SQL插入表不工作

我试图插入状态更新到我的数据库表中称为blabbing，但它不工作。SQL插入表不工作

我的PHP代码

$thisRandNum = rand(9999999999999,999999999999999999); 
$_SESSION['wipit'] = base64_encode($thisRandNum); // Will always overwrite itself each time this script runs 

// ------- POST NEW BLAB TO DATABASE --------- 
$blab_outout_msg = ""; 
if (isset($_POST['status']) && $_POST['status'] != "" && $_POST['status'] != " "){ 

    $blabWipit = $_POST['blabWipit']; 
    $sessWipit = base64_decode($_SESSION['wipit']); 
    if (!isset($_SESSION['wipit'])) { 

    } else if ($blabWipit == $sessWipit) { 

      // End Delete any blabs over 20 for this member 
     $status = $_POST['status']; 
     $status = stripslashes($status); 
     $status = strip_tags($status); 
     $status = mysql_real_escape_string($status); 
     $status = str_replace("'", "&#39;", $status); 
     $sql = mysql_query("INSERT INTO blabbing (mem_id, profile_id, the_blab, blab_date) VALUES('$logOptions_id','$logOptions_id','$status', now())"); 
     $blab_outout_msg = ""; 
     } 

}

我的HTML代码

<div style="background-color:#f2f2f2; border:#ebebeb 1px solid; padding:8px;"> 
    <form action="home.php" method="post" enctype="multipart/form-data" name="blab_from"> 
    <textarea name="status" id="status" rows="3" style="width:99%;"></textarea> 
    <input name="submit" type="submit" class="btn" value="Blab" /> &nbsp; &nbsp; Limit: <script>displaylimit("","status",255)</script> 
    <input name="blabWipit" type="hidden" value="<? print $thisRandNum;?>" /> 
    </form> 
    </div>

任何帮助赞赏

来源

2012-04-01 UnknownUser

什么是e恐怖消息？ – slash197 2012-04-01 09:39:14

没有错误信息，只是因为某种原因没有插入。 – UnknownUser 2012-04-01 09:40:48

您是否打开了错误报告？在mysql调用之前添加'error_reporting（E_ALL）'。 – slash197 2012-04-01 09:42:07

$thisRandNum = rand(9999999999999,999999999999999999); 
$_SESSION['wipit'] = base64_encode($thisRandNum); // Will always overwrite itself each time this script runs 

// ------- POST NEW BLAB TO DATABASE --------- 
$blab_outout_msg = ""; 
if (isset($_POST['status']) && $_POST['status'] != "" && $_POST['status'] != " "){ 

    $blabWipit = $_POST['blabWipit']; 
    $sessWipit = base64_decode($_SESSION['wipit']); 
    // if (!isset($_SESSION['wipit'])) { 

    //} else if ($blabWipit == $sessWipit) { 

      // End Delete any blabs over 20 for this member 
     $status = $_POST['status']; 
     $status = stripslashes($status); 
     $status = strip_tags($status); 
     $status = mysql_real_escape_string($status); 
     $status = str_replace("'", "&#39;", $status); 
     $sql = mysql_query("INSERT INTO blabbing (mem_id, profile_id, the_blab, blab_date) VALUES('$logOptions_id','$logOptions_id','$status', now())"); 
     $blab_outout_msg = ""; 
     // } 

}

试试这个，如果它的工作原理则else if ($blabWipit == $sessWipit) {这个条件是不正确的

来源

2012-04-01 10:19:07 sujal

SQL插入表不工作

回答

相关问题