SQL join`inimas po vieną naujausią eulutę

SELECT c.*, p1.*
FROM customer c
JOIN purchase p1 ON (c.id = p1.customer_id)
LEFT OUTER JOIN purchase p2 ON (c.id = p2.customer_id AND 
    (p1.date < p2.date OR p1.date = p2.date AND p1.id < p2.id))
WHERE p2.id IS NULL;

source: http://stackoverflow.com/questions/2111384/sql-join-selecting-the-last-records-in-a-one-to-many-relationship

jQuery perdavima x-auth-token

$.ajax({
   url : myurl,
   headers: {
        'X-Auth-Token' : token
   });

PHP / ajax apsauga, su kintamu raktu ir laiko žyma

$time = time();
$hash = md5($key . $time);
$url =  "http://domain.com/iframe?hash=$hash&time=$time";

On the other side you should first check if the timestamp is in the limits (e.g. not older than five minutes) and than rehash with the key and the submitted timestamp. If you get the same hash the request is valid.

Unikalūs SEO URL / SLUG

tarkime, jūs formuojate straipsnių URL iš straipsnių antraščių. Kyla poreikis išvengti vienodų URL jei bus du tokie pat straipsniai, pvz.: 'test-slug-1', 'test-slug-2' and 'test-slug-3'

Tam pagelbės funkcija

$slug = preg_replace("/-$/","",preg_replace('/[^a-z0-9]+/i', "-", strtolower($title)));

Just hit the database once, grab everything at once, chances are that's the biggest bottleneck.

$query = "SELECT * FROM $table_name WHERE  $field_name  LIKE '".$slug."%'";

Then put your results in an array (let's say $slugs)

//we only bother doing this if there is a conflicting slug already
if(mysqli_num_rows($result) !== 0 && in_array($slug, $slugs)){
    $max = 0;

    //keep incrementing $max until a space is found
    while(in_array( ($slug . '-' . ++$max ), $slugs) );

    //update $slug with the appendage
    $slug .= '-' . $max;
}

We use the in_array() checks as if the slug was my-slug the LIKE would also return rows such as

my-slug-is-awesome
my-slug-is-awesome-1
my-slug-rules

šaltinis: http://stackoverflow.com/questions/15971685/creating-unique-page-title-slugs-php

Mobile app per Web-View

Jei reikia sukurti appsą, nesunkiai galima integruoti į appsą web-appą, veikiantį serveryje, pasinaudojant WEB-VIEW: https://developer.chrome.com/multidevice/webview/gettingstarted ir
http://stackoverflow.com/questions/14670638/webview-load-website-when-online-load-local-file-when-offline

php image classes

https://github.com/claviska/SimpleImage     php 5.3 + GD
http://imagine.readthedocs.org/en/latest/index.html php 5.3 GD/imagick
http://phpthumb.sourceforge.net/ php 4+/5 GD/imagick

imagick pamokos: http://www.phpro.org/tutorials/Imagick.html

PHP URL shortening functions

Integer konvertavimas.

Jigu tinka - ir _ simboliai

function encode($number) {
    return strtr(rtrim(base64_encode(pack('i', $number)), '='), '+/', '-_');
}
function decode($base64) {
    $number = unpack('i', base64_decode(str_pad(strtr($base64, '-_', '+/'), strlen($base64) % 4, '=')));
    return $number[1];
}

jeigu norima tik raidyno ir skaiciu

1. function encode($val, $base=62, $chars='0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ') {
2. // can't handle numbers larger than 2^31-1 = 2147483647
3. $str = '';
4. do {
5. $i = $val % $base;
6. $str = $chars[$i] . $str;
7. $val = ($val - $i) / $base;
8. } while($val > 0);
9. return $str;
10. } 
12. function decode($str, $base=62, $chars='0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ') {
13. $len = strlen($str);
14. $val = 0;
15. $arr = array_flip(str_split($chars));
16. for($i = 0; $i < $len; ++$i) {
17. $val += $arr[$str[$i]] * pow($base, $len-$i-1);
18. }
19. return $val;
20. }

paprastam url konvertavimui

function base64url_encode($data) {
  return rtrim(strtr(base64_encode($data), '+/', '-_'), '=');
}

function base64url_decode($data) {
  return base64_decode(str_pad(strtr($data, '-_', '+/'), strlen($data) % 4, '=', STR_PAD_RIGHT));
}