8

I have a function for creating unique slug for a page title. It checks if the slug is available in the pages table then creates a unique slug by adding a '-int' accordingly. The function works fine for the first three entries eg for 'test slug' entered three time will create 'test-slug-1', 'test-slug-2' and 'test-slug-3'. Then after that I get an error "Fatal error: Maximum execution time of 30 seconds exceeded" for the fourth entry. There should be some problem with the logic, can anyone help me find it please.Below is the code:

function createSlug($title, $table_name, $field_name) { global $db_connect; $slug = preg_replace("/-$/","",preg_replace('/[^a-z0-9]+/i', "-", strtolower($title))); $counter = 1; do{ $query = "SELECT * FROM $table_name WHERE $field_name = '".$slug."'"; $result = mysqli_query($db_connect, $query) or die(mysqli_error($db_connect)); if(mysqli_num_rows($result) > 0){ $count = strrchr($slug , "-"); $count = str_replace("-", "", $count); if($count > 0){ $length = count($count) + 1; $newSlug = str_replace(strrchr($slug , "-"), '',$slug); $slug = $newSlug.'-'.$length; $count++; }else{ $slug = $slug.'-'.$counter; } } $counter++; $row = mysqli_fetch_assoc($result); }while(mysqli_num_rows($result) > 0); return $slug; 

}

2
  • 1
    Instead of repeatedly hitting the database, you can just use LIKE to return all matching results then if the number of results is greater than 0 append the count of the number of the results to the slug. Commented Apr 12, 2013 at 12:46
  • Append count+1 is not the best idea as a record #3 could have been deleted but record 4 was not, that leaves 3 records, the next one you add is going to conflict Commented Apr 12, 2013 at 13:18

8 Answers 8

33

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 

etc which would cause issues, the in_array() checks ensure that we are only checking against the exact slug that was entered.

Why don't we just count the results and +1?

This is because if you had multiple results, and deleted a few, your next slug could well conflict.

E.g.

my-slug my-slug-2 my-slug-3 my-slug-4 my-slug-5 

Delete -3 and -5 leaves us with

my-slug my-slug-2 my-slug-4 

So, that gives us 3 results, the next insert would be my-slug-4 which already exists.

Why don't we just use ORDER BY and LIMIT 1?

We can't just do an order by in the query because the lack of natural sorting would make my-slug-10 rank lower than my-slug-4 as it compares character by character and 4 is higher than 1

E.g.

m = m y = y - = - s = s l = l u = u g = g - = - 4 > 1 !!! < 0 (But the previous number was higher, so from here onwards is not compared) 
Sign up to request clarification or add additional context in comments.

3 Comments

You can order by length of field, then by field itself to achieve natural sorting :) sort of.
This is not working when below case. 1) My post -> my-post 2) My post -> my-post-1 3) My -> my-3 This conversion is going to wrong.
Hello, Hailwood how you going? could you take a look in my answer its similar than that one.. stackoverflow.com/questions/70906354/… Thank you soo much
15

Just use a single query to do all the heavy lifting for you...

$slug = preg_replace("/-$/","",preg_replace('/[^a-z0-9]+/i', "-", strtolower($title))); $query = "SELECT COUNT(*) AS NumHits FROM $table_name WHERE $field_name LIKE '$slug%'"; $result = mysqli_query($db_connect, $query) or die(mysqli_error($db_connect)); $row = $result->fetch_assoc(); $numHits = $row['NumHits']; return ($numHits > 0) ? ($slug . '-' . $numHits) : $slug; 

5 Comments

Much Thanks, that works perfect. Thanks to everyone who contributed, you have all helped in some way. Much appreciated.
Excellent! Don't forget to mark one of the answers as accepted as this turned into a huge list of code-heavy answers!
bad example. it will add double slugs after a row has been deleted
I dont thik this is not good idea. assume there is slug_2 in a 2nd row, but it never check for it. if there is same title used for 3 times there will be 1 title and 2 title_2 s
This is not working when below case. 1) My post -> my-post 2) My post -> my-post-1 3) My -> my-3 This conversion is going to wrong.
3

You can just select slug with the biggest number and increase it with 1:

$query = "SELECT $field_name FROM $table_name WHERE $field_name LIKE '".$slug."-[0-9]*' ORDER BY $field_name DESC LIMIT 1"; 

[0-9]* in query means any count of numbers.

