时间:2021-07-01 10:21:17 帮助过:160人阅读
$result = mysql_query("SELECT * FROM table");$fields = mysql_num_fields($result);for ($i=0; $i < $fields; $i++) { $names[] = mysql_field_name($result, $i);}print_r($names);
try { $dbh = new PDO('mysql:host=localhost;dbname=test', 'root', ''); $stmt = $dbh->query("select * from tbl_name"); for($i=0; $i<$stmt->columnCount(); $i++) { echo $stmt->getColumnMeta($i)['name'], PHP_EOL;//php5.4适使用,之前的版本需拆成两句 }} catch (PDOException $e) { die("Error!: " . $e->getMessage());}
$sqlstr="select * from table"
$sqlstr="select * from table"
$sqlstr="select * from table"; $rerult=mysql_quer($sqlstr); while($field=mysql_fetch_field($result){ echo "$field->name";}
谢谢版主,终于搞定了,php实在不熟!!!
本帖最后由 xuzuning 于 2013-03-11 13:52:28 编辑
PHP code?123456789try { $dbh = new PDO('mysql:host=localhost;dbname=test', 'root', ''); $stmt = $dbh->query("select * from tbl_na……