php - if else statement problem in phpmysql

650

Could you help me with my script.

I have made an phpmysql script in which username and email is taken from a form and run it through the database. If the username and email exists in the database it will show 'already exist' other wise it will add to the database.

This is my script.The script is working (not showing any errors) but in the if else statement there is some problem (I think a syntax problem), it only showsalready member exist even if it is not in database. I am notable to identify the problem. I am a newbe in phpmysql

if(isset($_POST['submit'])) {
$name=mysql_real_escape_string($_GET['name']);
$tes=mysql_real_escape_string($tes);
print $name;
print $tes;

$sel=mysql_query("SELECT * FROM member WHERE email='$tes' AND usrename='$name'");
if(!isset($sel)) {
 echo('There is already a member');
}
else
{

//insert

 $sql = "INSERT INTO member SET username='$name',email='$tes', date=CURDATE()";
 if (@mysql_query($sql)) {
 echo('<p>You have been added to database.</p>');
 } 
 }
 }

564

Answer

Solution:

instead of

if(!isset($sel))

you can to use

if(mysql_num_rows($sel) > 0)

check mysql_num_rows

716

Answer

Solution:

On your mysql_query line there is a spelling error: AND usrename should be AND username

592

Answer

Solution:

The following portion of code is not doing what you think it is :

$sel=mysql_query("SELECT * FROM member WHERE email='$tes' AND usrename='$name'");
if(!isset($sel)) {
    echo('There is already a member');
}

, as long as the query doesn't fail (because of an error in the SQL), will return a resource allowing you to access the resultset.

And even if the query fails, it'll return false -- with means the variable will still be set.

What you probably want is to try fetching some data from the $set resource, to see if your expected (or not) data is there -- see for example .

And if you only want to check if a user exists, no need for aselect * query -- just use aselect count(*), to see how many users correspond to your criteria, without loading their data.

335

Answer

Solution:

you need to add some code..

<?php    
$sel=mysql_query("SELECT * FROM member WHERE email='".$tes."' AND usrename="'.$name.'");

$data = mysql_fetch_array($sel)

if($data[0]['email']!=''){

    echo 'not in database..';

}

else{

    echo 'in database....';
}
?>

942

Answer

Solution:

try this solution

        <?php
    error_reporting(E_ALL);
    ini_set('display_errors', 1);
    $mysqli = new mysqli("localhost", "root", "", "articles");

    /* check connection */
    if ($mysqli->connect_errno) {
        printf("Connect failed: %s\n", $mysqli->connect_error);
        exit();
    }

    /* Select queries return a resultset */
    $con = 1;
    $sql = sprintf("SELECT * FROM users WHERE user_name='%s'", mysql_real_escape_string($user));    

    if ($result = $mysqli->query($sql)) {
        printf("Select returned %d rows.\n", $result->num_rows);
        if($result->num_rows == 0) {
            //TODO :: insert new record
        }
        else {
            //TODO:: USER already exists
        }

        /* free result set */
        $result->close();
    }


    $mysqli->close();
    ?>

380

Answer

Solution:

login in core php

<?php
$usename = $_POST['username'];
$pass = $_POST['password'];

$rowcount = mysqli_num_rows($result);
if($rowcount > 0)
{
  echo "Usename already exits";
}
else
{
  $sql = mysql_query('INSERT INTO `table_name`');
  echo (!empty($sql)) ? "User added." : "Error occured.";
}
?>

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