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?

@bono $dates[0] should output 1424995200, $dates[1] should output 1424822400, and $dates[2] should output 1424908800 — Yash
– Yash, Commented Feb 23, 2015 at 13:45
Just don't do that. mysql is deprecated as you can see in the red box. You'll have to switch to PDO or mysqli. — Elias Van Ootegem
– Elias Van Ootegem, Commented Feb 23, 2015 at 14:02
After the while loop, add `echo '<pre>' . print_r($dates);echo '</pre>'; and add the results to this post. — TMH
– TMH, Commented Feb 23, 2015 at 14:23
PHP is removing mysql extension at the end of the year start learning mysqli or PDO — Junius L
– Junius L, Commented Feb 23, 2015 at 14:29

Carlos Baston · Accepted Answer · 2015-02-24 17:46:26Z

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>";
        };
    };

 ?>

Rimble · Accepted Answer · 2015-02-23 15:21:34Z

1

$CheckTime = mysqli_query($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/>';
}

edited Feb 23, 2015 at 15:21

answered Feb 23, 2015 at 15:16

Rimble

89310 silver badges22 bronze badges

1 Comment

Rimble Over a year ago

@Yash was this answer useful to you? If so please mark it as the answer. If not tell us how to further assist in solving your problem.

Angga Lisdiyanto · Accepted Answer · 2015-02-23 14:19:47Z

0

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.

answered Feb 23, 2015 at 14:19

Angga Lisdiyanto

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3 Comments

tadman Over a year ago

You don't need to apologize. If you think you have a good answer, post it. If you change your mind about that, delete it.

Angga Lisdiyanto Over a year ago

I just feeling if the stactoverflow now is changed into 'mad' people if we post a wrong answer or question.

tadman Over a year ago

Most people are just trying to keep the answers section tidy by removing answers that are confusing, misleading or incomplete. Unfortunately some do not have a whole lot of diplomacy and come across as quite abrasive.

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