Add element and key to array php

Question

I'm trying to add an element to array, but I get a weird output. The code is the following:

$getalltokens = $db->query("SELECT * FROM Tables WHERE available = '$comp'");
    while ($row = $getalltokens->fetch(PDO::FETCH_ASSOC))
    {
        $fid = $row['FID'];
        $tok = $row['token'];
        $sql = $db->query("SELECT Firstname,Lastname FROM Users WHERE Token = '$tok'");
        $rez = $sql->fetch(PDO::FETCH_ASSOC);
        $names[] = $rez;
        $fidzy = array(
            'FID' => $fid
        );
        array_push($names, $fidzy);
    }
    $getalltokens = $db->query("SELECT FID FROM Tables WHERE available = '$comp'");
    $tokenz = $getalltokens->fetchAll(PDO::FETCH_ASSOC);
    echo json_encode($names);

And the output I get is:

[{"Firstname":"Test","Lastname":"Test"},{"FID":"5"},
{"Firstname":"Test2","Lastname":"Test2"},{"FID":"4"}]

While what I need is the FID to be inside the $names array, so it would be more like:

[{"Firstname":"Test","Lastname":"Test","FID":"5"}]

You don't need $fidzy there. Just do $names['FID'] = $fid; and get rid of array_push(). — Ulver
– Ulver, Commented Jun 16, 2015 at 2:49
@Ulver, you did same mistake like me. $names is array of arrays. — Shamil Yakupov
– Shamil Yakupov, Commented Jun 16, 2015 at 2:50
This is an XY Problem. You shouldn't be performing iterated queries here. Use a JOIN clause in a properly constructed query to relate related data and avoid abusing system resources. — mickmackusa
– mickmackusa ♦, Commented Feb 20 at 9:37

Shamil Yakupov · Accepted Answer · 2015-06-16 02:49:10Z

2

$rez['FID'] = $fid; /* Added */
$names[] = $rez;
/* $fidzy and array_push removed */

edited Jun 16, 2015 at 2:49

answered Jun 16, 2015 at 2:35

Shamil Yakupov

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4 Comments

Predrag Beocanin Over a year ago

The result of that is the following: {"0":{"Firstname":"Vukasin","Lastname":"Vulovic"},"FID":"9"} Do you think if I change the query to grab all the data at once, it would help? I'm gonna try that now

Predrag Beocanin Over a year ago

No worries, however, the new snippet results in the following: Notice: Undefined variable: names in /home/user/domain/api/getusers.php on line 46 null Line 46 is the echo json_encode($names); If I understood correctly, you're trying to add key and value to $rez which isn't an array? Since I'm doing the while for every token separately?

Shamil Yakupov Over a year ago

PDO::FETCH_ASSOC returning an array, so $rez is an array. Did you declare your $names before while? Also, I dont know what are you trying to do, but better to make it in one query like SELECT t.*, u.Firstname, u.Lastname FROM Tables as t, Users as u WHERE u.Token = t.token AND t.available = '$comp'.

Predrag Beocanin Over a year ago

I didn't delcare, that's what I'm looking to do but I'm pretty stuck here:

$sql = $db->query("SELECT Users.Firstname,Users.Lastname FROM Users LEFT JOIN Tables.FID ON Users.Token = Tables.token WHERE Users.Token = '$tok'");

not the best left joiner I'll admit

Bruce · Accepted Answer · 2015-06-16 06:19:02Z

1

You can use instead of array_push() like

$arrayname[indexname] = $value;

if you use array_push()

<?php
$array[] = $var;
?>

Note: If you use `array_push()` to add one element to the array it's
better to use$array[] = because in that way there is no overhead of calling a function.
Note: `array_push()` will raise a warning if the first argument is not an array. This differs from the `$var[]` behavior where a new array
is created.

Reference Array push

edited Jun 16, 2015 at 6:19

answered Jun 16, 2015 at 4:47

Bruce

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1 Comment

Predrag Beocanin Over a year ago

This is true, however, in this very specific case, a different solution is the best. Your suggestion does give the desired output, but changes formating or adds the FID without FID:1, just adds 1. I'll post the answer in a second

Predrag Beocanin · Accepted Answer · 2015-06-16 06:24:51Z

0

The solution to the specific problem at hand is selecting all the necessary data in a single query, removing the need to add elements to any array. This is done in the following fashion:

$sql = $db->query("SELECT
Users.Firstname,Users.Lastname,Tables.FID
FROM Users,Tables 
WHERE Users.Token = Tables.token");
$rez = $sql->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($rez);

answered Jun 16, 2015 at 6:24

Predrag Beocanin

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1 Comment

mickmackusa Feb 20 at 9:38

I recommend not using old-skool comma joins. Modern programming leans toward JOIN with an ON or USING clause.

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