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我正在繪製應用程序。我有2個工具(鉛筆和橡皮)和一些顏色。 現在,當我檢查鉛筆我可以選擇一種顏色,但當我檢查橡皮擦,我想選擇一種顏色,我想再次檢查鉛筆工具,但橡皮擦工具保持選中狀態。更改後返回到默認複選框
任何指導都非常appriciated。
這裏有一個例子,代碼:
代碼:
HTML:
<!-- Pencil & Eraser -->
<div class="graphic-tools">
<!-- <div id="pencil" class="pencil"></div> -->
<input checked="checked" id="pencil" type="radio" name="tool" value="pencil" />
<label class="tool pencil" for="pencil"></label>
<p class="">Potlood</p>
<!-- <div id="eraser" class="eraser"></div> -->
<input id="eraser" type="radio" name="tool" value="eraser" />
<label class="tool eraser" for="eraser"></label>
<p>Gum</p>
</div>
<!-- colors -->
<div id="colors" class="colors"></div>
的JavaScript:
// Color swatches
var colors = ['#000000', '#274e8d', '#6bb44b', '#e5cd46', '#e98b3a', '#d83538'];
for (var i = 0, n=colors.length; i<n; i++) {
var swatch = document.createElement('div');
swatch.className = 'swatch';
swatch.style.backgroundColor = colors[i];
swatch.addEventListener('click', setSwatch);
document.getElementById('colors').appendChild(swatch);
}
function setColor(color) {
context.fillStyle = color;
context.strokeStyle = color;
var active = document.getElementsByClassName('activeSwatch')[0];
if (active) {
active.className = 'swatch';
}
}
function setSwatch(e) {
//identify swatch
var swatch = e.target;
//set color
setColor(swatch.style.backgroundColor);
//give active class
swatch.className += ' activeSwatch';
}
setSwatch({target: document.getElementsByClassName('swatch')[0] });
// Determine Tool
document.getElementById('pencil').onchange = function() {
if (this.checked) {
tool = 'pencil';
setSwatch({target: document.getElementsByClassName('swatch')[0] });
}
};
document.getElementById('eraser').onchange = function() {
if (this.checked) {
tool = 'eraser';
setColor('#FFFFFF');
}
};
是的,這是非常非常! –