Question

这个Html编码与css和php.I我得到一个错误，你结果变量未声明。但我已经在我的form.please检查并告诉我有关此错误。计算器

this style .css file


        <style>
        form {
    display:block; 
    background-color: #333399;
    width:300px;
    height:500px;
    border:thick;
    border: #330000;

    color: #FFCC00;
    }


h1 {
    text-align:center;
    z-index: 2px;
    }
    </style>

这是Php编码

if(isset($_POST['add'])){
    $first_value = $_POST['f_value'];
    $sec_value = $_POST['s_value'];

    //--calculation variables---//

         $result = $first_value + $sec_value;           


        }
        ?>  

    </head>

Html表单从这里开始

<body>
        <form  method="post" action="new.php" name="calculator">

    &nbsp;<h1> calculator</h1> 
    <p>
    <strong>Frsit value</strong>&nbsp;&nbsp;&nbsp;&nbsp;
    <input  type="text" name="f_value" >

    <p><strong>Second value</strong> <input type="text" name="s_value" maxlength="50">
    <p>
    &nbsp;<input name="add" type="submit" value="add" >
    <!--&nbsp;<input name="sub" type="submit" value="sub">
    &nbsp;<input name="sub" type="submit" value="multiply">
    &nbsp;<input name="sub" type="submit" value="divide">-->
    `enter code here`<p>

    <h2 style="border:thick">Result
      <input type="text" maxlength="50" value="<?php echo $result ; ?>" Name='result' >
    </h2>


        </form>

</body>

</html>

Answer 1

在这里使用isset

<input type="text" maxlength="50" value="<?php if(isset($result)) { echo $result; } ?>" Name='result' >

Answer 2

if(isset($_POST['add'])){
    $first_value = $_POST['f_value'];
    $sec_value = $_POST['s_value'];
    $result = $first_value + $sec_value;  
}
else{
    $result= '';
}

Answer 3

您的$result超出了范围：

<?php
if(isset($_POST['add'])) {  //scope begins here
  //php omitted for brevity

  //$result is declared within this scope
  $result = $first_value + $sec_value;

} // scope ends here - after this point, $result no longer exists!
?>
<!-- html omitted for brevity -->
<!-- This is OUTSIDE the scope where $result was declared - we can't get it any more! -->
<input type="text" maxlength="50" value="<?php echo $result ; ?>" Name='result' >

要解决此问题，首先在与您想要回显的范围相同的范围内声明$ result：

<?php
$result = 0;

if(isset($_POST['add'])) {
  //php omitted for brevity

  //change $result's value
  $result = $first_value + $sec_value;
}
?>
<!-- html omitted for brevity -->
<input type="text" maxlength="50" value="<?php echo $result ; ?>" Name='result' >

Futher information about variable scope

我在获取Html表单时出错，同时给出PHP代码的值

3 个答案: