Sql result object to string in php

Question

Im creating a forum where user can create a poll by giving the question and options. Then I send these to a database using Ajax and PHP and create a table with the name PollTable0. And as new ones get added the number increases s.a. "PollTable1", "PollTable2" etc. I get these numbers by counting the number of tables in the database. My code looks like this:

<?php
if (isset($_POST['NewPoll'])) {
    $PollArray = $_POST['NewPoll'];
}
$con = mysqli_connect("localhost", "root", "123", "polldatabase") or die("DIE");

$getnumber = "SELECT COUNT(*) FROM information_schema.TABLES
            WHERE TABLE_SCHEMA='polldatabase'";
$number = mysqli_query($con, $getnumber);

$sql = "CREATE TABLE PollTable" . $number . " (" . $PollArray[0] . "VarChar(255))";
mysqli_query($con, $sql);

for ($i = 1; $i < count($PollArray); $i++) {
$target =   "ALTER TABLE PollTable" . $number . " ADD " . $PollArray[i] . " VARCHAR(255)";
mysqli_query($con, $target);
}
echo "Thank you for submitting your poll";
mysqli_close($con);
?>

Now the problem is $number is an I believe sql result object. Therefore when adding up the string with $number it gives the error:

Object of class mysqli_result could not be converted to string

If I am right, is there a way of converting result objects to strings. I have tried strval() and some other methods, which didnt work. If I am wrong, what do you think is the reason I am getting this error? Thanks in advance.

Marcin Nabiałek · Accepted Answer · 2014-07-01 07:28:04Z

1

You should add after line:

$number = mysqli_query($con, $getnumber);

the following lines:

$number = mysqli_fetch_row($number);
$number = $number[0];

mysqli_query only query database but if you want to fetch results you need to use one of the functions: mysli_fetch_row, mysqli_fetch_array, mysqli_fetch_assoc

answered Jul 1, 2014 at 7:28

Marcin Nabiałek

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Comments

Abhik Chakraborty · Accepted Answer · 2014-07-01 07:33:06Z

1

The issue is here

$number = mysqli_query($con, $getnumber);

Mysqli::query will return

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

And you are using $number in the next line and hence its issuing the error.

You will need to fetch the result before using it

http://www.php.net/manual/en/mysqli-result.fetch-assoc.php

Something as

$getnumber = "SELECT COUNT(*) as total FROM information_schema.TABLES
            WHERE TABLE_SCHEMA='polldatabase'";
$result = mysqli_query($con, $getnumber);
$data = mysqli_fetch_assoc($result);
$number = $data['total'];

answered Jul 1, 2014 at 7:33

Abhik Chakraborty

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Comments

Lewis · Accepted Answer · 2014-07-01 07:34:47Z

0

You need to change

$getnumber = "SELECT COUNT(*) FROM information_schema.TABLES
            WHERE TABLE_SCHEMA='polldatabase'";
$number = mysqli_query($con, $getnumber);

to

$getnumber = "SELECT COUNT(*) as number FROM information_schema.TABLES
            WHERE TABLE_SCHEMA='polldatabase'";
$query = mysqli_query($con, $getnumber);
$result = mysqli_fetch_object($query);
$number = $result->number;

edited Jul 1, 2014 at 7:34

answered Jul 1, 2014 at 7:28

Lewis

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4 Comments

Marcin Nabiałek Over a year ago

But you also need to fetch result. Don't you?

let_the_coding_begin Over a year ago

@MarcinNabiałek what do you mean?

let_the_coding_begin Over a year ago

Its not giving that error anymore but apparently I still have mistakes... Thank you though

let_the_coding_begin Over a year ago

Undefined property: mysqli_result::$number This is on the line I added: $number = $result->number;

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