这是我的源代码,我在PHP源代码中使用了sqlite数据库.当我运行脚本时,它给了我这个错误
- <?PHP
- $db = $_SERVER['DOCUMENT_ROOT']."umer.db";
- $handle = sqlite_open($db) or die("Could not open database".sqlite_error_string(sqlite_last_error($handle)));
- $query = "SELECT * FROM books";
- $result = sqlite_query($handle,$query) or die("Error in query: ".sqlite_error_string(sqlite_last_error($handle)));
- if (sqlite_num_rows($result) > 0) {
- echo "<table cellpadding=10 border=1>";
- while($row = sqlite_fetch_array($result)) {
- echo "<tr>";
- echo "<td>".$row[0]."</td>";
- echo "<td>".$row[1]."</td>";
- echo "<td>".$row[2]."</td>";
- echo "</tr>";
- }
- echo "</table>";
- }
- sqlite_close($handle);
- ?>