Question

var script = document.createElement('script');
script.src = "//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js";
script.type='text/javascript';
var done = false;
script.onload = script.onreadystatechange = function() {

if (!done && (!this.readyState || this.readyState == "loaded" || this.readyState == "complete")) {
            done = true;
    promptForUserEntries();
}
};  
document.getElementsByTagName("head")[0].appendChild(script);

这会在第一个Head中注入一个脚本。我有三个其他脚本，我以相同的方式插入主题，现在我正在寻找如何一次在一个标签中插入多个脚本。

谢谢

Answer 1

var scripts = ["/path/to/script.js", "/path/to/script2.js"];

for (index = 0; index < scripts.length; ++index) {
    var script = document.createElement('script');
    script.src = scripts[index];
    script.type='text/javascript';
    var done = false;
    script.onload = script.onreadystatechange = function() {
        if (!done && (!this.readyState || this.readyState == "loaded" || this.readyState == "complete")) {
            done = true;
            //promptForUserEntries();
        }
    };  
    document.getElementsByTagName("head")[0].appendChild(script);
}

这将循环遍历您的脚本数组，并附加它们

Answer 2

这里是根据Ole的回答修改的解决方案。它同时允许带有src标签和脚本块的脚本-例如，对于在获得Cookie同意后加载Google Analytics（分析）代码很有用。

/**
 * Append a block of script code to the head
 * Code can contain scripts with src attribute
 * and script blocks
 */
function appendScriptBlocks (code) {

    let tempEl = document.createElement('div')
    tempEl.innerHTML = code
    let scripts = tempEl.getElementsByTagName('script')

    for (let index = 0; index < scripts.length; ++index) {
        var script = document.createElement('script');
        script.type= 'text/javascript';

        if(scripts[index].src) {
            script.src = scripts[index].src;
        }
        else {
            script.innerHTML = scripts[index].innerHTML
            eval(scripts[index].innerHTML)
        }
        
        document.getElementsByTagName('head')[0].appendChild(script);
    }
}

如何一次附加多个脚本？ jQuery的

2 个答案: