PHP json_encode 在 MySQL 表的第一个索引上返回空数组

Question

嗨，我是 PHP 新手，我正在尝试从 MySQL 表中获取 JSON 数组形式的数据。我确实从 PHP 文件中获取了 JSON 数组，但第一个索引上有一个空数组。

这是我在 MySQL 上创建的表结构:

CREATE TABLE IF NOT EXISTS `tbl_employee` (
  `employee_id` int(4) NOT NULL AUTO_INCREMENT,
  `employee_name` varchar(60) NOT NULL,
  `designation` varchar(30) NOT NULL,
  `hired_date` date NOT NULL,
  `salary` int(10) NOT NULL,
  PRIMARY KEY (`employee_id`),
);

INSERT INTO `tbl_employee` (`employee_id`, `employee_name`, `designation`, `hired_date`, `salary`) VALUES
(1, 'Steve', 'VP', '2013-08-01', 60000),
(2, 'Robert', 'Executive' '2014-10-09', 20000),
(3, 'Luci', 'Manager', '2013-08-20', 40000);
(4, 'Joe', 'Executive', '2013-06-01', 25000);
(5, 'Julia', 'Trainee', '2014-10-01', 10000);

这是我的 PHP 文件:

<?php
    //open connection to mysql db
    $dbhost = 'clvm.ddns.net:8026';
    $dbuser = 'root';
    $dbpass = '';
    $dbname = 'ta_trial';
    //$conn = mysql_connect($dbhost, $dbuser, $dbpass);

    $connection = mysqli_connect($dbhost,$dbuser,$dbpass,$dbname) or die("Error " . mysqli_error($connection));

    //fetch table rows from mysql db
    $sql = "select * from tbl_employee";
    $result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));

    //create an array
    $emparray[] = array();
    while($row =mysqli_fetch_assoc($result))
    {
        $emparray[] = $row;
    }

    echo json_encode($emparray);
?>

这就是我得到的:

[[],{"employee_id":"1","employee_name":"Steve","designation":"VP","hired_date":"2013-08-01","salary":"60000"},{"employee_id":"2","employee_name":"Robert","designation":"Executive","hired_date":"2014-10-29","salary":"20000"},{"employee_id":"3","employee_name":"Luci","designation":"Manager","hired_date":"2013-08-20","salary":"40000"},{"employee_id":"4","employee_name":"Joe","designation":"Executive","hired_date":"2013-06-01","salary":"25000"},{"employee_id":"5","employee_name":"Julia","designation":"Trainee","hired_date":"2014-10-01","salary":"10000"}]

我不明白为什么第一个索引上会有一个空数组？谁能指出错误或我在这里做错了什么？有没有更好的解决方案来做到这一点？谢谢您，任何帮助将不胜感激。

Answer 1

你错过了一些东西 $emparray[] = array();到 $emparray = array();

<?php
    //open connection to mysql db
    $dbhost = 'clvm.ddns.net:8026';
    $dbuser = 'root';
    $dbpass = '';
    $dbname = 'ta_trial';
    //$conn = mysql_connect($dbhost, $dbuser, $dbpass);

    $connection = mysqli_connect($dbhost,$dbuser,$dbpass,$dbname) or die("Error " . mysqli_error($connection));

    //fetch table rows from mysql db
    $sql = "select * from tbl_employee";
    $result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));

    //create an array
    $emparray = array();
    while($row =mysqli_fetch_assoc($result))
    {
        $emparray[] = $row;
    }

    echo json_encode($emparray);
?>

检查结果是否正常