[Create form send to database]

You should use prepared statements and bind user input as parameters rather than using mysqli_real_escape_string().

Better still, as you are still learning, put your efforts into learning PDO instead of mysqli.

If $con is a PDO connection, the above code becomes

$stmt = $con->prepare("INSERT INTO clients2 (fullname, company, email, serial_no)
                       VALUES (?,?,?,?)
                       ";
$stmt->execute( [ $_POST['fullname'], $_POST['company'], $_POST['email'], $_POST['serial_no'] ]  );                

IE prepare the query putting placeholders for the user inputs then execute it passing the values in an array.

[Create form send to database]

30 minutes ago, phppup said:

I also don't know that you need $con in these lines

$fullname = mysqli_real_escape_string($con, $_POST['fullname']);

A look at the manual would confirm that the $con IS required

[HTML form select multiple -> MySQL / PHP query...]

I would recomend that you don't prepopulate the "square" and then update.

If you start with an empty table ...

CREATE TABLE `square` (
  `square_id` int(11) NOT NULL ,
  `user_id` int(11) NOT NULL,
  PRIMARY KEY (`square_id`),
  KEY `idx_square_user_id` (`user_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

... and add squares as they are selected then it becomes impossible for a user to poach someone else's square - the primary key will make it a strictly "first come first served" situation. Should cut down the bar-room brawls.

[Need help finding the right query (basic)]

If your db date fields are DATETIME or TIMESTAMP then they will not match your $message_date because of the time values.

IE "2019-06-09" <> "2019-06-09 10:30:57"

You need to extract the date portions

... WHERE DATE(message_date) = ?

Also , to make the date calculations easier

echo "<select>";                                    

$d1 = new DateTime('next sunday');
$d2 = new DateTime("last sunday of december");
$d2->modify('+1 day');
$sevendays = new DateInterval('P7D');
$dp = new DatePeriod($d1, $sevendays, $d2);
foreach ($dp as $d) {
    echo '<option value="' . $d->format('Y-m-d').'">' . $d->format('M d, Y') . '</option>';
}
                                                                                        
echo "</select>";

 

[HTML form select multiple -> MySQL / PHP query...]

20 minutes ago, Jim R said:

(It’s just a bunch of friends choosing up squares for sporting events.

In that case, doesn't your query need to to check if a square is already taken? In between you displaying available squares and a user picking his squares, someone else might have picked one or more of the same squares.

[how to setup and run a mysql-server - remotley]

9 hours ago, dil_bert said:

well - to be honest - i really do not want to do any harm - so i don’t recommend granting ALL permissions. 
For standard users, i now do recommend granting GRANT SELECT,INSERT,UPDATE,DELETE permissions.

this is what i tell everybody...

Then stop telling them.

The last thing you want is for standard users to be able to grant privileges. That is an admin function.

[HTML form select multiple -> MySQL / PHP query...]

Why the hell would you want to put the same user's names and email address into multiple records? Think "normalization"!. Database tables are not spreadsheets.

[php getting min values index]

Having got the keys with the min value you then need to get those items from the original array

$a = array(-15, 3, 5, -1, -15);
$b = array(1, 3, 5);

function lowest_int ($c){
    $result = [];
    foreach (array_keys($c, min($c)) as $k) {
            $result[$k] = $c[$k];
    }
    return $result;    
}          

echo '<pre>';
print_r(lowest_int($a));
print_r(lowest_int($b));
echo '</pre>';

output

Array
(
    [0] => -15
    [4] => -15
)
Array
(
    [0] => 1
)

An alternative solution could be to use array_filter() in your function to remove items not equal to the minimum value

function lowest_int ($c){
      $min = min($c);
      return array_filter( $c, function($v) use($min) { return $v==$min; } );
}          

 

[how to retrive userName,comment, and createdAt from users and comments table while I have used userId as a Foreign key on the comment table]

FROM A,B WHERE A.xxx = B.xxx

Much better to use explicit joins ...

FROM A INNER JOIN B ON A.xxx = B.xxx

• The joins then reflect the structure of the data.
• The WHERE clause isn't cluttered with these structure details
• Implicit syntax can only be used for INNER JOINS so no code consistency with other join types
• Optimization seems improved now but there used to be a distinct speed advantage when using explicit joins in older versions of mysql

[how to retrive userName,comment, and createdAt from users and comments table while I have used userId as a Foreign key on the comment table]

12 minutes ago, mahenda said:

#1054 - Unknown column 'c.comment' in 'field list'

If you are getting that error then the create statements you posted earlier must be incorrect. It says your "comments" table does not have a column "comment".

I created tables using your create code and ginerjm's query (although I don't agree with with his implicit join syntax) doesn't give an error.

[Problem adding user to mysql with php?]

55 minutes ago, gw1500se said:

but you are missing the 'dbname=...' parameter in the connect.

Connections are to a server - specifying a default database is optional.

