Converting Array String to Int always got the return int(1)

Question

I've tried to convert my string array to INT by using intval But the result / return always int(1) while the database value was 14698

my database :

+-------+---------+------+-----+---------+----------------+
| Field | Type    | Null | Key | Default | Extra          |
+-------+---------+------+-----+---------+----------------+
| id    | int(11) | NO   | PRI | NULL    | auto_increment |
| curr  | char(3) | NO   |     | NULL    |                |
| rate  | int(11) | NO   |     | 0       |                |
+-------+---------+------+-----+---------+----------------+

+----+------+-------+
| id | curr | rate  |
+----+------+-------+
|  1 | USD  |     1 |
|  2 | IDR  | 14698 |
|  3 | JPY  |   112 |
|  4 | EUR  |     1 |
|  5 | THB  |    33 |
+----+------+-------+

Here's my crudQuery.php page :

    <?php

require 'config.php';

if (isset($_POST['save'])) {

    require_once 'functions.php';

    $qty = $_POST['qty'];
    $qty = (int) $qty;
    $curr = $_POST['curr'];
    $price = $_POST['price'];
    $price = (int) $price;

    $rate = queryInt("SELECT rate FROM rate_master WHERE curr = '$curr'");

    var_dump($rate);echo $rate[0];

here's my functions.php page :

<?php
require 'config.php';

    function queryInt($query) {
        global $conn;
        $result = mysqli_query($conn, $query);
        $rows = [];
        while ($row = mysqli_fetch_assoc($result)) {
            $rows [] = intval($row); 
        }
        return $rows;
    }

Then, the result of var_dump was :

array(1) { [0]=> int(1) } 1

What I suppossed to do, in order to return the value from database imstead of 1 ?

change var_dump($rate);echo $rate[0]; to var_dump($rate);echo $rate[1]; — Alex Maina
– Alex Maina, Commented Dec 5, 2019 at 13:00

04FS · Accepted Answer · 2019-12-05 13:02:13Z

1

while ($row = mysqli_fetch_assoc($result)) {
    $rows [] = intval($row); 
}

mysqli_fetch_assoc returns an array, you can not cast an array “into int”, and expect that to have a meaningful result.

intval($row) needs to be intval($row['rate'])

answered Dec 5, 2019 at 13:02

04FS

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