Mysql插入不起作用，也没有给出错误 - Mysql insert not working and not giving errors

Mysql insert not working and not giving errors

我不知道为什么下面的代码不能将数据插入mysql。

if (!$link = mysql_connect('server', 'user', 'password')) {
    echo '700';
    exit;
}
if (!mysql_select_db('vendors', $link)) {
    echo '701';
    exit;
}
$sql2 = "INSERT INTO transactions (TransID, payment_status, last_name, first_name, payer_email, address_name, address_state, address_zip, address_country, verify_sign, payment_gross, ipn_track_id, business, reciver_email) VALUES ('kris', 'kris', 'kris', 'kris', 'kris','kris', 'kris', 'kris', 'kris', 'kris', 'kris', 'kris', 'kris', 'kris')";
$result2 = mysql_query($sql2, $link);

代码出了什么问题？php没有给出任何错误。

请尽量不要使用mysql_connect，而是使用mysqli_connect或PDO_mysql读取此

还可以使用die来查找代码中是否存在任何错误

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}

否则（推荐方式）-

程序风格

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO Persons (firstname, lastname, email)
VALUES ('Happy', 'John', 'john@example.com')";
if (mysqli_query($conn, $sql)) {
    echo "New Person created successfully";
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);

MySQLi面向对象风格

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$sql = "INSERT INTO Persons (firstname, lastname, email)
VALUES ('Happy', 'John', 'john@example.com')";
if ($conn->query($sql) === TRUE) {
    echo "New Person created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();

尝试更改此

 $result2 = mysql_query($sql2, $link);

进入这个

$result2 = mysql_query($sql2, $link)or die(mysql_error());

您必须编写如下代码才能在代码中获得错误

$result = mysql_query($sql2,$link) or die(mysql_error());

该or die(mysql_error())将在查询时出错