Question

在同一目录中使用以下文件我在运行XAMPP 1.8.3和[PHP：5.5.15。

时遇到了一个无法POST错误

<html>
<head>
    <title> IsEven </title>
</head>
<body>

    <form action= "IsEven.php" method="post">
    Enter a number: <input type="text" name="number" />
                                    <input type="submit" value="submit" />
    </form>

</body>
</html>

这是我的PHP文件

<?php

//Assign Values to the variable.
$number = $_POST['number'];

//If statements to check Even number or not.

//Is_numeric, Isset() and round() functions.
if (is_numeric($number) && isset($number)){
    if (round($number, 0) % 2 == 0){
           echo "The number " .$number. " entered is a even number<br>";
       }
    else
   {
    echo "The number " . $number. " is a non-even number";
   } //End round if statement.
} //End is numeric if statement
else
{
    echo "Please enter a numeric value.";
}

?>

Answer 1

这是我今天在一些代码中使用的技术，它回答了我的问题。

<!DOCTYPE HTML>
<html>
    <head>
        <title> Temperature Conversion </title>
    </head>
    <body>

        <form method="post" action= "<?php echo $_SERVER["PHP_SELF"]; ?>">
            Fahrenheit: <input type="radio" name="temp_metric" value="Fahrenheit" />
            Celcius: <input type="radio" name="temp_metric" value="Celcius" />
            <input type="submit" name="submit" value="submit" />
        </form>

    </body>
</html>


<?php

//VARIABLE DECLARATIONS
$degrees = NULL;
$temp_degrees = NULL;
$curr_metric = NULL;
$converted_metric = NULL;
$converted_temp = NULL;

//LOGIC
if ('POST' === $_SERVER['REQUEST_METHOD'] and isset($_POST['temp_metric'])) {
    $curr_metric = $_POST['temp_metric'];
    for ($degrees = 0; $degrees <= 100; $degrees++) {
        if($curr_metric == "Fahrenheit") {
            $converted_metric = "Celcius";
            $temp_degrees = ($degrees - 32) * (5/9);
            $converted_temp = round( $temp_degrees, 1);
        } 

        else {
            $curr_metric = "Celcius";
            $converted_metric = "Fahrenheit";
            $temp_degrees = $degrees * 9/5 + 32;
            $converted_temp = round($temp_degrees , 1);
        }

        //output the result of the users selection
        echo "<div> $degrees $curr_metric is $converted_temp     $converted_metric. </div>";

    }//end of for loop
}//end of if current metric is not null


?>

我遗失的部分位于我最初的HTML中，以便“”＆gt;使我能够将表单数据发送回与IF结合的PHP文件（'POST'=== $ _SERVER ['REQUEST_METHOD']）允许我排除PHP，直到我点击提交按钮。

这可能不是正确的方法，但因为我使用XAMPP它的工作原理。但是我知道如果使用传统的Web服务器使用单独的PHP和HTML文件效果很好。到目前为止，我还没有在托管公司完全建立开发环境。暂时不得不这样做。

PHP无法运行XAMPP的POST错误

1 个答案: