PHP mysqli準備語句，如我如何使用mysqli進行類似的查詢并獲得所有結果？這是我的代碼，但不起作用：$param = "%{$_POST['user']}%";$stmt = $db->prepare("SELECT id,Username FROM users WHERE Username LIKE ?");
$stmt->bind_param("s", $param);$stmt->execute();$stmt->bind_result($id,$username);$stmt->fetch();這段代碼似乎不起作用。我已經搜了很多遍了。此外，它可能返回超過1行。那么，即使它返回超過1行，我如何才能得到所有的結果呢？
3 回答
MMMHUHU
TA貢獻1834條經驗 獲得超8個贊
$param = "%{$_POST['user']}%";$stmt = $db->prepare("SELECT id,Username FROM users WHERE Username LIKE ?");
$stmt->bind_param("s", $param);$stmt->execute();$stmt->bind_result($id,$username);while ($stmt->fetch()) {
echo "Id: {$id}, Username: {$username}";}$param = "%{$_POST['user']}%";$stmt = $db->prepare("SELECT id,Username FROM users WHERE Username LIKE ?");
$stmt->bind_param("s", $param);$stmt->execute();$result = $stmt->get_result();while ($row = $result->fetch_array(MYSQLI_NUM)) {
foreach ($row as $r) {
print "$r ";
}
print "\n";}
慕雪6442864
TA貢獻1812條經驗 獲得超5個贊
CONCATLIKE$stmt=$db->prepare("SELECT suggestioncustomername$key='J%';$key='';), $stmt->bind_param('s', $key);$stmt->execute();_%$key='J\%';
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