Javascript复选框检查

Question

我想使用Javascript来检查是否已选中复选框,如果未选中该复选框,则提交表单在检查之前不会继续.以下是我的代码.

<SCRIPT language=javascript>
function checkAcknowledgement(checkbox){
    alert(checkbox.toString());
    if (checkbox.checked == false){
        alert('Please read through the acknowledgement and acknowledge it.');
        return false;
    } else {
        return true;
    }
}
</script>

<form action="ioutput.php" method="POST">
    <input name="form" type="hidden" id="form" value="true">
    ... some html form ...
    <input type="checkbox" id="acknowledgement" value="1" /><br><br>
    <input type="submit" value="submit" onclick="return checkAcknowledgement(this)"/>
</form>

无论何时检查表单,它都会返回警告警告,即尽管我手动检查了表单,但仍未检查表单.我该如何解决？

Answer 1

您必须将事件绑定到表单而不是提交按钮.此外,如果要提交复选框输入,请附加名称而不是ID:

<input type="checkbox" name="acknowledgement" value="1" /><br><br>
<form action="ioutput.php" method="POST" onsubmit="return checkAcknowledgement(this)">

然后,修改功能:

function checkAcknowledgement(form){
    var checkbox = form["acknowledgement"];
    alert(checkbox); //shows [HTMLInputElement]
    if (!checkbox.checked){ //A shorter method for checkbox.checked == false
        alert('Please read through the acknowledgement and acknowledge it.');
        return false;
    } else {
        return true;
    }
}