我如何在另一个JavaScript文件中添加一个JavaScript文件,类似于CSS中的 @import?


当前回答

我有要求无同步加载一系列的JavaScript文件,最终进行呼叫回复。

// Load a JavaScript file from other JavaScript file
function loadScript(urlPack, callback) {
    var url = urlPack.shift();
    var subCallback;

    if (urlPack.length == 0) subCallback = callback;
    else subCallback = function () {
        console.log("Log script: " + new Date().getTime());
        loadScript(urlPack, callback);
    }

    // Adding the script tag to the head as suggested before
    var head = document.getElementsByTagName('head')[0];
    var script = document.createElement('script');
    script.type = 'text/javascript';
    script.src = url;

    // Then bind the event to the callback function.
    // There are several events for cross browser compatibility.
    script.onreadystatechange = subCallback;
    script.onload = subCallback;

    // Fire the loading
    head.appendChild(script);
}

例子:

loadScript(
[
    "js/DataTable/jquery.dataTables.js",
    "js/DataTable/dataTables.bootstrap.js",
    "js/DataTable/dataTables.buttons.min.js",
    "js/DataTable/dataTables.colReorder.min.js",
    "js/DataTable/dataTables.fixedHeader.min.js",
    "js/DataTable/buttons.bootstrap.min.js",
    "js/DataTable/buttons.colVis.min.js",
    "js/DataTable/buttons.html5.min.js"
], function() { gpLoad(params); });

第二个脚本不会充电,直到第一个完全充电,因此...

结果:

此分類上一篇

其他回答

var s=["Hscript.js","checkRobert.js","Hscript.js"];
for(i=0;i<s.length;i++){
  var script=document.createElement("script");
  script.type="text/javascript";
  script.src=s[i];
  document.getElementsByTagName("head")[0].appendChild(script)
};

我基本上做如下,创建一个新的元素,并添加它来主导:

var x = document.createElement('script');
x.src = 'http://example.com/test.js';
document.getElementsByTagName("head")[0].appendChild(x);

在 jQuery 中:

// jQuery
$.getScript('/path/to/imported/script.js', function()
{
    // Script is now loaded and executed.
    // Put your dependent JavaScript code here.
});
var js = document.createElement("script");

js.type = "text/javascript";
js.src = jsFilePath;

document.body.appendChild(js);

我有要求无同步加载一系列的JavaScript文件,最终进行呼叫回复。

// Load a JavaScript file from other JavaScript file
function loadScript(urlPack, callback) {
    var url = urlPack.shift();
    var subCallback;

    if (urlPack.length == 0) subCallback = callback;
    else subCallback = function () {
        console.log("Log script: " + new Date().getTime());
        loadScript(urlPack, callback);
    }

    // Adding the script tag to the head as suggested before
    var head = document.getElementsByTagName('head')[0];
    var script = document.createElement('script');
    script.type = 'text/javascript';
    script.src = url;

    // Then bind the event to the callback function.
    // There are several events for cross browser compatibility.
    script.onreadystatechange = subCallback;
    script.onload = subCallback;

    // Fire the loading
    head.appendChild(script);
}

例子:

loadScript(
[
    "js/DataTable/jquery.dataTables.js",
    "js/DataTable/dataTables.bootstrap.js",
    "js/DataTable/dataTables.buttons.min.js",
    "js/DataTable/dataTables.colReorder.min.js",
    "js/DataTable/dataTables.fixedHeader.min.js",
    "js/DataTable/buttons.bootstrap.min.js",
    "js/DataTable/buttons.colVis.min.js",
    "js/DataTable/buttons.html5.min.js"
], function() { gpLoad(params); });

第二个脚本不会充电,直到第一个完全充电,因此...

结果:

此分類上一篇

雖然這些答案很棒,但有一個簡單的「解決方案」已經存在,因為脚本充電存在,它將覆蓋大多數人使用案例的99.999%。

<!DOCTYPE HTML>
<html>
    <head>
        <script src="script1.js"></script>
        <script src="script2.js"></script>
    </head>
    <body></body>
</html>

如果脚本2需要脚本1,这实际上是做这样的事情的绝对最简单的方式,我很惊讶,没有人带来了这一点,因为这是最明显和最简单的答案,几乎每个案例都会适用。