php - mysql 不会取 null，而是设置 0

Question

我试图在我的数据库中插入一个 NULL 值，我尝试了不同的示例，但结果一直显示为 0 所以我强制它，注意下面的行//********** $gefid = null; yet 仍然显示为 0，已尝试使用和不使用引号、大写字母，随你便。 该字段绝对允许空值，空值是默认值 代码:

<?php

DEFINE ('DBUSER', 'xxxxxxxxx');
DEFINE ('DBPW', 'xxxxxxxxxxxxxxxxxxxxxxx');
DEFINE ('DBHOST', 'xxxxxxxxxxxxxxxxxxxxxxxxxxxxx.com');
DEFINE ('DBNAME', 'Mydb');


$dbc = mysqli_connect(DBHOST,DBUSER,DBPW);
if (!$dbc) {
die("Database connection failed really badly: " . mysqli_error($dbc));
exit();
}

$dbs = mysqli_select_db($dbc, DBNAME);
if (!$dbs) {
die("now the dam Database selection bit failed: " . mysqli_error($dbc));
exit();
}
$lat = mysqli_real_escape_string($dbc, $_GET['lat']);
$lng = mysqli_real_escape_string($dbc,$_GET['lng']);

$prox = mysqli_real_escape_string($dbc,$_GET['prox']);
$description = mysqli_real_escape_string($dbc,$_GET['description']);
$id = mysqli_real_escape_string($dbc,$_GET['id']);
$direction = mysqli_real_escape_string($dbc,$_GET['direction']);
$avoiddays = mysqli_real_escape_string($dbc,$_GET['avoiddays']);
$validfrom = mysqli_real_escape_string($dbc,$_GET['validfrom']);
$validto = mysqli_real_escape_string($dbc,$_GET['validto']);



if (isset($_GET['gefid'])) {
  $gefid = mysqli_real_escape_string($dbc,$_GET['gefid']);
} else {
  $gefid = null;
}
//***********************************
$gefid = null;
$expiry = mysqli_real_escape_string($dbc,$_GET['expiry']);

$query = "INSERT INTO realtime (rt_lat,rt_lng,rt_prox,rt_description,rt_direction,rt_avoiddays,rt_validto,rt_validfrom,rt_gefid,rt_expiry) VALUES ('$lat', '$lng', '$prox', '$description','$direction','$avoiddays','$validto','$validfrom','$gefid','$expiry')"; 

$result = mysqli_query($dbc, $query) or trigger_error("Query MySQL Error: " . mysqli_error($dbc));

mysqli_close($dbc);
?>

Answer 1

问题在于您的INSERT 语句，而不是变量赋值。要插入 NULL 值，您的语句应如下所示(注意，NULL 周围没有引号):

INSERT INTO table (column) VALUES (NULL);

您可以创建一些额外的代码来添加或省略这些引号，但我强烈建议使用 prepared statements ，您不必再担心转义和引号。