Question

我有这个PHP代码。当我尝试手动导航到此PHP脚本而不提交时，未显示未设置帖子的错误。我做错了什么？

<?php
if (!isset($_POST['submit']));
 {
    $productname=$_POST['productname'];
    //$conn string will go here 
    $result=$conn->query("INSERT INTO products(pname)VALUES('$productname')");
    if($result)
    {
       echo "<font color=\"green\">";
       echo("Successfully Inserted new Products");
       echo"</font>!";
    }
    else
     {
       echo("error");
     }
}
?>

Answer 1

<?php
// semicolon removed
if (isset($_POST['submit'])) {
    $productname=$_POST['productname'];
    //$conn string will go here 
    $result=$conn->query("INSERT INTO products(pname)VALUES('$productname')");
    if($result) {
        echo "<font color=\"green\">";
        echo("Successfully Inserted new Products");
        echo"</font>!";
    }
} else { // else placed correctly ...
    echo("error");
}
?>

Answer 2

通过缩进，我们可以看到您的其他条件位于＆＃34; if (!isset($_POST['submit']))＆＃34; 另外，我认为在if (!isset($_POST['submit']))

之后你不需要分号

<?php
if (!isset($_POST['submit'])); // <== This semicolon shouldn't exist
{
    $productname=$_POST['productname'];
    //$conn string will go here 
    $result=$conn->query("INSERT INTO products(pname)VALUES('$productname')");
    if($result)
    {
        echo "<font color=\"green\">";
        echo("Successfully Inserted new Products");
        echo"</font>!";
    }
    else
    {
        echo("error");
    }
} // <== Need an "else" here
?>

以下是代码应该如何

<?php
if (isset($_POST['submit']))
{
    $productname=$_POST['productname'];
    //$conn string will go here 
    $result=$conn->query("INSERT INTO products(pname)VALUES('$productname')");
    if($result)
    {
        echo "<font color=\"green\">";
        echo("Successfully Inserted new Products");
        echo"</font>!";
    }
    else
    {
        echo "<font color=\"red\">";
        echo("Error when inserting");
        echo"</font>!";
    }
} 
else
{
    echo "error";
}
?>

编辑：

你需要做＆＃34; if (isset($_POST['submit']))＆＃34;而不是＆＃34; if (!isset($_POST['submit']))＆＃34;测试表单是否已提交。我修改了上面的代码。

ISSET检查PHP无法正常工作

2 个答案: