这是我的目标:
copy_python:
if test -d $(VIRTUAL_ENV)/lib; then \
cp -a $(VIRTUAL_ENV)/lib/python2.7/site-packages/. ./package/tmp/; \
fi
if test -d $(VIRTUAL_ENV)/lib64; then \
cp -a $(VIRTUAL_ENV)/lib64/python2.7/site-packages/. ./package/tmp/; \
fi
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这是错误:
/bin/sh: 2: Syntax error: end of file unexpected (expecting "fi")
Makefile:28: recipe for target 'copy_python' failed
make: *** [copy_python] Error 2
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为什么会出现此错误?
我在models文件夹(artisans,stores,items和itemStores)中有4个模型,但它只将items模型加载到db变量中.为什么会这样?
server.js
var express = require('express');
var app = express();
var bodyParser = require('body-parser');
var db = require('./models');
var artisans = require('./app/routes/artisans.js');
var stores = require('./app/routes/stores.js');
var items = require('./app/routes/items.js');
var itemStores = require('./app/routes/itemStores.js');
var PORT = process.env.PORT || 3000;
// Set up the express app to handle data parsing
app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended : true }));
app.use(bodyParser.text());
app.use(bodyParser.json({ type : "application/vmd.api-json"}));
// Static directory
app.use(express.static("app/public"));
artisans(app, db);
stores(app, db);
items(app, db);
itemsStores(app, db);
console.log(Object.keys(db));
db.sequelize.sync().then(function () {
app.listen(PORT, function () …Run Code Online (Sandbox Code Playgroud)