startA; i < a.length; i++) { while (k > 0 && a[i] != b[k]) { k = map[k]; } if (a[i] == b[k]) { k++; } } return k; } /** * Returns true if the string consistently uses Windows line endings. */ export function hasOnlyWinLineEndings(string) { return string.includes('\r\n') && !string.startsWith('\n') && !string.match(/[^\r]\n/); } /** * Returns true if the string consistently uses Unix line endings. */ export function hasOnlyUnixLineEndings(string) { return !string.includes('\r\n') && string.includes('\n'); } export function trailingWs(string) { // Yes, this looks overcomplicated and dumb - why not replace the whole function with // return string.match(/\s*$/)[0] // you ask? Because: // 1. the trap described at https://markamery.com/blog/quadratic-time-regexes/ would mean doing // this would cause this function to take O(n²) time in the worst case (specifically when // there is a massive run of NON-TRAILING whitespace in `string`), and // 2. the fix proposed in the same blog post, of using a negative lookbehind, is incompatible // with old Safari versions that we'd like to not break if possible (see // https://github.com/kpdecker/jsdiff/pull/550) // It feels absurd to do this with an explicit loop instead of a regex, but I really can't see a // better way that doesn't result in broken behaviour. let i; for (i = string.length - 1; i >= 0; i--) { if (!string[i].match(/\s/)) { break; } } return string.substring(i + 1); } export function leadingWs(string) { // Thankfully the annoying considerations described in trailingWs don't apply here: const match = string.match(/^\s*/); return match ? match[0] : ''; }
Save