无法为Angularjs,PHP MySQL获取数据和插入数据

时间:2016-09-20 15:19:46

标签: php angularjs

有两个表,我可以从一个表(客户)获取数据。但是我无法使用此代码从第二个表中获取数据。

简而言之,我们的系统从send.html获取所有数据并使用下表中的代码插入数据库。我希望从Login表中获取id数据,之后我将插入此id和其他客户数据。

$app->post('/send', function() use ($app) {
    $response = array();
    $qwr = json_decode($app->request->getBody());
    verifyRequiredParams(array('email', 'name', 'password'),$r->customer);
    require_once 'passwordHash.php';
    $db = new DbHandler();
    $phone = $qwr->customer->phone;
    $name = $qwr->customer->name;
    $email = $qwr->customer->email;
    $address = $qwr->customer->address;
    $password = $qwr->customer->password;
    $isUserExists = $db->getOneRecord("select * from **login** where phone='$phone' or email='$email'");
    if(!$isUserExists){
        $qwr->customer->password = passwordHash::hash($password);
        $tabble_name = "**customers**";
        $column_names = array('uid', 'phone', 'name', 'email', 'password', 'city', 'address', 'rname', 'rmail', 'rbcode', 'tfor', 'cnumber', 'cname', 'edate', 'cwc', 'badress', 'hname', 'sstreet', 'scity', 'scountry', 'scode', 'toValue', 'amount');
        $result = $db->insertIntoTable($r->customer, $column_names, $tabble_name);
        if ($result != NULL) {
            $response["status"] = "success";
            $response["message"] = "successfully";
            $response["uid"] = $result;
            if (!isset($_SESSION)) {
                session_start();
            }
            $_SESSION['uid'] = $response["uid"];
            $_SESSION['phone'] = $phone;
            $_SESSION['name'] = $name;
            $_SESSION['email'] = $email;
            $_SESSION['rname'] = $rname;
            echoResponse(200, $response);

你的建议是什么?

0 个答案:

没有答案