Question

我试图运行此代码，但我遇到了错误。有什么想法吗？

ON

Answer 1

由于你只能有一个var http = require('http'); var path = require('path'); var socketio = require('socket.io'); var express = require('express'); var router = express(); var server = http.createServer(router); var io = socketio.listen(server); router.use(express.static(path.resolve(__dirname, 'public'))); var clients = []; io.on('connection', function (socket) { console.log("ID: "+socket.id.substring(2)+"has connected"); clients[socket.id] = { vote:'' }; io.sockets.emit('count', socket.client.conn.server.clientsCount); socket.on('vote', function(data){ clients[socket.id].vote = data; // good or bad var result = calcVote(clients); io.sockets.emit('vote', result); }); socket.on('disconnect', function() { delete clients[socket.id]; io.sockets.emit('count', socket.client.conn.server.clientsCount); }); }); var calcVote = function(calcclients){ var sums = { good:0, bad:0 }; calcclients.forEach(function(client){ switch (client.vote) { case 'good' : sums.good += 1; break; case 'bad' : sums.bad += 1; break; } }); return sums; }; server.listen(process.env.PORT || 3010, process.env.IP || "0.0.0.0", function(){ var addr = server.address(); console.log("Chat server listening at", addr.address + ":" + addr.port); }); node.js socket.io子句，你可以使用这样的内连接：

WHERE

Answer 2

应该有单WHERE条款。

SELECT notes.project_id, notes.title, notes.description 
FROM notes, project_members 
WHERE 
(notes.title LIKE '%second%' OR notes.description LIKE  '%second%')
AND notes.project_id = project_members.project_id AND project_members.user_id = '7';

查询会在您的 LIKE 条件之间看到 AND 。

您也可以考虑使用JOIN，像这样。

SELECT notes.project_id, notes.title, notes.description 
FROM notes
JOIN project_members
ON notes.project_id=project_members.project_id
WHERE 
    (notes.title LIKE '%second%' OR notes.description LIKE  '%second%')
AND project_members.user_id = '7';

Answer 3

您有两个WHERE子句，但单SELECT只允许一个。但是，如果您使用正确的，明确的JOIN语法，这将轻松修复：

SELECT n.project_id, n.title, n.description 
FROM notes n JOIN
     project_members  p
     ON n.project_id = p.project_id
WHERE (n.title LIKE '%second%' OR n.description LIKE '%second%') AND
      p.user_id = '7';

注意：

始终使用正确的，明确的JOIN语法。从不在FROM子句中使用逗号。
表别名使查询更易于编写和阅读。
我非常确定您的WHERE条件缺少括号，符合您的预期逻辑。
如果user_id是任意类型的数字，则不要在“7”周围使用单引号。这可能会使编译器和人员感到困惑。

使用LIKE条件加入两个mysql表。

3 个答案: