javascript - php mysqli fetch_all()正常返回json， 前端ajax却拿不到json？

Question

php代码

<?php
//include "db_connect.php";
$db = new MySQLi("localhost", "root", "password", "paginate");
$db->set_charset('utf8mb4');

header('Content-type: application/json;charset=utf-8');

$total            = 0;   // 表格数据总行数
$per_page         = 6;   // 每页行数
$current_page     = 1;   // 当前显示第几页
$total_pages      = 0;   // 总页数

// 获取当前页数
if (!empty($_GET['current_page'])) {
  $current_page = $_GET['current_page'];
}

// 获取表格数据总行数
$query1 = 'select count(*) from user';
$row1   = $db->query($query1);
if ($row1 && $rowd = $row1->fetch_row()) {
  $total = $rowd[0];
}

// 获取总页数
$total_pages = ceil($total / $per_page);

// 分页
$pre = ($current_page - 1) * $per_page;
$query2 = "select * from user limit $pre,$per_page";
$rows = $db->query($query2);

$arr = [];
if($rows) {
  $arr = $rows->fetch_all(MYSQLI_ASSOC);
}
$pagination = [
  'total'         => $total,
  'per_page'      => $per_page,
  'current_page'  => $current_page,
  'total_pages'   => $total_pages
];

array_push($arr, $pagination);
echo json_encode($arr, true);

postman截图：

js代码：

 window.onload = function () {
        getData(1);
    }

    function addURLParam(url, name, value) {
        url += (url.indexOf("?") == -1 ? "?" : "&");
        url += encodeURIComponent(name) + "=" + encodeURIComponent(value);
        return url;
    }
    
    function getData(page) {
        var data = document.getElementById("data");

        var xhr = new XMLHttpRequest();
        xhr.onreadystatechange = function () {
            if (xhr.readyState == 4 && xhr.status == 200) {
                var jsonText = xhr.responseText;
                console.log(JSON.parse(jsonText));
            }
        }
        var url = "paginate.php";
        url = addURLParam(url, "current_page", "page");
        xhr.open("get", url, true);
        xhr.send();
    }

chrome console前面的数据库的内容均没有

应该是

$query2 = "select * from user limit $pre,$per_page";
$rows = $db->query($query2);

$arr = [];
if($rows) {
  $arr = $rows->fetch_all(MYSQLI_ASSOC);
}

这里的sql出了问题，因为没加if($rows)之前会有错，然而我很奇怪为什么php可以正常运行，php版本是7.0.8

Answer 1

It’s a problem that js does not correctly pass parameters into addURLParam()

url = addURLParam(url, "current_page", "page");

changed to

url = addURLParam(url, "current_page", page);

It’s really shameful to say that I discovered this bug after I posted the question. I edited and modified the question once, but the school had a power outage and the computer ran out of power, so I couldn’t test it. . . . . .

---

But the important point is that I added the addURLParam method last. It turned out to be direct

xhr.open("get",  "paginate.php?current_page="+page, true);

addURLParam() is a method in js advanced programming, it says

An error that often occurs when using GET requests is that there is a problem with the query string format. The name and value of each parameter in the query string must be encoded using encodeURIComponent() before being placed at the end of the URL

Friends who know the reason can answer this point, I will search it first~