PHP Adding values to an array using while(mysql_fetch_array)?

Question

I'm trying to add values to an array after getting data from a mysql query, this obviously involved a while ($x = mysql_fetch_array($MysqlQuery)) {} as seen below:

      $CheckTime = mysql_query("SELECT * FROM cp11641_timetable.booking");
      $dates = array();
      while ($date = mysql_fetch_array($CheckTime)) {
        $DateInt = strtotime($date['Date']);
        //echo $DateInt . " ";
        $dates[] = $DateInt;
        echo $dates[1] . " ";
      }

However when I echo $dates[x], it'll display the value in the x position of the array, but it'll show it by (x+1) times (i.e. $dates[0] will show 'a' once, $dates[1] will show 'b' twice, and $dates[2] will show 'c' thrice)

How do I fix this? What's causing the problem?

Answer 1

This works, make sure to put in your correct credentials to connecting to your database on step #1. everything else false in place.

<?php 

    /* ==============================================
    This is the new way of connecting to database 
    using mysqli
    ================================================*/

    // Step #1 create credentiasl for database connection
    $host = ""; //type your host ex. localhost between the quotes
    $user = ""; //your username between the quotes
    $pass = ""; //your password between the quotes
    $db   = ""; //your database you are connecting to between the quotes

    // step #2 create connection to database
    $conn = new mysqli($host, $user, $pass, $db);

    //step #3 check and see if connection is working and error free
    if ($conn->error) {

        die("Could not connect to the database");

    } else{

        // create array dates
        $dates = array();

        // create select statement
        $CheckTime = ("SELECT * FROM cp11641_timetable.booking");

        // query the the database using the connection
        $sql_CheckTime = $conn->query($CheckTime);

        // if rows available in table add them to array dates
        while ($row = mysqli_fetch_assoc($sql_CheckTime)) {

            $dates[] = $row;

        }

        //optional uncomment bottom line to check if dates array has data will display as array on webpage
        // var_dump($dates); 

        // loop through array 
        foreach ($dates as $date){

            // echo out data you want to display. 'Date' = column name
            echo strtotime($date['Date']) . "<br>";
        };
    };

 ?>

Answer 2

$CheckTime = mysqli_query($mysql_connection, "SELECT * FROM cp11641_timetable.booking");
$dates = array();
while ($date = mysqli_fetch_assoc()($CheckTime)){ // Use mysqli_* for queries.
    $DateInt = strtotime($date['Date']); // This will show an UNIX timestamp
    $dates[] = $DateInt; // Fills the array with the timestamp.
}

Your problem is that you use mysql_fetch_array. But then try to use $date['Date']. If you want to use the column names as indices in the $date array. You need to use mysql_fetch_assoc().

On a different note and as mentioned in the comments use the mysqli_* extension or PDO. In this answer I've used mysqli_*

Please note the $mysql_connection in the mysqli_query function. MySqli Query Doc

Most likely if you use the code below it should work as intended. Still strongly advise to switch to mysqli_*

$CheckTime = mysql_query("SELECT Date FROM cp11641_timetable.booking");
$dates = array();
while ($date = mysql_fetch_array($CheckTime)){
    $DateInt = strtotime($date[0]);
    $dates[] = $DateInt;
}

foreach($dates as $timestamp){
    echo $timestamp . '<br/>';
}

Answer 3

I recommend you to using mysql_fetch_assoc() and then display the dates horizontally or vertically with html/css style. Sorry if my answer is bad choose.