Question

我在将变量传递到第二个php文件时遇到问题。

这是我的checkboxtestbackend.php，我需要从index.php传递两个变量。当我尝试在index.php中使用全局$ variable时，此操作无济于事，如果我尝试制作包含与checkboxtestbacked.php中的变量相同信息的新变量，则结果是相同的：

$checkboxstatus = $_POST['checkboxstatus'];

$host = "localhost";
$user = "root";
$password = "";
$dbname = "odcinki";


$cxn = mysqli_connect($host,$user,$password,$dbname);
if (mysqli_connect_errno()) {echo "Brak polaczenia" . mysqli_connect_error();}





$query = " UPDATE `1`
           SET lastname = '$checkboxstatus'
           WHERE id = '$postID' ";
           
$result = mysqli_query($cxn, $query) or die ("zesrało sie i chuj");

index.php：

<?php
$host = "localhost";
$user = "root";
$password = "";
$dbname = "odcinki";

$conn = mysqli_connect($host,$user,$password,$dbname);
if (mysqli_connect_errno()) {echo "Brak polaczenia" . mysqli_connect_error();}

$postID = get_queried_object_id();
   
   $current_user =  wp_get_current_user();
   
   $current= $current_user->ID ;

$sql = "SELECT lastname FROM `$current` WHERE id = '$postID'";

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
$keepchecked= $row['lastname'];

}







?>

 <?php if ( is_user_logged_in() ) {     ?> Status: <input type="checkbox" name = "seks" value ="1" id="checkboxtest" <?php echo ($keepchecked == "YES") ? 'checked="checked"' : ''; ?> >
<?php
}
?>
<script type="text/javascript">
$(document).ready(function(e) {
$('#checkboxtest').change(function(){
    if( $('#checkboxtest').prop('checked') )
       {checkboxstatus = "YES";}
       else
       {checkboxstatus = "NO";}
   
$.ajax({
        type: "POST",
        url: "checkboxtestbackend.php",
        data: {checkboxstatus: checkboxstatus},
        success: function(result) {
                console.log('the data was successfully sent to the server');
            }
        })
        
});
});
    ?>

Answer 1

我认为您的代码应该是这样的

JsonSerializationReplacer

Y0u需要制作JavaScript变量并将其发布到checkboxtestbackend.php页面

$checkboxstatus = $_POST['checkboxstatus'];
$postID = $_POST['postID'];


$host = "localhost";
$user = "root";
$password = "";
$dbname = "odcinki";


$cxn = mysqli_connect($host,$user,$password,$dbname);
if (mysqli_connect_errno()) {echo "Brak polaczenia" . mysqli_connect_error();}





$query = " UPDATE `1`
           SET lastname = '$checkboxstatus'
           WHERE id = '$postID' ";
           
$result = mysqli_query($cxn, $query) or die ("zesrało sie i chuj");

将变量传递到第二个文件时遇到问题

1 个答案: