MYSQL not receiving data from PHP

Question

I am trying to send data from input to a my sqldata base. Here is the coding for it trying to send the information to the database. It doesn't appear in the database; what is wrong with the coding?

    <?php
$con=mysqli_connect(/*hostname*/"localhost",
                    /*username*/"user",
                    /*password*/"pass",
                    /*database name*/"dbase");

// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$a =  mysqli_real_escape_string($con, $_POST['a']);
$b =  mysqli_real_escape_string($con, $_POST['b']);
$c =  mysqli_real_escape_string($con, $_POST['b']);
$d =  mysqli_real_escape_string($con, $_POST['d']);
$e =  mysqli_real_escape_string($con, $_POST['e']);
$f =  mysqli_real_escape_string($con, $_POST['f']);
$g = 10 + ($e - $f);

mysql_query("INSERT INTO 'mensscore', (Name, Club, Level, App, Score);
    VALUES ('".$a."',
            '".$b."',
            '".$c."',
            '".$d."'
            '".$g."')");
    ?>

Remove ',' after table name. It should be `INSERT INTO mensscore (Name, Club.. ` instead of `INSERT INTO 'mensscore', (Name, Club.. `. And also use `mysqli_query` instead of `mysql_query` — Lepanto, Apr 28 '14 at 09:10

score 1 · Accepted Answer · edited May 23 '17 at 11:57

1

You are mixing mysql and mysqli. Change:

mysql_query("INSERT INTO 'mensscore', (Name, Club, Level, App, Score);
    VALUES ('".$a."',
            '".$b."',
            '".$c."',
            '".$d."'
            '".$g."')");

To:

mysqli_query($con, "INSERT INTO `mensscore` (Name, Club, Level, App, Score)
    VALUES ('$a',
            '$b',
            '$c',
            '$d',
            '$g'
    );");

Also see How can I prevent SQL injection in PHP and PHP: mysqli_stmt - Manual because using POST is not always safe as people can do SQL injection.

edited May 23 '17 at 11:57

Community

1
1

answered Apr 28 '14 at 09:09

I hate Stack Overflow

11
9

Warning: mysqli_query() expects at least 2 parameters, 1 given in C:\xampp\htdocs\judge\MG\score.php on line 47 – The Okay Man Apr 28 '14 at 09:14
@TheOkayMan But it has two parameters. Try again because edited – I hate Stack Overflow Apr 28 '14 at 09:16
OK now please select as working answer by clicking on the tick. And also please upvote to set as worked answer. @TheOkayMan – I hate Stack Overflow Apr 28 '14 at 09:22

score 0 · Answer 2 · answered Apr 28 '14 at 09:09

0

You have some errors in the query. Firstly use mysqli_query() as here, you are using mysqli. Remove the unwanted , and ; from the query and try with this,

mysqli_query($con,"INSERT INTO 'mensscore' (Name, Club, Level, App, Score)
    VALUES ('$a',
            '$b',
            '$c',
            '$d'
            '$g')");

answered Apr 28 '14 at 09:09

Jenz

8,280
7
44
77

include `$con` as parameter in the mysqli_query – gbestard Apr 28 '14 at 09:11

AVM · Answer 3 · 2014-04-28T10:01:06.933

0

Try this

mysqli_query($con,"INSERT INTO mensscore (Name, Club, Level, App, Score)VALUES ('".$a."',
        '".$b."',
        '".$c."',
        '".$d."',
        '".$g."')");

edited Apr 28 '14 at 10:01

answered Apr 28 '14 at 09:09

AVM

592
2
11
25

score 0 · Answer 4 · answered Apr 28 '14 at 09:10

0

Try this

mysqli_query("INSERT INTO mensscore  (Name, Club, Level, App, Score)
    VALUES ('{$a}','{$b}','{$c}','{$d}','{$g}'");

answered Apr 28 '14 at 09:10

sshet

1,152
1
6
15

score 0 · Answer 5 · answered Apr 28 '14 at 09:12

try like this

<?php
    $con=mysqli_connect("localhost","user","pass","dbase");

    // Check connection
    if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $a =  mysqli_real_escape_string($con, $_POST['a']);
    $b =  mysqli_real_escape_string($con, $_POST['b']);
    $c =  mysqli_real_escape_string($con, $_POST['b']);
    $d =  mysqli_real_escape_string($con, $_POST['d']);
    $e =  mysqli_real_escape_string($con, $_POST['e']);
    $f =  mysqli_real_escape_string($con, $_POST['f']);
    $g = 10 + ($e - $f);
    $sql = "INSERT INTO mensscore(Name, Club, Level, App, Score)VALUES('".$a."','".$b."','".$c."','".$d."''".$g."')";
    $result = $con->query($sql) or die("Error executing query" . mysqli_error($conn));
        ?>

score 0 · Answer 6 · answered Apr 28 '14 at 09:12

0

don't single quote the table name mensscore
don't use semicolon before VALUES keyword

answered Apr 28 '14 at 09:12

Xing Fei

287
1
6

MYSQL not receiving data from PHP

6 Answers6

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