This query will select row with $slug at start and the bigest number.

After that you can parse result get number and increase it.

In this case you will have only one query and lot of unused performance.

UPDATE

This will not work, because slug-8 will be "bigger" than slug-11. But no idea how to fix it. maybe ORDER BYidDESC ?

UPDATE 2

Query can be ordered by length too and it will work right. Thanks to Jack:

$query = "SELECT $field_name FROM $table_name WHERE $field_name LIKE '".$slug."-[0-9]*' ORDER BY LENGTH($field_name), $field_name DESC LIMIT 1"; 

UPDATE 3

Also added check for original slug. Thanks to Hailwood.

$query = "SELECT $field_name FROM $table_name WHERE $field_name = '".$slug."' OR $field_name LIKE '".$slug."-[0-9]*' ORDER BY LENGTH($field_name), $field_name DESC LIMIT 1"; 

6 Comments

Have you tried ORDER BY LENGTH($field_name) DESC, $field_name DESC?
@Jack hmm, probably will work. Thanks, will update my answer.
Although this doesn't check for the existence of the original slug (aka no -int) so all slugs will end up having a -int
Two more things, 1) you have two sets or ORDER BY, 2) you want OR not AND, no string could ever be itself and itself + -[digit]
@Hailwood made more updates on this answer than in my rest answers :)
|
2

Why don't you just create a slug and leave the rest of the job that involves indexing to MySQL. Here is a slugify function ( it is a slightly modified version used by Symfony framework ).

function slugify( $text ) { $text = preg_replace('~[^\\pL\d]+~u', '-', $text); $text = trim($text, '-'); $text = iconv('utf-8', 'ASCII//IGNORE//TRANSLIT', $text); $text = strtolower(trim($text)); $text = preg_replace('~[^-\w]+~', '', $text); return empty($text) ? substr( md5( time() ), 0, 8 ) : $text; } 

And the MySQL part can be solved with trigger ( change the table and column names ).

BEGIN declare original_slug varchar(255); declare slug_counter int; set original_slug = new.slug; set slug_counter = 1; while exists (select true from post where slug = new.slug) do set new.slug = concat(original_slug, '-', slug_counter); set slug_counter = slug_counter + 1; end while; END 

MySQL Insert row, on duplicate: add suffix and re-insert

1 Comment

Combining this with @Narek's solution is probably the most elegant solution
1

For the one part I would create an object that is dealing with the part creating the slug and handling the number:

// generate new slug: $slug = new NumberedSlug('Creating Unique Page Title Slugs in PHP'); echo $slug, "\n", $slug->increase(), "\n"; // read existing slug: $slug = new NumberedSlug('creating-unique-page-title-slugs-in-php-44'); echo $slug->getNumber(), "\n"; 

Output:

creating-unique-page-title-slugs-in-php creating-unique-page-title-slugs-in-php-1 44 

For the other part, the database, this already greatly simplifies your code (please double check, I've done this quick). Also see how you can benefit from the Mysqli object you actually have (but not use as is):

function createSlug($title, $table_name, $field_name, Mysqli $mysqli = NULL) { $mysqli || $mysqli = $GLOBALS['db_connect']; $slug = new NumberedSlug($title); do { $query = "SELECT 1 FROM $table_name WHERE $field_name = '" . $slug . "'"; if (!$result = $mysqli->query($query)) { throw new RuntimeException(var_export($mysqli->error_list, true)); } if ($result->num_rows) { $slug->increase(); } } while ($result->num_rows); return $slug; } 

