着插入從形式組ID 2，但其在組ID插件1

    <form method="post" action="" class="form-horizontal no-margin well"> 
       <div class="control-group"> 
        <input type="text" name="post" id="post" style=" 
width: 98%; 
height: 50px;"/>     </div> 
       <div style="padding-left:95%!important" class="form-actions no-margin"> 
        <button onClick="window.location.reload(true);" style="height:30px" name="submit" type="submit" class="btn btn-info"> 
        <p>Post</p></button> 
       </div> 
       </form> 
<?php 
$link=mysql_connect("localhost","root","")or die("Can't Connect..."); 
mysql_select_db("project") or die("Can't Connect to Database..."); 
$q="SELECT * FROM post p , member m , groupp g WHERE p.group_id='".$dbgroupid."' and p.m_id=m.m_id ORDER BY p_id DESC"; 
    $queryRessource = mysql_query($q,$link) or die(mysql_error()); 
    $numrows=mysql_num_rows($queryRessource); 
    if($numrows > 0) 
    { 
    while($row = mysql_fetch_array($queryRessource)) { 
    echo"<br><table border=\"1\" width=\"100%\"><tr><td valign=\"top\" style=background:#f4f4f4;;height:100px><b>".$row['username']."</b><br />".$row['post']."</td></tr><tr><td valign=\"top\" style=background:#f4f4f4;;height:100px> 
    <form method=\"post\" action=\"\" class=\"form-horizontal no-margin well\"> 
    <input type=\"text\" name=\"comment".$row['p_id']."\" id=\"comment\" /> 
    <button onClick=\"window.location.reload(true);\" style=\"height:30px\"   name=\"submit".$row['p_id']."\" type=\"submit\" class=\"btn btn-info\">      Comment</button></form></td></tr></table>"; 


$qq="SELECT * FROM comment c , member m WHERE c.p_id='".$row['p_id']."'and c.m_id=m.m_id"; 
$queryRessource1 = mysql_query($qq,$link) or die(mysql_error()); 
$numrows1=mysql_num_rows($queryRessource1); 
if($numrows1 > 0) 
{ 
    echo "<table border=\"1\" width=\"100%\">"; 
while($row1 = mysql_fetch_array($queryRessource1)) 
{echo"<tr><td valign=\"top\" style=background:#f4f4f4;;height:100px>" 
     ."&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<b>".$row1['username']."</b><br />"."&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;".$row1['comment']."</td></tr>"; 
     } 
     echo "</table>"; 
     } 
     if(isset($_POST['submit'.$row['p_id']])) 
    { 
     if(empty($_POST['comment'.$row['p_id']])) 
     { 
      echo '<script language="javascript">'; 
      echo 'alert("You Cant Add Empty Comment")'; 
      echo '</script>'; 
     } 
     else { 
      $p_id=$row['p_id']; 
    $dbmemberid=$_SESSION['Member_ID']; 
    $comment=$_POST['comment'.$row['p_id']]; 
    $link=mysql_connect("localhost","root","")or die("Can't Connect..."); 
    mysql_select_db("project") or die("Can't Connect to Database..."); 
    $Query="INSERT INTO `comment`(`comment_id`,`comment`,`p_id`,`group_id`,`m_id`) VALUES ('','$comment','$p_id','$dbgroupid','$dbmemberid')"; 
    mysql_query($Query,$link)or die (mysql_error()); 
    echo "<meta http-equiv='refresh' content='0'>"; 

    }} 

      } 

    if(isset($_POST['submit'])) 
    { 
     if(empty($_POST['post'])) 
     { 
      echo '<script language="javascript">'; 
      echo 'alert("You Cant Add Empty Post")'; 
      echo '</script>'; 
     } 
     else { 
    $post=$_POST['post']; 
    $link=mysql_connect("localhost","root","")or die("Can't Connect..."); 
    mysql_select_db("project") or die("Can't Connect to Database..."); 
    $query="INSERT INTO `post` VALUES ('','$post','$dbmemberid','$dbgroupid')"; 
    mysql_query($query,$link)or die (mysql_error()); 
    echo "<meta http-equiv='refresh' content='0'>"; 

    }} 
    ?>

我的問題，我用$ _GET獲得小組會話ID，但是當我試圖在第2組中插入（GROUP_ID = 2）它的在數據庫didnt插入和它只是插入數據庫是在組ID時1（GROUP_ID = 1）着插入從形式組ID 2，但其在組ID插件1

我的會話設置爲

if(isset($_SESSION['UserName']) && isset($_SESSION['Member_ID'])) 
{ 
    $dbusername=$_SESSION['UserName']; 
    $dbmemberid=$_SESSION['Member_ID']; 
    $dbgroupid=$_GET['group_id']; 

    }

所以我需要做的，讓我的形式在第二個職位組當我插入帖子或評論在組組ID = 1 nu m行回聲每個帖子兩次...

來源

2016-12-02 Sultan Selawe

您確實應該使用mysqli_，因爲mysql已被棄用 –

您也應該清理您插入到數據庫中的變量。 mysqli_real_escape_string（）或addslashes至少。 –

嘗試給你的表單一個動作。即。

<?php $id = $_GET['id']; ?> 

<form method = "POST" action = "current_page.php?id=<?php echo $id; ?>">

來源

2016-12-03 18:58:01

group.php中的相同頁面？id = 1它插入到數據庫，但在id = 2的同一頁面中沒有插入 –

着插入從形式組ID 2，但其在組ID插件1

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