Question

我想根据来自PHP脚本的结果调用javascript函数。

if ($verify == 'Y' and $approve== 'Y' and $approve_2=='Y') { 
        $state = "disable";
    } else { 
        $state = "undisable"; 
    }

以下是我的javascript函数

function disable() {
                document.getElementById("approvedd1").disabled = true;
                document.getElementById("approvedd2").disabled = true;
}

function undisable() {
                document.getElementById("approvedd1").disabled = false;
                document.getElementById("approvedd2").disabled = false;
}

document.onreadystatechange = <?php echo $state; ?>()

这不起作用。我希望在页面加载时调用可以禁用单选按钮的功能。

Answer 1

您的java脚本的当前输出将是

function disable() {
            document.getElementById("approvedd1").disabled = true;
            document.getElementById("approvedd2").disabled = true;
}

function undisable() {
            document.getElementById("approvedd1").disabled = false;
            document.getElementById("approvedd2").disabled = false;
}

document.onreadystatechange = disabled/undisabled()

您可以尝试更好

<强> PHP

if ($verify == 'Y' and $approve== 'Y' and $approve_2=='Y') { 
    $state = "true";
} else { 
    $state = "false"; 
}

Java脚本

 document.onreadystatechange = MyFun() {
    document.getElementById("approvedd1").disabled = <?php echo $state; ?>;
    document.getElementById("approvedd2").disabled = <?php echo $state; ?>;
 }

禁用基于PHP功能的单选按钮

1 个答案: