Javascript 如何在Java脚本中 Select 该元素Chrome扩展

我是个新手，我已经花了一个小时的时间从一个网络应用程序中 Select 了这个:

<button type="button" class="zp-button zp_zUY3r zp_B5hnZ zp_rhXT_ zp_FVbJk finder-select-multiple-entities-button" fdprocessedid="jowqnd"><div class="zp_kxUTD" data-elem="button-label"><div class="zp_fwjCX" data-input="checkbox" data-cy-status="unchecked"></div></div><i class="zp-icon apollo-icon apollo-icon-caret-down-small zp_dZ0gM zp_j49HX zp_sGyhf"></i></button>

我一直想把它选到"zp_FVbJk"级.

const dropdownSelector = 'zp_FVbJk'
// Handles button and has to be the same in popup.js and content.js until I find a better solution
function sharedTempHandleButton() {
    console.log("sharedTempHandleButton() started.");

    let buttonToClick;

    buttonToClick = document.getElementsByClassName(dropdownSelector)[0];

    console.log(buttonToClick)

    if (buttonToClick) {
        console.log("Button found: ", buttonToClick);
        buttonToClick.click();
    } else {
        console.error('Button not found using the provided CSS selector');
    }
}

它将ButtonToClick返回为未定义.

我试过了:

• 将 Select 器更改为".zp_FVbJk"
• 使用queryselector()