How to make it insert info from session

[Jedijon]

$result = mysqli_query($con,"SELECT * FROM $_SESSION['datakey']";

[Barand]

You would use an INSERT query.

A SELECT query retrieves data.

[Jedijon]

Like this $result = mysqli_query($con,"INSERT * FROM $_SESSION['datakey']";

[Barand]

No, not like that. You need the correct insert query syntax

or there is tutorial here

Edited November 21, 2018 by Barand

[Jedijon]

I don't wan't it to be inserted into the database but use the session info as the name of the table to grave data from for the html table

[Barand]

you need to put {..} around the array element inside a string

$result = mysqli_query($con,"SELECT * FROM {$_SESSION['datakey']}"; 

 

[Jedijon]

$result = mysqli_query($con,"SELECT * FROM table name"); but where the table name goes I wan't it use $_SESSION['datakey'] to get the table name

[mac_gyver]

this sounds like a bad design. 

[Jedijon]

Didn't work 

$result = mysqli_query($con,"SELECT * FROM {$_SESSION['datakey']}";

[Jedijon]

It will grave the info from the login SESSION

[Barand]

Define "didn't work" . That tells us nothing.

What does $_SESSION['datakey'] contain? Is it the name of a table?

What is your code after you get the $result?

We are not telepathic!

[Jedijon]

 $_SESSION['datakey'] contains the name of the table

[Barand]

24 minutes ago, mac_gyver said:

this sounds like a bad design. 

I couldn't agree more. Variable tables names always rings alarm bells

[Barand]

1 minute ago, Jedijon said:

$_SESSION['datakey'] contains the name of the table

OK, so that leaves

• Define "didn't work" . That tells us nothing.
• What is your code after you get the $result?

[benanamen]

I know OP, tell us what the Real problem is you are trying to solve rather than asking about the poorly implemented attempt at solving the problem. What is the overall task at hand?

Edited November 21, 2018 by benanamen

[Jedijon]

<?php
$con=mysqli_connect("IP","Username","Password","Database name");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM {$_SESSION['datakey']}"; 

echo "<table border='1'>
<tr>
<th></th>
<th></th>
<th></th>
<th></th>
<th></th>
</tr>";

while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['column1'] . "</td>";
echo "<td>" . $row['column2'] . "</td>";
echo "<td>" . $row['column3'] . "</td>";
echo "<td>" . $row['column4'] . "</td>";
echo "<td>" . $row['column5'] . "</td>";
echo "</tr>";
}
echo "</table>";

mysqli_close($con);
?>

[Barand]

Try putting session_start() right at the top of your code.

You need that to access session data

[Jedijon]

I have that 

<?php
// Initialize the session
session_start();
 
// Check if the user is logged in, if not then redirect him to login page
if(!isset($_SESSION["loggedin"]) || $_SESSION["loggedin"] !== true){
    header("location: ../login");
    exit;
}
?>

[Jedijon]

Error message 

Parse error: syntax error, unexpected ';', expecting ',' or ')' in /srv/disk1/2648096/www/site/test/index.php on line 20

[Barand]

PS Before you connect to mysql, call

mysqli_report(MYSQLI_REPORT_ERROR|MYSQLI_REPORT_STRICT);

That will force it to report errors.

[benanamen]

Notice anything wrong here?

$result = mysqli_query($con,"SELECT * FROM {$_SESSION['datakey']}"; 

 

[ginerjm]

You should write query statements into a variable to make it easier to debug should you need to, as in this case.

	$q = "SELECT * FROM {$_SESSION['datakey']}"; 
	echo "Query looks like: $q";
	 $results = mysqli_query($con, $q);
	

Run this code and look at the actual query you created to ensure that your assumptions are correct.

[Barand]

3 minutes ago, Jedijon said:

I have that 

Not in the code you showed us

[Jedijon]

Query looks like: SELECT * FROM datakey
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in /srv/disk1/2648096/www/site/test/index.php on line 35

 

It shows the session data

Edited November 21, 2018 by Jedijon

[Jedijon]

It works