Php 将数组字符串转换为Int始终得到返回Int（1）

我尝试使用intval将字符串数组转换为INT 但是结果/返回总是int（1） 而数据库值为14698

我的数据库：

+-------+---------+------+-----+---------+----------------+
| 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 |
+----+------+-------+

    <?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];

<?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;
    }

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

这是我的crudQuery.php页面：

+-------+---------+------+-----+---------+----------------+
| 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 |
+----+------+-------+

    <?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];

<?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;
    }

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

为了从数据库im而不是1返回值，我支持做什么

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

mysqli\u fetch\u assoc

返回一个数组，您不能将数组强制转换为int，并期望得到有意义的结果

intval（$row）

需要是

intval（$row['rate']）

change

var\u dump（$rate）；echo$rate[0]至
var\u转储（$rate）；echo美元汇率[1]