我如何修改此代码以写入Laravel控制器

发布于 2025-01-20 14:40:12 字数 306 浏览 1 评论 0原文

我的问题是，当我通过ESP8266从RFID卡发送帖子时，数据是RFID号码，但是数据无法显示，因为我给出的代码尚未写入Laravel中。

此PHP代码：

<?php
    if(isset($_POST['uid'])) {
        $uid = $_POST["uid"];
        $sql = mysqli_query($dbconnect, "INSERT INTO tb_entry VALUES ('$uid')");
    }
?>

如何将此代码写入Laravel控制器？

原文

My problem is that when I send a post from an rfid card via esp8266, the data is an rfid number, but the data can't appear because the code I gave has not been written into Laravel.

This PHP code :

<?php
    if(isset($_POST['uid'])) {
        $uid = $_POST["uid"];
        $sql = mysqli_query($dbconnect, "INSERT INTO tb_entry VALUES ('$uid')");
    }
?>

How can I write this code into Laravel controller?

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丶视觉 2025-01-27 14:40:12

应该是这样的

if ($request->has('uid')) {
    DB::table('tb_entry')->insert([
        'uuid' => $request->uuid
    ]);
}

it should be like this

if ($request->has('uid')) {
    DB::table('tb_entry')->insert([
        'uuid' => $request->uuid
    ]);
}

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筱武穆 2025-01-27 14:40:12

创建一个控制器： php artisan make:controller MyController
向控制器添加一个方法，例如 public function test(Request $request) {}
稍微重写您的代码并粘贴：

public function test(Request $request) {

    if (null !== $request->post('uid')) {
        $uid = $request->post('uid');
        // Note: you can use Laravel model instead plain mysqli query
        // ** Beware of SQL injection vulnerabilities! **
        $sql = mysqli_query($dbconnect, "INSERT INTO tb_entry VALUES ('$uid')");
    }
}

create a controller: php artisan make:controller MyController
add a method to the controller like public function test(Request $request) {}
rewrite your code a little bit and pasted in:

public function test(Request $request) {

    if (null !== $request->post('uid')) {
        $uid = $request->post('uid');
        // Note: you can use Laravel model instead plain mysqli query
        // ** Beware of SQL injection vulnerabilities! **
        $sql = mysqli_query($dbconnect, "INSERT INTO tb_entry VALUES ('$uid')");
    }
}

回复收藏 0 原文

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