angular post to MySQL error

Question

I am working on an ionic app where the user needs to post data to a MySQL database. Due to some very helpful answers on SO I have gotten around the CORS issues however I have now run into another problem. The submit.php file looks like this:

<?php
    if (isset($_SERVER['HTTP_ORIGIN'])) {
        header("Access-Control-Allow-Origin: {$_SERVER['HTTP_ORIGIN']}");
        header('Access-Control-Allow-Credentials: true');
        header('Access-Control-Max-Age: 86400');    // cache for 1 day
    }

    // Access-Control headers are received during OPTIONS requests
    if ($_SERVER['REQUEST_METHOD'] == 'OPTIONS') {

        if (isset($_SERVER['HTTP_ACCESS_CONTROL_REQUEST_METHOD']))
            header("Access-Control-Allow-Methods: GET, POST, OPTIONS");         

        if (isset($_SERVER['HTTP_ACCESS_CONTROL_REQUEST_HEADERS']))
            header("Access-Control-Allow-Headers:        {$_SERVER['HTTP_ACCESS_CONTROL_REQUEST_HEADERS']}");

        exit(0);
    }

    $data = json_decode(file_get_contents("php://input"));
    $celeb = $data->celeb;
    $camp = $data->camp;
    $spirit = $data->spirit;

    $sport = $data->sport;
    $bizs = $data->bizs;
    $entrep = $data->entrep;

    $young = $data->young;
    $conser = $data->conser;
    $saty = $data->saty;

    $name = $data->name;
    $surname = $data->surname;
    $email = $data->email;

    $contacts = $data->contacts;

    $con = mysql_connect('localhost', 'root', 'thePssWrd');
    mysql_select_db('theDB', $con);

    $qry = 'INSERT INTO test 
                    (celeb,camp,spirit,sport,bizs,
                     entrep,young,conser,saty,name,
                     surname,email,contacts) 
             values ("' . $celeb . '","' . $camp . '",' .$spirit . 
                     ','.$sport. ','.$bizs. ','.$entrep. ','.$young. 
                     ','.$conser. ','.$saty. ','.$name. ','.$surname. 
                     ','.$email. ','.$contacts. ')';
    $qry_res = mysql_query($qry);

    if ($qry_res) {

        if ($qry_res) {
            $arr = array('msg' => "Submitted Successfully!!!", 'error' => '');
            $jsn = json_encode($arr);
            print_r($jsn);
        } else {
            $arr = array('msg' => "", 'error' => 'Error In Submit');
            $jsn = json_encode($arr);
            print_r($jsn);
        }
    } else {
        $arr = array('msg' => "", 'error' => 'This is the big error thing...');
        $jsn = json_encode($arr);
        print_r($jsn);
    }
?>

and my controller:

.controller('FrmController', function ($scope , $http) {
    $scope.errors = [];
    $scope.msgs = [];

    $scope.vote = function() {

        $scope.errors.splice(0, $scope.errors.length); 
        $scope.msgs.splice(0, $scope.msgs.length);

        $http.post('http://www.ann7.com/saty/submit.php', {

            'celeb'     : $scope.celeb, 
            'camp'      : $scope.camp, 
            'spirit'    : $scope.spirit,
            'sport'     : $scope.sport,
            'bizs'      : $scope.bizs, 
            'entrep'    : $scope.entrep, 
            'young'     : $scope.young,
            'conser'    : $scope.conser,
            'saty'      : $scope.saty, 
            'name'      : $scope.name, 
            'surname'   : $scope.surname,
            'email'     : $scope.email,
            'contacts'  : $scope.contacts

        }
        ).success(function(data, status, headers, config) {
            if (data.msg != '')
            {

              console.log(data.msg);
                $scope.msgs.push(data.msg);
            }
            else
            {
              console.log(data.error);
                $scope.errors.push(data.error);
            }
        }).error(function(data, status) { 
            $scope.errors.push(status);
        });
    }
});

If you look at the PHP it is outputting error messages and I am receiving the "This is the big error thing..." error message in my console which I am taking as a successful connection to the DB but an error in inserting the values... my html file looks like as an example:

<form  name="nominationForm" ng-controller="FrmController" class="falecomigo novalidate form-manager" >
    <div ng-controller="VoteCtrl" class="list list-inset" >
    <div class="styled-select">
           <select ng-model="bizs">
              <option value="e">option1</option>
              <option value="1">option1</option>
              <option value="2">option1</option>
              <option value="3">option1</option>
              <option value="4">option1</option>
              <option value="5">option1</option>
           </select>
        </div>
        <label class="item item-input">
            <input ng-model="name" name="fieldName" type="text" placeholder="Name" required ng-minlength="2" ng-maxlength="70">
        </label>

