有一个代码
$(document).ready(function() {
let loaded = false; // Индикатор загруженного графика
let originalContent = $("#div").html(); // Заглушка графика
$("button").click(function() {
if (loaded) {
$("#div").html(originalContent);
loaded = false;
this.innerText = 'Get Dynamic chart';
} else {
this.disabled = true; // Исключаем возможность повторных запросов
$.get("https://stackoverflow.getsandbox.com:443/dashboard", function(html) {
$("#div").html(html);
$("button").prop('disabled', false).text('Close Dynamic chart');
loaded = true;
});
}
});
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="div1">
<h2>Dynamic chart1</h2>
</div>
<button>Get Dynamic chart1</button>
<div id="div2">
<h2>Dynamic chart2</h2>
</div>
<button>Get Dynamic chart2</button>
<div id="div3">
<h2>Dynamic chart3</h2>
</div>
<button>Get Dynamic chart3</button>
<div id="div4">
<h2>Dynamic chart4</h2>
</div>
<button>Get Dynamic chart4</button>
<div id="div5">
<h2>Dynamic chart5</h2>
</div>
<button>Get Dynamic chart5</button>
网站页面上有 5 个这样的按钮,如何使当第一个按钮被按下时,它的内容打开,而不是一次按下所有按钮?
1 个回答