[php getting min values index]

Have your function return the keys instead of printing them, then print the function results

function lowest_int ($c){
   return array_keys($c, min($c));
}

 

[Computed Column Does Not Update And Unable To Insert Row]

I have added a couple more records by manually editing the table (MySql Workbech) and, again, no issues. (Note - generated values only appear when the output is refreshed)

[Computed Column Does Not Update And Unable To Insert Row]

I have just run this sequence of SQL and no issues

CREATE TABLE `foo` (
  `foo_id` int(11) NOT NULL AUTO_INCREMENT,
  `number` int(11) DEFAULT NULL,
  PRIMARY KEY (`foo_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;

INSERT INTO foo (number) VALUES (5),(10);

SELECT * FROM test.foo;
+--------+--------+
| foo_id | number |
+--------+--------+
|      1 |      5 |
|      2 |     10 |
+--------+--------+

--
--  add generated columns
--

ALTER TABLE foo add number2 DECIMAL GENERATED ALWAYS AS (number*2) STORED, add nombre3 DECIMAL GENERATED ALWAYS AS (number*3) STORED;

--
-- check new table
--
SELECT * FROM test.foo;
+--------+--------+---------+---------+
| foo_id | number | number2 | nombre3 |
+--------+--------+---------+---------+
|      1 |      5 |      10 |      15 |
|      2 |     10 |      20 |      30 |
+--------+--------+---------+---------+

--
--  insert a couple more records
--

INSERT INTO foo (number) VALUES (15),(20);

--
--  check again
--

SELECT * FROM test.foo;
+--------+--------+---------+---------+
| foo_id | number | number2 | nombre3 |
+--------+--------+---------+---------+
|      1 |      5 |      10 |      15 |
|      2 |     10 |      20 |      30 |
|      3 |     15 |      30 |      45 |
|      4 |     20 |      40 |      60 |
+--------+--------+---------+---------+

Perhaps you need to give us more details on exactly what you are doing

[Illegal string offset 'user_id']

Try this

[Illegal string offset 'user_id']

P.S.

If you just want the user_name

echo $_SESSION['test']['user_name'];

 

[Illegal string offset 'user_id']

11 minutes ago, Barand said:

$value is not an array, specifically not an array with a key of 'user_id'

 

1 hour ago, I-AM-OBODO said:

/* echo $value //works  */

It works because it is the  correct code.

[Illegal string offset 'user_id']

The usual advice I would give for this problem would be for you to echo $value within the loop so you can see what the variable contains.

In your case you have already done this and you will have seen the values "5" and "obodo".

As it is then blatantly obvious that $value is not an array, specifically not an array with a key of 'user_id', then it's hard to see why you even need to ask the question.

[Best way to store telephone numbers in database]

1. Don't put them in there in the first place.

or

2.

$phone = "(123) 456-7890" ;
$justdigits = '';

for ($i=0, $k=strlen($phone); $i<$k; $i++) {
    $justdigits .= (ctype_digit($phone[$i])) ? $phone[$i] : ''; 
}
echo $justdigits ;     //--> 1234567890

 

When it comes to formatting the output I find  a custom formatting function of use;
 

$phone = '1234567890234'
$format = '(###) ###-#### ext ###';

echo formatIt ($format, $phone);                      // -> (123) 456-7890 ext 234


function formatIt($format,$str)
{
    $i = $j = 0;
    $res = '';
    $kf = strlen($format);
    $str = str_replace(' ','',$str);
    $ks = strlen($str);
    while ($i < $kf  && $j < $ks) {
        $res .= $format[$i]=='#' ? $str[$j++] : $format[$i];
        ++$i;
    }
    if ($j<$ks) $res .= substr($str,$j);
    return $res;
}

 

[Check if User and Email already exists]

Same goes for user login. Don't tell them if they got the username wrong or the password wrong, just tell them it's an invalid login.

[Telephone input mask]

Something like this?

<input type="text" name="phone" id="phone" value="(___) ___ - ____" >

and

        $("#phone").keydown( function(e) {
            var output = $("#phone").val()
            if (output == '') output = "(___) ___ - ____"
            if (output.indexOf('_') != -1) {
                e.preventDefault()
                if (48 <= e.keyCode && e.keyCode <= 57) {         // keyboard top row
                    output = output.replace('_', e.keyCode-48)
                }
                else if (96 <= e.keyCode && e.keyCode <= 105) {   // numeric keypad
                    output = output.replace('_', e.keyCode-96)
                }
                $("#phone").val(output)
            }
       })

 

[Validation With Jquery Not Working]

Your browser's developer tools

[is_logged_in not working on one page]

If your url is that then you should be passing $_GET['id'] to the query, not $_GET['postid']

 

EDIT - and you should get a message telling you index postID is not defined.

[is_logged_in not working on one page]

7 minutes ago, Barand said:

Did you check if $_GET['postID'] contains a valid id?

Sorry - misread the list above.

If there is nothing in $_GET['postID'] how do you expect to find the record with that id?

[is_logged_in not working on one page]

Did you check if $_GET['postID'] contains a valid id?