当我在php块中编写切换代码时，它无法正常工作

i am new in php. i just want to make a button for hide or un-hide the div
and i placed a code in php block when i write code in php block.button not work properly; when i code it outside the php block then it run. why it not run in php block?

<!DOCTYPE html>
    <html>
    <head>
        <title></title>
    </head>
    <body>
    <?php
    echo"<button onclick='myFunction('confession13')'>Comment</button>
    <div id='confession13'>
    <p>hlo world</p>
    </div>";
    ?>
    <script type="text/javascript">
    function myFunction(p1) 
    {   
        console.log(p1);
        var x = document.getElementById(p1);
        if (x.style.display === "none") 
        {
            x.style.display = "block";
        } 
        else {
            x.style.display = "none";
        }
    }
    </script>
    </body>
    </html>

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1条回答默认最新

douliang4858 2017-12-28 13:12

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<!DOCTYPE html>
    <html>
    <head>
        <title></title>
    </head>
    <body>
     <button onclick='myFunction("confession13")'>Comment</button>
     <div id='confession13'>
     <p>hlo world</p>
    </div>
    <script type="text/javascript">
    function myFunction(p1) 
    {   
        console.log(p1);
        var x = document.getElementById(p1);
        if (x.style.display === "none") 
        {
            x.style.display = "block";
        } 
        else {
            x.style.display = "none";
        }
    }
    </script>
    </body>
    </html>