编码问题-PHP，MySQL-奇怪的行为

如何解决编码问题-PHP，MySQL-奇怪的行为

我正在尝试进行Ajax调用并获取JSON中的响应。问题在于JSON获取的字符编码不正确。因此，JSON格式不正确，获取并出错。

这仅发生在我的本地主机中，我已经在服务器中尝试过，并且可以正常工作。我不明白发生了什么。

这是Ajax调用：

var getEmployees = function () {
    var name;
    var code;
    var nameCode;
    var result = [];
    $.ajax({
        type: "POST",url: "getEmployees.PHP",dataType: 'json',success: function (data) {
            var len = data['data'].length;
            for( var i = 0; i<len; i++) {
                name = data['data'][i]['name'];
                code = data['data'][i]['code'];
                nameCode = name + ' - ' + code;
                result.push(nameCode);
            }
            renderEmployees(name);
        },error: function (xhr,status,textResponse) {
            console.log(xhr.responseText);
            console.log(status);
            console.log(textResponse);
        }
    });
}

这是$ getEmployees.PHP的PHP代码

-- Validate database

$getEmployeesQuery = "SELECT e.id,e.name,e.code,e.email,e.active,e.updated_at ";
$getEmployeesQuery .= "FROM emp e ";
$getEmployeesQuery .= "LEFT JOIN ecredit ec ON e.id = ec.employee_id ";
$getEmployeesQuery .= "LEFT JOIN credit c ON c.id = ec.credit_id ";
$getEmployeesQuery .= "LEFT JOIN product p ON p.id = c.product_id ";
$getEmployeesQuery .= "WHERE e.company_id = " .$companyId. " ";
if (isset($searchQuery)) {
    $getEmployeesQuery .= "and (e.name LIKE '%".$searchQuery."%' ";
    $getEmployeesQuery .= "or e.code LIKE '%".$searchQuery."%' ";
    $getEmployeesQuery .= "or e.email LIKE '%".$searchQuery."%') ";
}
if (isset($status)) {
    $getEmployeesQuery .= "and e.active = ".$status." ";
}
$getEmployeesQuery .= "group by e.id ";
$getEmployeesQuery .= "order by e.updated_at desc";

$getEmployees = MysqLi_query($dbConnection,$getEmployeesQuery);


if ($getEmployees) {
    while ($row = MysqLi_fetch_array($getEmployees)) {
        $getEmployeeCreditsQuery = "SELECT sum(ec.quantity) as quantity,c.expiration_date ";
        $getEmployeeCreditsQuery .= "FROM ecredit ec ";
        $getEmployeeCreditsQuery .= "JOIN credit c ON c.id = ec.credit_id ";
        $getEmployeeCreditsQuery .= "WHERE c.expiration_date > '$currentDate' and ec.employee_id = ".$row['id']." ";

        $getEmployeeCredits = MysqLi_query($dbConnection,$getEmployeeCreditsQuery);

        if (MysqLi_num_rows($getEmployeeCredits)>0) {
            while ($result = MysqLi_fetch_array($getEmployeeCredits)){
                $quantity = $result['quantity'];
            }
        } else {
            $quantity = 0;
        }

        $getEmployeesInfo['data'][] = array(
            'id' => (int)$row['id'],'name' => $row['name'],'code' => $row ['code'],'email' => $row ['email'],'status' => (int)$row['active'],'quantity' => $quantity
        );

    }

    foreach ($getEmployeesInfo['data'] as $row => $id) {
        $rowIds['Meta']['rowIds'][] = $id['id'];
    }

    $mergedArray = array_merge($rowIds,$getEmployeesInfo);

    return print_r(json_encode($mergedArray));

}

以下是数据示例： data sample

当数据较小时，它可以正常工作，但是当JSON传递超过65528个字符时，始终在该位置获得一个无法识别的字符，这是控制台中的Ajax响应。

parsererror
tagify-employees.js:26 SyntaxError: Unexpected token  in JSON at position 65528
    at parse (<anonymous>)
    at ajaxConvert (plugins.bundle.js:9013)
    at done (plugins.bundle.js:9483)
    at XMLHttpRequest.<anonymous> (plugins.bundle.js:9785)

我检查了所有数据，其中没有任何特殊字符。同样，为UTF-8指定了Meta和header标签。这是服务器的配置。

Server info

请，我不知道为什么会这样，并且我无法在计算机上尝试某些功能。如果有人可以指导我，我将不胜感激。另外，如果您需要更多信息，我可以提供。