Javascript。如何停止执行脚本，直到所有脚本都加载完毕

window.onload = function() {
    // 所有资源加载完毕后的代码
};

<script src="script1.js" defer></script>
<script src="script2.js" defer></script>

<script src="script1.js" async></script>
<script src="script2.js" defer></script>

function loadScript(src, callback) {
    var script = document.createElement('script');
    script.src = src;
    script.onload = callback;
    document.head.appendChild(script);
}

loadScript('script1.js', function() {
    loadScript('script2.js', function() {
        // 所有脚本加载完毕后的代码
    });
});

function loadScript(url) {
    return new Promise((resolve, reject) => {
        const script = document.createElement('script');
        script.src = url;
        script.onload = resolve;
        script.onerror = reject;
        document.head.appendChild(script);
    });
}

Promise.all([
    loadScript('script1.js'),
    loadScript('script2.js')
]).then(() => {
    // 所有脚本加载完毕后的代码
}).catch(error => {
    console.error('脚本加载失败:', error);
});