let dbclickre = true;
function flipped() {
if (dbclickre) {
document.querySelector(".linkrec").setAttribute("Id", "flipped");
} else {
document.querySelector(".linkrec").removeAttribute("Id", "flipped")
}
dbclickre = !dbclickre;
}
// const dbclickre = document.querySelector(".reverse");
// function flipped() {
// if (dbclickre.style.backgroundColor = 'white') {
// document.querySelector(".linkrec").setAttribute("Id", "flipped");
// } else {
// document.querySelector(".linkrec").removeAttribute("Id", "flipped")
// }
// }
Instead of relying on background color for checking the state of flip, you could check for existence of Id
attribute. Here the the changed code:
const dbclickre = document.querySelector(".reverse");
function flipped() {
if ( document.querySelector(".linkrec").getAttribute("Id") == undefined ) {
document.querySelector(".linkrec").setAttribute("Id", "flipped");
} else {
document.querySelector(".linkrec").removeAttribute("Id", "flipped")
}
}
Why element.style
would not work?
From MDN Web Docs :
The
style
property is used to get as well as set the inline style of an element.
Hence, the style
property would not work with embedded or external CSS.
Also, it may not be a good idea to use hard-coded colors as the condition, because changing colors in the respective CSS classes would completely break the functionality.
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