如何在PHP中使用MySQL NOT NULL？

Question

我已经使用PHP在MySQL数据库中插入值（以下示例代码）：

<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
<style>
label{text-align:left;
       width:150px;
       display:block;
           float:left;
           clear:right;
       font-size:18;}

form {
    width: 30%;
    margin: 0 auto;
}
</style>
<label>Name:</label><input name=name autocomplete=off><br>
<label>Date of Birth:</label><input size=2 name=dd placeholder=dd> / <input size=2 name=mm placeholder=mm> / <input size=6 name=yyyy placeholder=yyyy><br>
<label>Time of Birth:</label><input name=tob placeholder="hh : mm : ss"><br>
<label>City:</label><input name=city><br><br>
<center><input type=submit value=Save></center>
</form>

<?php
if ($_SERVER["REQUEST_METHOD"] == "POST"){
$servername = "localhost";
$username = "userame";
$password = "password";
$dbname = "myDB2";
if(empty($_POST["name"])){ echo "Name is Required<br>";}
else {
$name = $_POST["name"];
}
var_dump($name);
$tob = $_POST["tob"];
$city = $_POST["city"];
$dd = $_POST["dd"];
$mm = $_POST["mm"];
$yyyy = $_POST["yyyy"];

$dob = "$yyyy-$mm-$dd";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

//Inserting Data
$sql = "INSERT INTO Person
VALUES('$name','$dob','$tob','$city')";
if ($conn->query($sql) === TRUE) {
    echo "<br>New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
}
?>

AND列Name设置为varchar(60) NOT NULL UNIQUE数据类型和约束。

---

尽管

• Name列的NOT NULL约束，以及

• if(empty($_POST["name"])){ echo "Name is Required";} else { $name = $_POST["name"]; }

当我插入提交表单而未在表单的“名称输入”中输入任何值时，出现以下消息（根据代码）

Name is Required
NULL
New record created successfully

和我的数据库输出像：

在这里，您可以看到var_dump($name)表示$name为NULL但MySQL数据库仍在Name Column中接受该值！

---

所以，我的问题是， NOT NULL的用途是什么？ 如何在PHP中使用MysQL NOT NULL ？ 在表格帮助下插入数据？

Answer 1

您需要适当地将PHP的null转换为SQL的NULL类型。 在PHP中进行简单的字符串插值时：

"INSERT INTO foo VALUES ('$bar')"

其中$bar为null ，结果最终为：

INSERT INTO foo VALUES ('')

MySQL将其视为空字符串，而不是NULL ，所以很好。

您必须执行以下操作：

$value = null;
if ($value === null) {
    $bar = 'NULL';
} else {
    $bar = "'$value'";
}
$query = "INSERT INTO foo VALUES ($bar)";

这里的结果是这两个之一：

INSERT INTO foo VALUES (NULL)
INSERT INTO foo VALUES ('baz')

然后MySQL将null视为NULL 。

但是，如果要使用PHP达到这样的长度，则很显然，如果您的值为空，则可以不触发查询，而不必触发查询并使MySQL因约束而失败。

您还需要了解有关SQL注入的知识 。

Answer 2

空字符串不是NULL 。 您要在表格中插入的是一个空字符串：

$sql = "INSERT INTO Person VALUES('$name','$dob','$tob','$city')";

如果$name在PHP中为NULL，则只会导致这样的字符串： INSERT INTO Person VALUES('','...','...','...')

如果要将NULL传递给MySQL，则必须发送NULL ，而不是"NULL"而不是"" 。

$nameForMysql = is_null($name) ? "NULL" : "'$name'";
$sql = "INSERT INTO Person VALUES($nameForMysql,'$dob','$tob','$city')";

Answer 3

因此，只要使if...else更宽：

if(!isset($_POST["name"]) || empty($_POST["name"])){ echo "Name is Required<br>";}
else {
$name = $_POST["name"];
$tob = $_POST["tob"];
$city = $_POST["city"];
$dd = $_POST["dd"];
$mm = $_POST["mm"];
$yyyy = $_POST["yyyy"];

$dob = "$yyyy-$mm-$dd";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

//Inserting Data
$sql = "INSERT INTO Person
VALUES('$name','$dob','$tob','$city')";
if ($conn->query($sql) === TRUE) {
    echo "<br>New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
}

Answer 4

在没有值的地方指定NULL，sql将您的数据视为空字符串而不是null，因此请使用以下代码

<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
<style>
label{text-align:left;
       width:150px;
       display:block;
           float:left;
           clear:right;
       font-size:18;}

form {
    width: 30%;
    margin: 0 auto;
}
</style>
<label>Name:</label><input name=name autocomplete=off><br>
<label>Date of Birth:</label><input size=2 name=dd placeholder=dd> / <input size=2 name=mm placeholder=mm> / <input size=6 name=yyyy placeholder=yyyy><br>
<label>Time of Birth:</label><input name=tob placeholder="hh : mm : ss"><br>
<label>City:</label><input name=city><br><br>
<center><input type=submit value=Save></center>
</form>

<?php
if ($_SERVER["REQUEST_METHOD"] == "POST"){
$servername = "localhost";
$username = "userame";
$password = "password";
$dbname = "myDB2";
if(empty($_POST["name"])){ echo "Name is Required<br>";
$name = "NULL";
}
else {
$name = $_POST["name"];
}
var_dump($name);
$tob = $_POST["tob"];
$city = $_POST["city"];
$dd = $_POST["dd"];
$mm = $_POST["mm"];
$yyyy = $_POST["yyyy"];

$dob = "$yyyy-$mm-$dd";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

//Inserting Data
$sql = "INSERT INTO Person
VALUES('$name','$dob','$tob','$city')";
if ($conn->query($sql) === TRUE) {
    echo "<br>New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
}
?>

希望这对您有帮助