PHP load emails from database to array

Question

Basically what I'm trying to do is load all the emails that are in a database and insert them into an array. But since printf outputs in length of the string I have no clue how to do it. I know this is probably something very basic but, yeah. I'm kindof new to php. Any idea's?

<?php

    $emails = array();

    mysql_connect("127.0.0.1", "root") or
        die("Could not connect: " . mysql_error());
    mysql_select_db("users");

    $result = mysql_query("SELECT email FROM users");

    while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
        $email = printf("%s", $row[0], $row[1]);
    }

    mysql_free_result($result);
    ?>

Every time you use [the `mysql_`](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) database extension in new code **[a Kitten is strangled somewhere in the world](http://2.bp.blogspot.com/-zCT6jizimfI/UjJ5UTb_BeI/AAAAAAAACgg/AS6XCd6aNdg/s1600/luna_getting_strangled.jpg)** it is deprecated and has been for years and is gone for ever in PHP7. If you are just learning PHP, spend your energies learning the `PDO` or `mysqli` database extensions and prepared statements. [Start here](http://php.net/manual/en/book.pdo.php) — RiggsFolly, Feb 01 '17 at 17:20
You're doing `SELECT email ...`, then `... $row[0], $row[1]);`. Can you spot the error? You're selecting one column but fetching data of two columns. — Rajdeep Paul, Feb 01 '17 at 17:21
The `mysql_connect()` is missing a parameter so the rest of the code is irrelevant. If you are learning PHP start with PDO dont waste time on the `mysql_` extension it is dead — RiggsFolly, Feb 01 '17 at 17:21

eetschel · Accepted Answer · 2017-02-01T21:01:58.073

Here's a way I would do it in connection with JS. May be useful to you. At first, connect to your database and get the table:

<?php
    $currConnection = new mysqli('127.0.0.1', 'root', 'users') or
        die('Connection failed: ' . $currConnection->connect_error);

    $selection = "SELECT email FORM users";

    $currSel = $currConnection->query($selection);
    $fetchedObj = $currSel->fetch_object();
?>

Now you have your data in an object. Now to the JS part (put this in your .html oder .php file):

<script type='text/javascript'> var email = <?php echo json_encode($fetchedObj); ?>;
</script>

This is a pretty neat way to get your PHP-variables inside JS btw. In JS you have a function that evaluates the array:

var emailInfo = [];

function getEmails(){
    var currentIteration = 0;
    for(var key in email){
        emailInfo[currentIteration] = email[key];
        currentIteration++;
    }
}

I know this is not exactly what you have asked for, but this is a pretty nice way to solve your problem, even if it involes JS and not purely PHP.

Now you can call your emailInfo-array just like you know it.

console.log(emailInfo[0]);
document.getElementById('emailBody').innerHTML = emailInfo[1];

Also might make sense to emit raw JSON from PHP and read that in with a `$.ajax` call. — tadman, Feb 01 '17 at 18:31

score 0 · Answer 2 · answered Feb 01 '17 at 17:26

This puts each row into a nested array of values for later retrieval:

$email = [];
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    array_push($email, array($row[0], $row[1]));
}

If you want to retrieve an element, use

 $email[0][0]

for example.

score 0 · Answer 3 · answered Feb 01 '17 at 17:39

0

use this code

<?php

    $emails = array();

    mysql_connect("127.0.0.1", "root") or
        die("Could not connect: " . mysql_error());
    mysql_select_db("users");

    $result = mysql_query("SELECT email FROM users");

    while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
        $email[] = $row[0];
    }

    mysql_free_result($result);
 ?>

answered Feb 01 '17 at 17:39

Think Big

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It's usually best to post only the code that changed, not the whole thing over again with no commentary as to what was fixed. – tadman Feb 01 '17 at 18:30

PHP load emails from database to array

3 Answers3