mySQL statement not running in PHP variable declaration - php

Question

In the following code I'm attempting to connect to my database, pull the maximum ID from my table and then generate a random number using the the rand() function. The code successfully connects me to the the database but when I try to call for the maximum ID it won't return a value.
When I try to echo the variable, it returns SELECT MAX(id) FROM 'file'.
<?php
// Connect to the database
$dbLink = new mysqli('localhost', 'username', 'password', 'database');
if(mysqli_connect_errno()) {
die("MySQL connection failed: ". mysqli_connect_error()); }
$amount = "SELECT MAX(id) FROM 'table'";
$rannmr = rand(1, $amount);
// Close the mysql connection
mysqli_close($dbLink);
?>
Any help in resolving this would be appreciated.

Answer 1

When I try to echo the variable, it returns SELECT MAX(id) FROM 'file'.
Firstly, you are using the wrong identifier for FROM 'table' being single quotes.
If table is indeed the table's name, wrap it in backticks, your question shows file.
$amount = "SELECT MAX(id) FROM `table`";
Either way, you cannot use quotes around a table name. It appears you are using file as your table name.
So if table is only an example and it is called file let's just say, you would do:
$amount = "SELECT MAX(id) FROM `file`";
or
$amount = "SELECT MAX(id) FROM file";
Then, you also need to query, using mysqli_query() which you are not doing.
$amount = mysqli_query($dbLink,"SELECT MAX(id) FROM `file`");
Or Object oriented style:
$amount = $dbLink->query("SELECT MAX(id) FROM `file`");
if($amount){
echo "Success!";
}else{
die('Error : ('. $dbLink->errno .') '. $dbLink->error);
}
See example #1 from http://php.net/manual/en/mysqli.query.php
Use or die(mysqli_error($dbLink)) to mysqli_query() which would have signaled the error.
http://php.net/manual/en/mysqli.error.php
Edit:
Try the following. You may need to modify $row[0] and rand(0,$count) as 1 depending on the column number.
$result = $dbLink->query("SELECT MAX(id) FROM mytable")
while ($row=$result->fetch_row()) { $count = $row[0]; }
$random = rand(0,$count);
echo $random;

Answer 2

use this:
$amount = "SELECT MAX(id) FROM table";

Answer 3

You forgot to execute the MySQL-query:
$amount = $dbLink->query("SELECT MAX(id) FROM table")->fetch_assoc();
$rannmr = rand(1, $amount[0]);

Answer 4

You never executed the query, you need more logic
if ($result = mysqli_query($dbLink, "SELECT MAX(id) as amount FROM `table`")) {
printf("Select returned %d rows.\n", mysqli_num_rows($result));
if ($row = mysqli_fetch_assoc($result)) {
$amount = $row['amount'];
$rannmr = rand(1, $amount);
}else{
echo 'no row found';
}
}
mysqli_close($dbLink);

Answer 5

I didn't seem to see the line of code which actually does the query:
Try this: Using the object-oriented mysqli approach
<?php
// Connect to the database
$dbLink = new mysqli('localhost', 'username', 'password', 'database');
if(mysqli_connect_errno()) {
die("MySQL connection failed: ". mysqli_connect_error()); }
$amount = "SELECT MAX(id) as max_id FROM 'table'";
// Do the actual query :
$run_query = $dbLink->mysql->query($amount);
// Retrieve the values:
$result = $run_query->fetch_array();
// Do the rand function together with the retrieved value
$rannmr = rand(1, $result['max_id']);
// Now you can echo the variable:
echo $rannmr;
// Close the mysql connection
mysqli_close($dbLink);
?>
Thanks!!