Question

请检查此代码，如果有任何错误，请告诉我。

// Create connection
$conn = mysqli_connect($servername, $username, $password, $shop_item);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$item_sql = "SELECT * FROM shop-items";
mysqli_query($conn, 'SET CHARACTER SET utf8;');
$result_item = mysqli_query($conn, $item_sql);

echo var_dump($result_item); //returns: bool(false)
if (mysqli_num_rows($result_item) > 0) { // doesn't execute. the error is "mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean";

Answer 1

正如我所提到的，shop-items被mysql视为shop minus items。如果你的表名真的是shop-items - 你应该使用反引号来逃避它：

$item_sql = "SELECT * FROM `shop-items`";

为了检查错误，您可以使用mysqli_error()功能：

$err = mysqli_error($conn);
echo $err;

Answer 2

使用类似于此的结构可能更好：

$conn = new mysqli('localhost', 'root', 'password', 'your_database');
// Check connection

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql="SELECT * FROM shop-items";

if ($result = $conn->query($sql)) {    
    // actions in case of success    
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

Answer 3

这可能是破折号的问题。你试过吗

SELECT * FROM `shop-items`

Answer 4

我会尝试使用以下代码来处理从PHP到您的数据库的所有查询，因为正如之前提到过的，您使用的错误可能来自非语法：

    $mysqli = new mysqli("localhost", "user", "password", "DB");

/* Check conection */
if (mysqli_connect_errno()) {
    printf("Conection error: %s\n", mysqli_connect_error());
    exit();
}

$query = "SELECT * FROM shop-items";

if ($stmt= $mysqli->prepare($query)) {

    /* execute the query */
    $stmt->execute();

    /* Bind the results to variables */
    $stmt->bind_result($col1, $col2); //as many variables as columns the query will return

    /* obtener los valores */
    while ($stmt->fetch()) {
        //do something with the results
    }

    /* close the query */
    $stmt->close();
}

/* close the DB connection */
$mysqli->close();
?>

希望这会有所帮助!!

sql query

4 个答案: