我正在嘗試使用此php文檔來(lái)使用一種形式將信息輸入數(shù)據(jù)庫(kù)。我不斷收到相同的錯(cuò)誤，列“ custID”不能為null。我不知道怎么了或怎么做。我可能需要為此任務(wù)加上L，但是如果以后能遇到同樣的問(wèn)題，我能得到一個(gè)答案的話，這將是有幫助的。我已經(jīng)嘗試在mysql代碼中執(zhí)行NOT NULL AUTO_INCREMENT。我還嘗試通過(guò)將NULL用于custID來(lái)做同樣的事情。兩者都不起作用。if(isset($_POST['submit'])){ $data_missing = array(); if(empty($_POST['custID'])){ $data_missing[] = 'Customer ID'; }else{ $custID = trim($_POST['custID']); } if(empty($_POST['custFirstName'])){ $data_missing[] = 'First Name'; }else{ $custFirstName = trim($_POST['custFirstName']); } if(empty($_POST['custLastName'])){ $data_missing[] = 'Last Name'; }else{ $custLastName = trim($_POST['custLastName']); } if(empty($_POST['address'])){ $data_missing[] = 'Address'; }else{ $address = trim($_POST['address']); } if(empty($_POST['city'])){ $data_missing[] = 'city'; }else{ $city = trim($_POST['city']); } if(empty($_POST['custstate'])){ $data_missing[] = 'State'; }else{ $custstate = trim($_POST['custstate']); } if(empty($_POST['custEmail'])){ $data_missing[] = 'Email'; }else{ $custEmail = trim($_POST['custEmail']); } if(empty($_POST['custPhone'])){ $data_missing[] = 'Phone'; }else{ $custPhone = trim($_POST['custPhone']); } if(empty($_POST['Password'])){ $data_missing[] = 'Password'; }else{ $Password = trim($_POST['Password']); } }
2 回答
慕村225694
TA貢獻(xiàn)1880條經(jīng)驗(yàn) 獲得超4個(gè)贊
您必須刪除，custID因?yàn)?code>AUTO_INCREMENT
$query = "INSERT INTO Customers (custFirstName, custLastName, address, city," . " custstate, custEmail, custPhone, Password) VALUES (?, ?, ?, ?, ?, ?, ?, ?)";
和這段代碼
mysqli_stmt_bind_param($stmt, "sssssssss", $custFirstName,$custLastName, $address, $city, $custstate, $custEmail, $custPhone, $Password);
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