Issue trying to output records stored in MySql

Question

I have a simple page which basically gets unique input text from user and queries the database to see if that any contact record belonging to that input. So far I know connection is established and it is trying to query the database. However, no record is retrieved. This is what the code looks like:

<form action="processAttendance.php" method="POST">
  <div class="input">
    <input type="text" class="button" id="inviteCode" name="inviteCode" placeholder="Enter invite code i.e. E2562SA">
    <input type="submit" class="button" id="submit" value="Register"> &nbsp;
  </div>
</form>

<?php  
include 'dbh.php';
$inviteCode = $_POST['inviteCode'];
echo $inviteCode;
$sql = "SELECT `FirstName`, `LastName`, `Confirmed`, `InviteCode`, `PlusOnes` FROM `Guest` WHERE 'InviteCode' = '$inviteCode'";
$result = $conn->query($sql);
$row = mysqli_fetch_assoc($result);
echo "my result" . $row."<br>";
echo "my result" . $row['classtype'];

The result printed back is as follows:

TEST

my resultArray

my result

Could anyone offer some insight as to why this could be the case?

**Danger**: You are **vulnerable to [SQL injection attacks](http://bobby-tables.com/)** that you need to [defend](http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php) yourself from. — Quentin, May 18 '17 at 13:54
Apparently you do have results, but they are in an array format due to `mysqli_fetch_assoc `, thus getting _"my resultArray"_ , you would need to iterate over your results to echo each row's content. — CodeGodie, May 18 '17 at 14:01

CodeGodie · Accepted Answer · 2017-05-18T14:20:06.680

Youre not fetching your rows correctly. Youre using mysqli OOP, so you should use ->fetch_assoc() and you need to iterate over your results to get your rows. Something like this:

$sql = "SELECT `FirstName`, `LastName`, `Confirmed`, `InviteCode`, `PlusOnes` FROM `Guest` WHERE 'InviteCode' = '$inviteCode'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {

        echo "<pre>";
        print_r($row);
        echo "</pre>";

        echo "my FirstName: " . $row['FirstName'];
    }
} else {
    echo "No results";
}
$conn->close();

BTW, you should use prepared statements to prevent sql injections.

score 0 · Answer 2 · answered May 18 '17 at 13:54

However, no record is retrieved

Incorrect. If you didn't get a record that you wouldn't get:

my resultArray

mysqli_fetch_assoc returns NULL if there are no results. You are getting an array.

$row['classtype'] is blank because you didn't include classtype in the list of fields you were selecting.

score 0 · Answer 3 · answered May 18 '17 at 13:54

Your sql statement is this:

SELECT FirstName, LastName, Confirmed, InviteCode, PlusOnes FROM Guest WHERE 'InviteCode' = '$inviteCode'

You are attempting to output $row['classtype'], but look at the SELECT part of your SQL, that column was not selected.

Use classtype to your sql statement to make it available, assuming that is a field in your table.

SELECT `classtype` FROM `Guest` WHERE 'InviteCode' = '$inviteCode'

After getting $row, you can run the statement var_dump($row) to output the entire object.

score -1 · Answer 4 · answered May 18 '17 at 13:50

-1

Here is a method I use:

for($row as $data) {

echo $data['classtype'];

}

Try this. I am assuming you will be dealing with the rows of all members you have there. It will go through all the rows with whatever info you want.

answered May 18 '17 at 13:50

abssse

1

That won't help. They are using `mysqli_fetch_assoc` not `mysqli_fetch_all` – Quentin May 18 '17 at 13:56

Issue trying to output records stored in MySql

4 Answers4