Why else if condition doesn't work?

Question

PHP

Why the else if conditions don't work
Here is the code itself

$x = "мой текст";
$search="http";
$dub = mysql_query("SELECT (text) FROM `biznes` WHERE 1", $db);
$ret = mysql_fetch_array($dub);
if (strstr($x, $search)) {
    echo "Рекламный пост, пропускаем";
  mysql_query("INSERT INTO `biznes`(reklama) VALUES ('1')");
}else if ($ret == $x){
  mysql_query("INSERT INTO `biznes`(reklama) VALUES ('1')");
  }else{
  mysql_query("INSERT INTO `biznes`(foto,text) VALUES ('/images/$imge','$x')");
  echo "Занесли пост в базу";
}

The conditions are as follows
By the $ret value, we get a selection from the database, with all the lines where the text needs to be checked And
so the condition itself
If the text contains http, then we make an entry in the database with the text "Advertising post, skip" Is there a text in the database that
we want to add now ? , skip" If not that, not that, then we add the text that we have in the value of $x, i.e. the second condition, it simply skips.

Answer 1

Why not just make the column unique? Why produce empty records in the table, if you can put 1 in the reklama field of a single message in case of an error adding a double ... or increase by 1 - then there will be a spam counter)

Answer 2

More or less like this:

$sql =mysql_query ("SELECT 'text' FROM 'biznes' WHERE 'text'='text')";
if(mysql_num_rows($sql) < 0)
{  

mysql_query("INSERT INTO `biznes`(foto,text) VALUES ('/images/$imge','$x')");
}else {
    mysql_query("INSERT INTO `biznes`(reklama) VALUES ('1')"); 
}

Answer 3

elseif is spelled together