But as others have already written you should first get all slugs that are numbered at once from the database and then pick a unique one if necessary. This will reduce the number of database calls. Also the code is much more compact:

function createSlug2($title, $table_name, $field_name, Mysqli $mysqli = NULL) { $mysqli || $mysqli = $GLOBALS['db_connect']; $slug = new NumberedSlug($title); $query = "SELECT $field_name FROM $table_name WHERE $field_name LIKE '$slug-_%'"; if (!$result = $mysqli->query($query)) { throw new RuntimeException(var_export($mysqli->error_list, true)); } $existing = array_flip(call_user_func_array('array_merge', $result->fetch_all())); $slug->increase(); while (isset($existing[$slug])) { $slug->increase(); } return $slug; } 

See it in action.

6 Comments

Whoops, I broke it :D eval.in/private/2759b55f18fd2f There is nothing to stop your users from manually creating a page with the slugs 'php-5', 'php-4', 'php-3', 'php-1' Which causes the code to fall over as it is expecting the numbers to be ordered ascendingly in the array, which nothing says they have to be.
@Hailwood: oh right. just array_merge the result via call_user_func_array and then sort with natsort. -- eval.in/private/3c8c6b27c5414d -- or even just use in_array or array_flip and isset
I think I must be tired, I'm a nit picking bastard tonight ;)
I'll upvote, but I broke it again :D eval.in/private/03780cbbf5d041 If $slug-[digit] exists, even if just $slug doesn't it's going to throw it's toys and assume that $slug is taken :D
I did understood OP wanted to have slugs numbered, always, that's also why I $slug->increase() before the check once to give it that number. If you don't want that, just remove the first increase.
|
1

I wasn't fully satisfied with the answers, so I came up with a slightly different approach.

(SELECT CONCAT({$slug}, '-', counter) FROM ( SELECT (@row_number:=@row_number + 1) AS counter, ev.* FROM ( SELECT REPLACE(slug, {$slug}-, '') AS remainder FROM products, (SELECT @row_number:=0) AS t WHERE slug LIKE '{$slug}%' ) ev ORDER BY LENGTH(remainder), remainder ) sr WHERE counter <> remainder) LIMIT 1 

What this basically does, it checks all the existing values in the DB that are similar to the new slug, and matches it against the row number to check for gaps, and if none is found, it uses the maximum number that's generated by the first identical slug that gets pushed to the end (note: we replace slug- and not slug)

Comments

0
$query = "SELECT * FROM $table_name WHERE $field_name LIKE '".$slug."%'"; $result = mysqli_query($db_connect, $query) or die(mysqli_error($db_connect)); //EDITED BASED ON COMMENT SUGGESTIONS //create array of all matching slug names currently in database $slugs = array(); while($row = $result->fetch_row()) { $slugs[] = $row['field_name']; } //test if slug is in database, append - '1,2,..n' until available slug is found if(in_array($slug, $slugs)){ $count = 1; do{ $testSlug = $slug . '-' . $count; $count++; } while(in_array($testSlug, $slugs)); $slug = $testSlug; } //insert slug 

You should be able to do this in a single database call with the LIKE keyword that will reduce your execution time.

6 Comments

This would break if you had slug "test-slug-1" and tried to enter slug "test"
what about if you have rows slug-1, slug-3, slug-5 and adding one more?
Good suggestions, will revise or defer to Hailwoods solution that submitted at a similar time
Shouldn't this be while(in_arr... instead of while(!in_arr...? otherwise you're looping until an item is found in the array? Although I do like this solution!
Also, by modifying $slug each loop, instead of incrementing the int you are just creating e.g. $slug-1-2-3-4-5
|
0

You could use the Fbeen/UniqueSlugBundle. This Bundle is lightweight and does what it needs to do.

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.