donglei7152 2013-09-12 12:04
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表单提交后,PHP会话数据清除

I'm storing session data in $relnum variable, as shown below:

if (!isset($_SESSION)) {
    session_start();
    $_SESSION['releasen'] =$_POST['release_no'];
    $relnum=  $_SESSION['releasen']; 
}

And displaying it in form text field, as shown below;

<input name="relnu" id="relnu" type="text" value="<?php if ($rel==''){ echo $relnum;} else echo $rel; ?>" readonly="true"/>

I am submitting above Form data in mysql, as shown below;

if (isset($_POST['submitM']))  {
    $faultd=$_POST['faultdistribution'];
    $faultdes=$_POST['faultdescription'];
    $faultsev=$_POST['faultseverity'];
    $faultt=$_POST['faulttype'];
    $faultn=$_POST['faultcmnt'];
    $rel=$_POST['relnu'];
    $query = mysql_query("INSERT INTO `fault` (`fault-cmnt`,`fault-type`, `release_no`, `fault-discription`, `fault-severity`, `fault-distribution`) VALUES ('$faultn','$faultt', '$rel', '$faultd', '$faultsev', '$faultdes')")
        or die(mysql_error());
    echo "Data Added sucessfully";
}

The data is successfully submitted, but after that $relnum variable displays nothing. I am unable to comprehend what is the reason, as i am storing the data in session.

P.S: I am not using unset.

Please help, what i am missing?

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5条回答 默认 最新

  • doujin4031 2013-09-12 12:18
    关注

    You have no need to post the session data in the form, else your code seems fine. You can do it easy way with posting it in form like below:

    if (isset($_POST['submitM']))  {
        session_start();
        $faultd=$_POST['faultdistribution'];
        $faultdes=$_POST['faultdescription'];
        $faultsev=$_POST['faultseverity'];
        $faultt=$_POST['faulttype'];
        $faultn=$_POST['faultcmnt'];
        $rel=$_SESSION['releasen'];
        $query = mysql_query("INSERT INTO `fault` (`fault-cmnt`,`fault-type`, `release_no`, `fault-discription`, `fault-severity`, `fault-distribution`) VALUES ('$faultn','$faultt', '$rel', '$faultd', '$faultsev', '$faultdes')")
            or die(mysql_error());
        echo "Data Added sucessfully";
    }
    

    it will work fine, please test it.

    Thank you

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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