我尝试了多种方法尝试并使其工作。已经尝试了以前关于此主题的所有答案,并且无法得到它。
如果$ _POST变量没有值,我试图插入NULL而不是数据库中的字符串NULL。它只是插入字符串'NULL'或只是一个空白列。以下是我尝试查询的所有方法。
我的数据库类有一个方法sql_prep
:
public function sql_prep($postVariable){
$output;
if(trim($postVariable) == ''){
$output = 'NULL';
}else{
$output = strval(mysqli_real_escape_string($this->connection, $postVariable));
};
return $output;
}
下面是该查询:
if(isset($_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;
该查询甚至不会任何数据推到数据库中,即使该字段填写。另一种方法我试过查询:
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())");
这将返回字符串“NULL”到数据库中,如果$ _POST变量是空的。我也试过我sql_prep
功能改成这样:
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比索导致查询只是一个空白列推入分贝。
无法想出这一个,真的想推SQL NULL如果没有值。
数据库字段是否设置为默认NULL? – caramba
我相信当你将NULL连接成一个字符串时,它将NULL转换为一个空字符串。这就是为什么你要为应该设置为NULL的字段获得“空白值”。 –
是将其设置为默认NULL –