Question

我知道这里被问了一百万次，但是我没有得到它的生命。我在php页面中有一个复选框，代码如下：

CODE UPDATED BELOW

这是一个简单的开/关开关。它应该将checked或not的值发送到php表单onff-switch.php，以便表单可以使用以下代码发布到MySql表：[UPDATED]

CODE UPDATED BELOW

第一个问题是，如果我使用$ con，我将通过终端收到错误。它告诉我有语法错误，即使我已经用w3schools.com进行了双重检查。

除此之外，当选中该框时，该值应为ON，如果未选中，则值为OFF，并且应该在数据库表中为UPDATEd。

更新：我已将所有元素移动到一个页面，如下所示，但仍然没有发布到数据库：

<?php
$con=mysqli_connect("localhost","USER","PASS","DB");
// Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if (isset($_POST['formSubmit']))
{
    // escape variables for security
    $scanningvalue = mysqli_real_escape_string($_POST['onoffswitch']);

    $sqlon="INSERT INTO configuration (scanning) VALUES ('$scanningvalue')";
    $result=mysqli_query($con, $sqlon);
    if($scanningvalue=='') $scanningvalue='off';
       if($result)
    {
       echo "The db operation done (1 record added) and switch value ".$scanningvalue;
    }
else
{
   echo "The db operation error and switch value ".$scanningvalue;
}
}
mysqli_close($con);
?> 

<form action="" method="post">
<div class="onoffswitch">
    <input type="checkbox" 
               name="onoffswitch" 
               class="onoffswitch-checkbox" 
               id="myonoffswitch" 
               value="on"
              checked
               >
    <label class="onoffswitch-label" for="myonoffswitch">
        <div class="onoffswitch-inner"></div>
        <div class="onoffswitch-switch"></div>
    </label>
</div>

</form>

Answer 1

改变这个 -

$scanningonoff = mysqli_real_escape_string($_POST['onoffswitch']);

$sqlon="INSERT INTO configuration (scanning) VALUES ('$scanningon')";

错别字 -

1) $con=mysqli_connect("localhost","config","password","mysql_db"));
                                                                  ^ extra bracket

2)   if (isset($_POST['onoffswitch'])) {
        {                              ^  extra bracket
         mysqli_query($con,$sqlon);
        }
    else {
        mysqli_query($con,$sqloff);
    }

Answer 2

对于Db Connection事项，你最后会放置双括号。删除它

$con=mysqli_connect("localhost","config","password","mysql_db");

您可以通过名称而不是通过其值来调用元素，因此它将是＆＃39; onoffswitch＆＃39;使用$ _POST调用

检索数据时

$scanningonoff = mysqli_real_escape_string($_POST['onoffswitch']);

Answer 3

试试这个

<?php
$con=mysqli_connect("localhost","config","password","mysql_db");
// Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}



if (isset($_POST['formSubmit']))
{
// escape variables for security
$scanningvalue = mysqli_real_escape_string($_POST['onoffswitch']);

$sqlon="INSERT INTO configuration (scanning) VALUES ('$scanningvalue')";
$result=mysqli_query($con,$sqlon);
if($scanningvalue=='') $scanningvalue='off';
if($result)
{
   echo "The db operation done (1 record added) and switch value $scanningvalue";
}
else
{
   echo "The db operation error and switch value $scanningvalue";
}
}
mysqli_close($con);
?>

复选框无法正常工作

3 个答案: