MySQL fetch数组期望参数1是资源 - MySQL fetch array expects parameter 1 to be resource

MySQL fetch array expects parameter 1 to be resource

<?php
include_once 'DB_Connect.php';
function getCategories() {
    $db = new DB_Connect();
    // array for json response
    $response = array();
    $response["Student"] = array();
    // Mysql select query
    $result = mysql_query("SELECT * FROM Student");
    while ($row = mysql_fetch_array($result)) {
        // temporary array to create single category
        $tmp = array();
        $tmp["id"] = $row["id"];
        $tmp["name"] = $row["name"];
        $tmp["number"] = $row["number"];
        $tmp["address"] = $row["address"];
        // push category to final json array
        array_push($response["Student"], $tmp);
    }
    // keeping response header to json
    header('Content-Type: application/json');
    // echoing json result
    echo json_encode($response);
}
getCategories();
?>

我有这个API为我的Android应用程序，但我有这个错误下面的问题。有什么想法吗?

<br />
<b>Warning</b>: mysql_fetch_array() expects parameter 1 to be resource, boolean given in <b>...'get_Student.php</b> on line <b>13</b><br />
{"Student":[]}

mysql_*已正式弃用。使用PDO或mysqli。这是你如何使用MySQLi库:

<?php
// Connect To Db
$mysqli = new mysqli("localhost", "my_user", "my_password", "my_database");
if (mysqli_connect_errno()) {
    printf("Connect failed: %s'n", mysqli_connect_error());
    exit();
}
// Query Function
function getCategories()
{
    // Init
    global $mysqli;
    $response = array('Student' => array());
    // Query Db
    $query = "SELECT * FROM Student";
    if ($result = $mysqli->query($query)) {
        while ($row = $result->fetch_assoc()) {
            $response['Student'][] = array(
                'id'      => $row['id'],
                'name'    => $row['name'],
                'number'  => $row['number'],
                'address' => $row['address'],
            );
        }
        $result->free(); // free up memory
    }
    // Finished
    header('Content-Type: application/json');
    exit(json_encode(response));
}
// Test
getCategories();
// Close Db Connect
$mysqli->close();

你可以有这样的代码:

// Connect To Db
$mysqli = new mysqli("localhost", "my_user", "my_password", "my_database");
if (mysqli_connect_errno()) {
    printf("Connect failed: %s'n", mysqli_connect_error());
    exit();
}

在一个名为db_connect.php的文件中，您可以将它包含在需要db连接的脚本中。因为变量$mysqli是在function() { ... }作用域之外初始化的;您需要使用global $mysqli;才能访问它。无论连接代码是在同一文件上还是从外部文件中包含，这都适用。快乐的编码。