Question

我正在尝试根据用户输入的内容对数据库进行简单搜索，因为我正在使用，javascript和PHP。

它几乎可以工作，但是我无法传递字符用户正在键入SQL查询。

我确信该变量在Ajax_request.php中到达了PHP POST方法，但是我不确定“functions.php”中的类是否已经被声明为能使它工作。

可能只是一个愚蠢的错误，但我是PHP和Web开发的新手

这些是我的档案：

的index.php

<!doctype html>

<html lang="en">
<head>
    <meta charset="utf-8" />

    <title>InChiaro Ticket Admin</title>
    <meta name="description" content="The HTML5 Herald" />
    <meta name="author" content="SitePoint" />
    <script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
</head>

<body>
    <div id="result"></div>
    <input type="text" class="search-filter" id="searchcodiceCliente" name="codiceCliente" /> <!-- text AREA CODICE CLIENTE-->
    <script type="text/javascript">
        $(function () {
            // We add the event on the class, which both inputs have
            $(".search-filter").keyup(function () {
                // Now we get the values from both inputs, using their ID's
                var codiceCliente = $("#searchcodiceCliente").val();
                //var fname = $("#searchfname").val();

                // Add both to the dataString (and URI encode the strings)

                var requestCodCliente = "get_codiceCliente_json"    
                // Check that at least one has any content
                if (codiceCliente != '')


                    $.ajax({
                        type: "POST",
                        url: "ajax_requests.php",
                        data: 'request='+ requestCodCliente +'&searchCliente='+ codiceCliente,
                        success: function (result) {
                            console.log(result);
                        }
                    });

            });
        });

    </script>

</body>
</html>

Ajax_request.php

<?php
if (!empty($_POST)) {
    $codCliente = $_POST['searchCliente'];
    $method = $_POST['request'];

    include 'Database.php';
    include 'functions.php';

    $db = new Database();
    $functions = new Functions($db, $codCliente);


    if (method_exists($functions, $method)) {
        $data = $functions->$method();
        header('Content-Type: application/json');
        echo json_encode($data);
    }
}
?>

的functions.php

<?php

/*function doLog($text)
{
    // open log file
    $filename = "log.txt";
    $fh = fopen($filename, "a") or die("Could not open log file.");
    fwrite($fh, date("d-m-Y, H:i")." - $text\n") or die("Could not write file!");
    fclose($fh);
}
doLog($codCliente);*/

class Functions
{
    private $db;
    private $codCliente;

    public function __construct(Database $db)
    {
        $this->db = $db;
    }
    public function setCodCliente($codCliente)
    {
        $this->codCliente = $codCliente;
    }


    public function get_codiceCliente_json()
    {
        $query = "SELECT * FROM clienti WHERE codiceCliente LIKE '%" . $this->codCliente . "%' LIMIT 15";
        $result = $this->db->dataQuery($query);
        return $result->fetchAll();
    }
}
?>

还有一个名为Database.php的文件只与数据库建立连接。

感谢您的帮助！

更新：

从mysql登录

2017-07-23T10:58:26.359942Z        65 Query     SELECT * FROM clienti WHERE codiceCliente LIKE '%%' LIMIT 15

Answer 1

我注意到的一些事情可能会导致您的问题......

1。由于日志中的部分SQL看起来像... LIKE '%%' ...，这意味着您的属性codCliente未在您的实例中正确设置。

修改此行代码......

$functions = new Functions($db, $codCliente);

对此...

$functions = new Functions($db);
$functions->setCodCliente($codCliente);

您的构造函数没有为codCliente属性定义参数...

2。还要检查ajax调用中的 data 属性它现在看起来像......

// ...
data: 'request='+ requestCodCliente +'&searchCliente='+ codiceCliente,
// ...

数据字段可以接受数组/对象作为值，因此如果您想确保正确发送和接收参数，请将其更改为类似的内容......

// ...
data: {
    request: requestCodCliente,
    searchCliente: codiceCliente
},
// ...

我希望这会有所帮助：）

Answer 2

我认为问题是当你调用它时，你没有在$codCliente构造函数中分配Functions参数，

$functions = new Functions($db, $codCliente);

您应该像这样更新Functions课程：

<?php

class Functions
{
    private $db;
    private $codCliente;

    public function __construct(Database $db, $codCliente)
    {
        $this->db = $db;
        $this->codCliente = $codCliente;
    }

    public function setCodCliente($codCliente)
    {
        $this->codCliente = $codCliente;
    }

    public function get_codiceCliente_json()
    {
        $query = "SELECT * FROM clienti WHERE codiceCliente LIKE '%" . $this->codCliente . "%' LIMIT 15";
        $result = $this->db->dataQuery($query);
        return $result->fetchAll();
    }
}

这一次，当您调用get_codiceCliente_json方法时，$this->codCliente属性不再为空;）

用于搜索的PHP查询SQL不在类方法中工作

2 个答案: