CODEWITHSUNDEEP
phpdatabase
Sp0T
Sp0T

Reputation: 284

Database connectivity error

In the below code when I pass $id_num to check the id field in database it accepts but when I want to pass user id to check with database it shows the following error;

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in line no 12

Can anyone tell me where I'm going wrong.

code:

if(isset($_POST['user_mail']) && isset($_POST['user_pass']))
{
    $var_1=$_POST["user_mail"];
    $var_2=$_POST["user_pass"];
    $result = mysqli_query($con,"SELECT * FROM jsrao_db2 WHERE user_mail=$var_1");
    while($row = mysqli_fetch_array($result))
    {
            if(($row['user_mail']==$var_1) && ($row['user_pass']==$var_2))//compare user name and password with database value
            echo "Welcome";
            else
            echo "Try Again";
    }

Upvotes: 0

Views: 61

Answers (2)

Rogue
Rogue

Reputation: 11483

Use Prepared Statements for cleaning up your code:

$result = false;
$stmt = $con->prepare("SELECT * FROM jsrao_db2 WHERE user_mail=?");
$stmt->bind_result($result);
$result = $stmt->bind_param("s", $var_1)->execute();
if ($result) {
    //work with $result
}

Upvotes: 1

Harish Singh
Harish Singh

Reputation: 3329

change your query

$result = mysqli_query($con,"SELECT * FROM jsrao_db2 WHERE user_mail=$var_1");<br>

should be

$result = mysqli_query($con,"SELECT * FROM jsrao_db2 WHERE user_mail='$var_1'");<br>

user_mail is an string so enclose $var_1 in '$var_1'

Upvotes: 2

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