我嘗試了多種方法嘗試並使其工作。已經嘗試了以前關於此主題的所有答案,並且無法得到它。
如果$ _POST變量沒有值,我試圖插入NULL而不是數據庫中的字符串NULL。它只是插入字符串'NULL'或只是一個空白列。以下是我嘗試查詢的所有方法。
我的數據庫類有一個方法sql_prep
:
public function sql_prep($postVariable){
$output;
if(trim($postVariable) == ''){
$output = 'NULL';
}else{
$output = strval(mysqli_real_escape_string($this->connection, $postVariable));
};
return $output;
}
下面是該查詢:
if(isset($_POST["createUserSubmit"])) :
$temp_connection = mysqli_connect(DB_SERVER, DB_USER, DB_PASS, DB_NAME);
$firstName = $db->sql_prep($_POST["firstName"]);
$lastName = $db->sql_prep($_POST["lastName"]);
$companyName = $db->sql_prep($_POST["companyName"]);
$streetAddress = $db->sql_prep($_POST["streetAddress"]);
$streetAddress2 = $db->sql_prep($_POST["streetAddress2"]);
$streetAddress3 = $db->sql_prep($_POST["streetAddress3"]);
$city = $db->sql_prep($_POST["city"]);
$state = $db->sql_prep($_POST["state"]);
$zip = $db->sql_prep($_POST["zipCode"]);
$country = $db->sql_prep($_POST["country"]);
$phone = $db->sql_prep($_POST["phone"]);
$fax = $db->sql_prep($_POST["fax"]);
$email = $db->sql_prep($_POST["email"]);
mysqli_query($temp_connection, "INSERT INTO Address(firstName, lastName, companyName, address1, address2, address3, city, state, zip, country, phone, fax, email, dateCreated, dateModified) VALUES (" . $firstName . ", " . $lastName . ", " . $companyName . ", " . $streetAddress . ", " . $streetAddress2 . ", " . $streetAddress3 . ", " . $city . ", " . $state . ", " . $zip . ", " . $country . ", " . $phone . ", " . $fax . ", " . $email . ", NOW(), NOW())");
mysqli_close($temp_connection);
redirect_to('./create-user.php');
endif;
該查詢甚至不會任何數據推到數據庫中,即使該字段填寫。另一種方法我試過查詢:
mysqli_query($temp_connection, "INSERT INTO Address(firstName, lastName, companyName, address1, address2, address3, city, state, zip, country, phone, fax, email, dateCreated, dateModified) VALUES ('{$firstName}', '{$lastName}', '{$companyName}', '{$streetAddress}', '{$streetAddress2}', '{$streetAddress3}', '{$city}', '{$state}', '{$zip}', '{$country}', '{$phone}', '{$fax}', '{$email}', NOW(), NOW())");
這將返回字符串「NULL」到數據庫中,如果$ _POST變量是空的。我也試過我sql_prep
功能改成這樣:
public function sql_prep($postVariable){
$output;
if(trim($postVariable) == ''){
$output = NULL; //returned PhP Null instead of string 'NULL'
}else{
$output = strval(mysqli_real_escape_string($this->connection, $postVariable));
};
return $output;
}
更改它返回,而不是「NULL」 NULL比索導致查詢只是一個空白列推入分貝。
無法想出這一個,真的想推SQL NULL如果沒有值。
數據庫字段是否設置爲默認NULL? – caramba
我相信當你將NULL連接成一個字符串時,它將NULL轉換爲一個空字符串。這就是爲什麼你要爲應該設置爲NULL的字段獲得「空白值」。 –
是將其設置爲默認NULL –