Question

<?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();

?>

我的Android应用程序有这个API，但我在下面遇到此错误的问题。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":[]}

Answer 1

mysql_*已被正式弃用。使用PDO或mysqli。这就是你使用MySQLi库的方法：

<?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();

您可以拥有以下代码：

// 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();
}

在名为 db_connect.php 的文件中，您可以将其包含在需要数据库连接的脚本中。因为变量$mysqli是在function() { ... }范围之外启动的;您需要使用global $mysqli;来访问它。无论连接代码在同一文件上还是包含在外部文件中，都适用。快乐的编码。

MySQL获取数组期望参数1是资源

1 个答案: