PHP/MySQL Database/sqlite current

Материал из Web эксперт
Перейти к: навигация, поиск

Using sqlite_current() and sqlite_next()

   <source lang="html4strict">

<?php

   function my_sqlite_fetch_array($result,$type = SQLITE_BOTH,$decode = true) {
      if(!sqlite_next($result)) {
           return false;
      } else {
           return sqlite_current($result, $type, $decode);
      }
   }
   $sqlite = sqlite_open(":memory:");
   sqlite_query($sqlite, "CREATE TABLE test(value INTEGER PRIMARY KEY)");
   for($count = 0; $count < 5; $count++) {
       sqlite_unbuffered_query($sqlite, "INSERT INTO test VALUES(NULL)");
   }
   $result = sqlite_query($sqlite, "SELECT * FROM test");
   while($row = my_sqlite_fetch_array($result)) {
       var_dump($row);
   }

?>

 </source>