Jer*_*ith 6 javascript requirejs
我想做这样的事情:
<script data-main="/js/D4/build/mainD4" src="/js/D4/build/require.js"></script>
<script data-main="/js/main" src="/js/require.js"></script>
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我可以构建第一个文件,并在/ js/main中包含build js文件,但是能够在两个项目上并行开发而不必一直构建它会快得多.现在当我尝试这个时,mainD4构建,然后js/main文件没有任何反应.
刚刚在这里找到了答案:https: //groups.google.com/forum/?fromgroups#!topic/simplyjs/YWFdgYSU2f4
<script src="scripts/require.js" data-main="scripts/main"></script>
<script>
require(['scripts/another/main']);
</script>
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要么
<script src="scripts/require.js" data-main="scripts/main"></script>
<script>
(function(){
var req = require.config({baseUrl:'scripts/another'});
req(['main']);
}());
</script>
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