我正在嘗試使用 PHP 通過查詢更新公司表，但數(shù)據(jù)庫(kù)沒有變化。MySQL 數(shù)據(jù)庫(kù)。$pncon->query("UPDATE `company` SET
`name` = '" . $companyName . "',
`parent_ID` = '" . $companyParent . "',
`address` = '" . $address . "',
`phone` = '" . $phone . "',
`fax` = '" . $fax . "',
`email` = '" . $email . "',
`remarks` = '" . $remarks . "',
`type` = '" . $system . "',
`status` = '". $status ."'
WHERE `id` = '" . $id . "' ");
2 回答
慕無忌1623718
TA貢獻(xiàn)1744條經(jīng)驗(yàn) 獲得超4個(gè)贊
您的查詢看起來不錯(cuò)，但錯(cuò)誤的數(shù)據(jù)類型、無效的 id 或變量或違反其他約束可能是問題所在。
使用 PDO 語句很好，因?yàn)樗踩?/p>
您可以將代碼修改為，
$sql = "UPDATE `company` SET?
? `name` = '" . $companyName . "',
? `parent_ID` = '" . $companyParent . "',
? `address` = '" . $address . "',
? `phone` = '" . $phone . "',
? `fax` = '" . $fax . "',
? `email` = '" . $email . "',
? `remarks` = '" . $remarks . "',
? `type` = '" . $type . "',
? `status` = '" . $status . "'
? WHERE `id` = '" . $id . "' " ;
$stmt = $pncon->prepare($sql); //prepare statement
$stmt->execute(); //execute the query
繁星coding
TA貢獻(xiàn)1797條經(jīng)驗(yàn) 獲得超4個(gè)贊
$pncon->query("UPDATE company SET
name = '$companyName',
parent_ID = '$companyParent',
address = '$address',
phone = '$phone',
fax = '$fax',
email = '$email',
remarks = '$remarks',
type = '$system',
status = '$status'
WHERE id = '$id'");
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