我想要一个简单的程序,当按下每个按钮时,基本上可以启用/禁用按钮。
因此,有三个按钮。左键在程序启动时启用,我想禁用它,然后单击onclick启用它旁边的按钮。当我按下第二个按钮时也是如此。
谁能告诉我我的代码出了什么问题?
<html>
<head>
<button id="func1" onclick="func(1)">func 1</button>
<button id="func2" disabled="false" onclick="func(2)">func 2</button>
<button id="func3" disabled="false" onclick="func()">func 3</button>
</head>
<body>
<script>
var number = '';
function func(number)
if(number == '1'){ //Sets button setting to disabled or enabled when wanted
after particular parts of the program are run.
document.getElementById('func1').disabled = true;
document.getElementById('func2').disabled = false;
document.getElementById('output').disabled = true;
}
else if(number == '2'){
document.getElementById('func1').disabled = true;
document.getElementById('func2').disabled = true;
document.getElementById('func3').disabled = false;
}
</script>
</body>
</html>
再次感谢 :)
最佳答案
首先,您需要将按钮从<head>
移到<body>
。
现在解决当前的问题:您的函数缺少其{}括号,并且您的注释分为两行,其中第二行引起语法错误。现在应该工作:
<button id="func1" onclick="func(1)">func 1</button>
<button id="func2" disabled="false" onclick="func(2)">func 2</button>
<button id="func3" disabled="false" onclick="func()">func 3</button><script>
function func(number){
if(number == '1'){ //Sets button setting to disabled or enabled when wanted after particular parts of the program are run.
document.getElementById('func1').disabled = true;
document.getElementById('func2').disabled = false;
document.getElementById('output').disabled = true;
}
else if(number == '2'){
document.getElementById('func1').disabled = true;
document.getElementById('func2').disabled = true;
document.getElementById('func3').disabled = false;
}
}
</script>