MySQL的获取数组预计参数1是资源 [英] MySQL fetch array expects parameter 1 to be resource

问题描述

 ＆LT; PHPinclude_once'DB_Connect.php';功能getCategories（）{
    $ DB =新DB_Connect（）;
    //数组JSON响应
    $响应=阵列（）;
    $响应[学生] =阵列（）;    // mysql的选择查询
    $结果= mysql_query（SELECT * FROM学生）;    而（$行= mysql_fetch_array（$结果））{
        //临时数组创建单一类别
        $ TMP =阵列（）;
        $ TMP [身份证] = $行[身份证];
        $ TMP [名称] = $行[名称];
        $ TMP [数量] = $行[数量];
        $ TMP [地址] = $行[地址];
        //类推到最终JSON数组
        array_push（$响应[学生]，$ TMP）;
    }    //保持响应头为JSON
    标题（内容类型：应用程序/ JSON'）;    //呼应JSON结果
    回声json_en code（$响应）;
}getCategories（）;？＆GT;
 

我有这个API为我的Andr​​oid应用程序，但我有下面这个错误的问题。任何想法了吗？

 ＆LT; BR /＆GT;
＆LT; B＆GT;警告＆LT; / B＆GT;：mysql_fetch_array（）预计参数1是资源，布尔中给出＆LT; B＆GT; ... \\＆get_Student.php LT; / B＆GT;上线474; B＆GT; 13 LT; / B＆GT;＆LT; BR /＆GT;
{学生：[]}
 

解决方案

的mysql _ * 正式去precated。请使用 PDO 或的mysqli 。这就是你如何与库MySQLi库做到这一点：

 ＆LT; PHP//连接到数据库
$库MySQLi =新的mysqli（本地主机，my_user，MY_PASSWORD，MY_DATABASE）;
如果（mysqli_connect_errno（））{
    的printf（连接失败：％S \\ n，mysqli_connect_error（））;
    出口（）;
}//查询功能
功能getCategories（）
{
    // 在里面
    全球$ mysqli的;
    $响应=阵列（'学生'=＆GT;阵（））;    //查询分贝
    $查询=SELECT * FROM学生;
    如果（$结果= $ mysqli-＆GT;查询（$查询））{
        而（$行= $ result-＆GT; FETCH_ASSOC（））{
            $响应['学生'] [] =数组（
                'ID'=＆GT; $行['身份证']，
                '名'=＆GT; $行[名称]
                '数'=＆GT; $行['号']，
                '地址'=＆GT; $行['地址']，
            ）;
        }
        $ result-＆GT;免费（）; //释放内存
    }    //完成
    标题（内容类型：应用程序/ JSON'）;
    出口（json_en code（响应））;
}//测试
getCategories（）;//关闭连接分贝
$ mysqli-＆GT;关闭（）;
 

---

您可以有这个code：

  //连接到数据库
$库MySQLi =新的mysqli（本地主机，my_user，MY_PASSWORD，MY_DATABASE）;
如果（mysqli_connect_errno（））{
    的printf（连接失败：％S \\ n，mysqli_connect_error（））;
    出口（）;
}
 

在一个名为 db_connect.php 并可以将其包含在您需要数据库连接的脚本。因为变量 $的mysqli 启动了函数之外（）{...} 范围;你需要使用全球$ mysqli的; 来访问它。无论连接code是在同一个文件或者被包括来自外部文件的适用。快乐编码。

<?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();

?>

I have this API for my Android app, but I have a problem with this error below. Any idea out there?

<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":[]}

解决方案

mysql_* is officially deprecated. Use either PDO or mysqli. This is how you do it with MySQLi library:

<?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();

---

You can have this code:

// 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();
}

In a file called db_connect.php and you can include it on the script you require db connection. Because the variable $mysqli is initiated outside of the function() { ... } scope; you need to use global $mysqli; to get access to it. This applies regardless of the connection code being on same file or being included from external file. Happy coding.

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