簡體   English   中英

如何從node.js中的php變量中獲取值

[英]How to get value from php variable in node.js

我正在開發一個項目考勤系統,我想訪問在 node.js 文件 custom.js 中的 php 文件 require.php 中聲明的變量,我應該如何實現? 到目前為止,我已經創建了與我的 sql server 的連接並硬編碼了

`var company_name="Aaqoo 2";`

一切正常,但我不想對其進行硬編碼我的js變量.....請幫助我然后我會問更多的問題

這是我的節點 js 代碼

var express = require('express');
var mysql = require('mysql');
var app = express();

var connection = mysql.createConnection({
    host:'localhost',
    user:'root',
    password:'',
    database:'attendance'
});

connection.connect(function(error) {
    if (!!error) {
    console.log('Error in connection');
    } else {
    console.log('Connected');  
    }
});
var company_name = "Aaqoo 2"; //here i want the value from php variable $sup_company_name;

connection.query("Select * from employee_leaves where employee_leave_company_name=?", [company_name], function(error,rows,fields) {
    if (!!error) {
        console.log("Error in the query");
    } else {
        console.log("succesfully done\n");
        console.log(rows);
    }
});

app.listen(1337);

我做到了,這是我的示例,只是我將它用於與 pgsql 的連接。

代碼節點js

var exec = require('child_process').exec;

exec(
    'php -r \'include("require.php"); echo $companyname."#@#".$hostname."#@#".$username."#@#".$database."#@#".$password;\'',

  function (err, stdout, stderr) {
    var [ companyname,hostname,username, database, password ] = stdout.split('#@#');
    const connectionData = {
      user: username,
      host: '',
      database: database,
      password: password,
      port: 5432,
    }
    const client = new Client(connectionData)

    client.connect()
    client.query('Select * from employee_leaves where employee_leave_company_name=?',companyname)
        .then(response => {
            //console.log(response.rows)
            client.end()
        })
        .catch(err => {
            client.end()
        })
  }
);

要求.php


$companyname= 'namecompany'; 
$hostname = 'localhost';
$database = 'namedatabase';
$username = 'root';
$password = '123';

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM