我的ajax調用返回帶有json數據的html

Question

我有一個奇怪的問題，我試圖使用對我的php程序的ajax調用來填充數據表，該php程序從內部從數據庫獲取數據。

的PHP：

<?php
require_once('config.php');
$query = mysql_query("select * from productdetails");
while($fetch = mysql_fetch_array($query))
{
        $output[] = array ($fetch[0],$fetch[1],$fetch[2],$fetch[3],$fetch[4],$fetch[5]);
}
echo json_encode($output, JSON_FORCE_OBJECT);
?>

HTML（ajax呼叫）：

$.ajax({
            url: 'process.php?method=fetchdata',
            data: "json",
            success: function(s){
            console.log($(s).text());
                    oTable.fnClearTable();
                        for(var i = 0; i < s.length; i++) {
                         oTable.fnAddData([
                                    s[i][0],
                                    s[i][1],
                                    s[i][2],
                                    s[i][3],
                                    s[i][4],
                                    s[i][5]
                                           ]);                                      
                                        } // End For

            },
            error: function(e){
               console.log(e.responseText); 
            }
            });

生成的輸出為

Call Stack #TimeMemoryFunctionLocation 10.0010242552{main}( )..\\process.php : 0 20.0010242840http://www.php.net/function.mysql-connect' target='_new'>mysql_connect ( )..\\process.php : 2 {"0":{"0":"1","1":"Iron","2":"AX12","3":"Google","4":"21.95","5":"HW"},"1":{"0":"2","1":"DartBoard","2":"AZ52","3":"Apple","4":"12.95","5":"SG"},"2":{"0":"3","1":"BasketBall","2":"BA74","3":"Microsoft","4":"29.95","5":"SG"},"3":{"0":"4","1":"Compopper","2":"BH22","3":"Google","4":"24.95","5":"HW"},"4":{"0":"5","1":"Gas Grill","2":"BT04","3":"Apple","4":"149.95","5":"AP"},"5":{"0":"6","1":"Washer","2":"BZ66","3":"Google","4":"399.99","5":"AP"}} （！）不建議使用：mysql_connect（）：不建議使用mysql擴展，並將在以后刪除：在第行的C：\\ wamp \\ www \\ datatableone \\ process.php中使用mysqli或PDO調用堆棧#TimeMemoryFunctionLocation 10.0010242552 {main }（）.. \\ process.php ： 0 20.0010242840http：//www.php.net/function.mysql-connect'target ='_ new'> mysql_connect（）.. \\ process.php ： 2 {“ 0”： {“ 0”：“ 1”，“ 1”：“ Iron”，“ 2”：“ AX12”，“ 3”：“ Google”，“ 4”：“ 21.95”，“ 5”：“ HW”}， “ 1”：{“ 0”：“ 2”，“ 1”：“ DartBoard”，“ 2”：“ AZ52”，“ 3”：“ Apple”，“ 4”：“ 12.95”，“ 5”：“ SG“}，” 2“：{” 0“：” 3“，” 1“：” BasketBall“，” 2“：” BA74“，” 3“：” Microsoft“，” 4“：” 29.95“，” 5“：” SG“}，” 3“：{” 0“：” 4“，” 1“：” Compopper“，” 2“：” BH22“，” 3“：” Google“，” 4“：” 24.95“，” 5“：” HW“}，” 4“：{” 0“：” 5“，” 1“：” Gas Grill“，” 2“：” BT04“，” 3“：” Apple“， “ 4”：“ 149.95”，“ 5”：“ AP”}，“ 5”：{“ 0”：“ 6”，“ 1”：“ Washer”，“ 2”：“ BZ66”，“ 3”： “ Google”，“ 4”：“ 399.99”，“ 5”：“ AP”}}

但是我所需的輸出應該是：（僅json）

{“ 0”：{“ 0”：“ 1”，“ 1”：“ Iron”，“ 2”：“ AX12”，“ 3”：“ Google”，“ 4”：“ 21.95”，“ 5”： “ HW”}，“ 1”：{“ 0”：“ 2”，“ 1”：“ DartBoard”，“ 2”：“ AZ52”，“ 3”：“ Apple”，“ 4”：“ 12.95”， “ 5”：“ SG”}，“ 2”：{“ 0”：“ 3”，“ 1”：“籃球球”，“ 2”：“ BA74”，“ 3”：“ Microsoft”，“ 4”： “ 29.95”，“ 5”：“ SG”}，“ 3”：{“ 0”：“ 4”，“ 1”：“ Compopper”，“ 2”：“ BH22”，“ 3”：“ Google”， “ 4”：“ 24.95”，“ 5”：“ HW”}，“ 4”：{“ 0”：“ 5”，“ 1”：“ Gas Grill”，“ 2”：“ BT04”，“ 3” ：“ Apple”，“ 4”：“ 149.95”，“ 5”：“ AP”}，“ 5”：{“ 0”：“ 6”，“ 1”：“洗衣機”，“ 2”：“ BZ66” ，“ 3”：“ Google”，“ 4”：“ 399.99”，“ 5”：“ AP”}}

任何建議，請。

謝賽

解：

PHP：

<?php
$servername = "localhost";
$username = "root";
$password = "123456";

try {
        $conn = new PDO("mysql:host=$servername;dbname=holt", $username, $password);
        $statement=$conn->prepare("SELECT * FROM productdetails");
        $statement->execute();
        $results=$statement->fetchAll(PDO::FETCH_ASSOC);
        $json=json_encode($results);
        echo $json;
    } catch(PDOException $e) {
        echo "Connection failed: " . $e->getMessage();
    }
?>

感謝您的建議。

Answer 1

如注釋中所述，您應該關閉警告，或者更好的方式是，以不產生警告的方式編寫代碼。

關於關閉警告： 關閉php / mysql上的警告和錯誤

您可以使用@符號抑制內聯錯誤，該符號是php中的錯誤控制運算符。 將@放在mysql_connect（）行的開頭應該可以擺脫它，但是您應該切換到PDO！

在PDO（我建議和使用）上： http : //code.tutsplus.com/tutorials/why-you-should-be-using-phps-pdo-for-database-access--net-12059

PDO保護您免受SQL注入的侵害，並允許將查詢發送到數據庫並事先進行構造，然后再通過“占位符”發送輸入。