尝试在 PHP 中从 MySQL 数据库生成 JSON 数组时出错 [重复]

Question

这是我的 get_categories.php 的 PHP 代码：

<?PHP
    require_once('connection.php');
    $query="SELECT * FROM categories";
    $result = mysqli_query($connection,$query); 
    $return_arr = array();
    while ($row = mysqli_fetch_array($result, mysqli_fetch_assoc) 
           {
        $row_array['category'] = $row['category'];
        $row_array['icon'] = $row['icon'];

        array_push($return_arr,$row_array);

    }

    echo json_encode($return_arr);

    ?>

和connection.php

<?php
$servername = "localhost"; //replace it with your database server name
$username = "root";  //replace it with your database username
$password = "password";  //replace it with your database password
$dbname = "db_client";
// Create connection
$connection = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$connection) {
    die("Connection failed: " . mysqli_connect_error());
}
?>

当我运行 get_categories.php 生成一个 json 数组时.. 出现此错误

解析错误：语法错误，意外';'在第 7 行的 C:\xampp\htdocs\get_categories.php 中

有人可以纠正我做错了什么吗？谢谢。

Answer 1

while ($row = mysqli_fetch_array($result, mysqli_fetch_assoc)
应该是：
while ($row = mysqli_fetch_array($result, mysqli_fetch_assoc))

你少了一个括号