        <label class="item item-input">
            <input ng-model="surname" name="fieldSurname" type="text" placeholder="Surname" required ng-minlength="2" ng-maxlength="70">
        </label>
        <label class="item item-input">
            <input ng-model="email" name="fieldEmail" type="email" placeholder="E-mail" required ng-maxlength="50">
        </label>

        <label class="item item-input">
            <input ng-model="contacts" name="fieldNumber" type="text" placeholder="Contact No." required ng-minlength="2" ng-maxlength="70">
        </label>
    </div>

    <div class="padding container">
        <div class="row">
            <button ng-click="vote()" class="button button-balanced  button-small col col-100"> Vote </button>
        </div>
    </div>
</form>

I am not sure where I am going wrong...

Could you post the output of mysqli_error() ?

Walid Ammar
– Walid Ammar

2015-08-19 08:32:40 +00:00
Commented Aug 19, 2015 at 8:32 — Walid Ammar
– Walid Ammar, Commented Aug 19, 2015 at 8:32
@MohammadWalid the error it is outputting undefined

letterman549
– letterman549

2015-08-19 08:45:45 +00:00
Commented Aug 19, 2015 at 8:45 — letterman549
– letterman549, Commented Aug 19, 2015 at 8:45

Community · Accepted Answer · 2017-05-23 12:06:17Z

2

Instead of sending yourself a fairly meaningless error message do this instead ALWAYS

} else {
    $arr = array('msg' => "", 'error' => mysql_error());
    $jsn = json_encode($arr);
    //print_r($jsn);
    echo $jsn;
}

This error will actually tell you what is wrong.

Of course in a live system you would not want to do this as the user would see error messages that are far to likely to help a hacker. So you would do something like this

} else {
    // log error for admins to check
    error_log( mysql_error(),3, '/some/path/to/error.log');
    $arr = array('msg' => "", 'error' => 'An error occured, see the log');

    $jsn = json_encode($arr);
    //print_r($jsn);
    echo $jsn;
}

ADDITIONAL CHECKS:

I now see you are not doing error checking on any mysql_ function calls, maybe the error is in the connection process, so add

$con = mysql_connect('localhost', 'root', 'thePssWrd');
if (!$con) {
    // log error for admins to check
    //error_log( mysql_error(),3, '/some/path/to/error.log');
    die('Could not connect: ' . mysql_error());
}


if (! mysql_select_db('theDB', $con) ) {
   die ('Can\'t select the database: ' . mysql_error());
}

Ahhhh

It look like you are not actually sending the json encoded data back to the browser as a simple string. Instead of print_r($jsn) do a simple echo $jsn; and that applies to all retuning of json data strings. At this point $json is not actually an array anyway, its a simple string variable.

You could also simplify the coding of that query

Remember that double quotes allow you to expand $variables so it might be this solves the underlying issue by just making the coding ot that query a bit more straight forward.

$qry = "INSERT INTO test 
                (celeb,camp,spirit,sport,bizs,
                 entrep,young,conser,saty,name,
                 surname,email,contacts) 
         values ('$celeb','$camp','$spirit','$sport','$bizs',
                 '$entrep','$young','$conser','$saty','$name',
                 '$surname','$email','$contacts')";

$surname was missing

I feel I should add this, as if I dont someone else will. Please if you are just learning PHP and its database access mechanisms, dont waste time learning the mysql_ extension. It is deprecated and is no longer available in the new PHP7 soon to be released. Look at this and learn either the mysqli_ extension or PDO. Have a read on this to help you pick one.

edited May 23, 2017 at 12:06

CommunityBot

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answered Aug 19, 2015 at 8:35

RiggsFolly

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6 Comments

letterman549 Over a year ago

Fair enough. console output is undefined

RiggsFolly Over a year ago

See additional checks, need to check almost all mysql_ functions return values

letterman549 Over a year ago

I have added the additional checks but I am still getting undefined in my console log. thanx for all the help so far.

letterman549 Over a year ago

haha I saw it the moment it was up. thanx! the error I am receiving is You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ',,,,,,,,,)' at line 1

RiggsFolly Over a year ago

Ok check the simplify query suggestion and correction

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