[請益] 簡單SQL 指令請教各位前輩

看板PHP作者worlds0731 (宅度還不夠)時間15年前 (2010/04/30 16:51)推噓0(0推 0噓 4→)

留言4則, 2人參與討論串1/1

想請教各位前輩,這是很簡單的問題卻困擾我很久蟲子一直抓不到... 當我對資料庫下SQL指令 $sql = "SELECT * FROM member"; 可以正確顯示4筆資料 $sql = "SELECT * FROM member WHERE sid = 1"; 這樣也可以正確顯示編號1的資料 $sql = "SELECT * FROM member WHERE name LIKE '%系統%'"; 可是當我下LIKE指令的時候就沒有資料顯示了 $sql = "SELECT * FROM member WHERE name = '系統管理員'"; 或者直接用 = 去比對資料也是沒有資料顯示我有用 phpMyAdmin 的SQL查詢視窗輸入3跟4的指令都是有正確回傳結果的... 想請教各位前輩是不是我在PHP腳本中哪邊有錯呢? 順便附上腳本內容: <?php require_once "./vo/__MemberVO.php"; class __SearchService { var $dbh; public function __construct() { $this->dbh = mysql_connect ("localhost", "root", "") or die ('I cannot connect to the database because: ' . mysql_error()); mysql_select_db ("beesdb"); } function getMemberVO() { $myMembers = array(); mysql_query("SET NAMES 'utf8'"); $sql = "SELECT * FROM member "; //$sql = "SELECT * FROM member WHERE name = '系統管理員'"; //$sql = "SELECT * FROM member WHERE name LIKE '%系統%'"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)) { $Member = new __MemberVO(); $Member->name = $row['name']; $Member->provide = $row['provide']; $myMembers[] = $Member; } return $myMembers; } } ?> -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.34.185.41