php sqlite_query函数的例子

2. /**
3. * sqlite_query函数简单示例
4. * by bbs.it-home.org
5. */
6. $sqldb = sqlite_open("mydatabase.db");
7. $results = sqlite_query($sqldb, "SELECT * FROM employee");
8. while (list($empid, $name) = sqlite_fetch_array($results)) {
9. echo "Name: $name (Employee ID: $empid)
   ";
10. }
11. sqlite_close($sqldb);
12. ?>

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例2，使用sqlite查询()函数

3. $db = sqlite_open(":memory:");

4. if(!$db) die("Could not create the temporary database");

5. $query = "CREATE TABLE cities(name VARCHAR(255), state VARCHAR(2))";

6. sqlite_query($db, $query);
7. $cities[] = array('name' => 'Chicago','state'=> 'IL');
8. foreach($cities as $city) {
9. $query = "INSERT INTO cities VALUES(" ."'{$city['name']}', '{$city['state']}')";
10. if(!sqlite_query($db, $query)) {
11. trigger_error("Could not insert city " . "'{$city['name']}, {$city['state']}");
12. }
13. }
14. sqlite_close($db);
15. ?>

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例3，sqlite_query查询测试

3. /**

4. * sqlite_query查询的测试用例
5. * by bbs.it-home.org
6. */
7. $dbconn = sqlite_open('phpdb');

8. if ($dbconn) {

9. script!
10. sqlite_query($dbconn, "INSERT INTO animal VALUES('a', 14)");
11. sqlite_query($dbconn, "INSERT INTO animal VALUES('b', 16)");
12. sqlite_query($dbconn, "INSERT INTO animal VALUES('c', 13)");
13. var_dump(sqlite_array_query($dbconn, "SELECT * FROM animal", SQLITE_ASSOC));
14. } else {
15. print "Connection to database failed!\n";
16. }
17. ?>

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