EXTJS 5和PHP使用准备好的语句创建

我正在使用extjs网格中的PHP进行CRUD。以下代码可以正常工作：

include("conect.php");

$contacts= $_POST['contacts'];

$data = json_decode($contacts);
$nome = $data->name;
$email = $data->email;
$phone = $data->phone;
$state = $data->state;

$sqlQuery = "INSERT INTO contact (name, email, phone,state) values ('%s', '%s', '%s', '%s')";
$sqlRes = $conn->query($sqlQuery);

echo json_encode(array(
    "success" => mysql_errno() == 0,

    "contatos" => array(
        "id" => mysql_insert_id(),
        "name" => $name,
        "email" => $email,
        "phone" => $phone,
        "state" => $state
    )
));

但是，我打算使用准备好的语句。

我进行了下一次尝试，但没有成功：

include("conect.php");

$statement = $conn->stmt_init();

$contacts = $_POST['contacts'];

$data = json_decode($contacts);

$name = $data->name;
$email = $data->email;
$phone = $data->phone;
$state = $data->state;


$sqlQuery = "INSERT INTO contact (name, email, phone, state, id) values (?, ?, ?, ?, ?)";

$statement = $conn->prepare($sqlQuery);

$statement->bind_param("ssssi", $name_val, $email_val, $phone_val, $state, $id_val);

$statement->execute();

$statement->bind_result($col1, $col2, $col3, $col4, $col5);

while($statement->fetch()){
    $output[] = array($col1, $col2, $col3, $col4, $col5);
};


$arr = array(
    'success' => mysql_errno() == 0,

    "contact" => array(
        "id" => mysql_insert_id(),
        "name" => $name,
        "email" => $email,
        "phone" => $phone,
        "state" => $state
    )
);


$statement->close();

echo json_encode($output, $arr);

$conn->close();

我究竟做错了什么？

extjs存储：

Ext.define('APP.store.StoreAPP', {
extend: 'Ext.data.Store',

model: 'APP.model.ModelAPP',

pageSize: 25,
autoLoad:true,
autoLoad: {start: 0, limit: 25},
autoSync: false, 

proxy: {
    type: 'ajax',
    api: {
        create: 'php/createContact.php',
        read: 'php/Contact.php',
        update: 'php/updateContact.php',
        destroy: 'php/deleteContact.php',
    },
    reader: {
        type: 'json',
//      root: 'contacts', //Extjs 4
        rootProperty: 'contacts', //Extjs 5
        successProperty: 'success'
    },
    writer: {
        type: 'json',
        writeAllFields: true,
        encode: true,
  //    root: 'contacts', //Extjs 4
        rootProperty: 'contacts' //Extjs 5
    }
}

});

提前致谢。

编辑

感谢回复。

我已经阅读了有关mysqli和准备好的语句的所有内容（互联网，书籍）。到目前为止，我找不到答案。

我认为问题出在json_decode和json_encode中。

以下与您的逻辑相同的代码效果很好：

include("conn.php");

$sqlQuery = "SELECT phone FROM contact WHERE name = ? AND email = ? ";

$statement = $conn->prepare($sqlQuery);

$name = "John";
$email = "[email protected]";

$statement->bind_param("s",$name, $email);

$statement->execute();

$statement->bind_result($col1, $col2);


while($statement->fetch()){
    $output[] = array($col1, $col2);
};

echo json_encode($output);

$statement->close();
$conexao->close();

我认为问题出在json_decode和json_encode中。

就像您说的那样，“联系人”是一个数组，它的extjs存储了rootProperty配置（=数据库名称）。它以文本字段形式（名称，电子邮件，电话，州）提供的数据。

知道有什么问题吗？

编辑

下一个代码有效，但不是我想要的。它似乎是mysql在json中的混合体，并使用mysqli准备好的语句进行编码。

mysqli_errno（）在json_encode中不起作用；只是mysql_errno（）。

如果您还有其他想法，我将不胜感激。

谢谢。

include("conect.php");

$contacts= $_POST['contacts'];

$data = json_decode($contacts);
$name = $data->name;
$email = $data->email;
$phone = $data->phone;
$state = $data->state;

$sqlQuery = "INSERT INTO contact (name, email, phone, state) values (?, ?, ?, ?)";
$statement= $conn->prepare($sqlQuery);
$statement -> bind_param ("ssss", $name_val, $email_val, $phone_val, $state_val);
$statement->execute();

echo json_encode(array(
    "success" => mysql_errno() == 0,

    "contatos" => array(
        "name" => $name,
        "email" => $email,
        "phone" => $phone,
        "state" => $state
    )
));