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up vote -1 down vote favorite

Friends,

I'm a newbie to PHP.

I've had a problem to deal with that I couldn't understand, so I posted it in this thread.

I've dynamically created 2 textboxes and a button.

  1. Question ID text field

  2. Question text field

  3. Change Button

for the change button I need to write a 'onclick' javascript to pass Question ID

and Question value to a PHP function (set_id) written inside the Same file. In fact that’s why i

Called Form action $_SERVER[“PHP_SELF”].

Here’s my code.

<html>
<head>
<script>


function getvalue(value)
{
    var qid_value = 'qid_'+value.substring(4);

    alert('QID = '+ document.getElementById(qid_value).value + ' QUESTION = ' + document.getElementById(value.substring(4)).value);

/*
I created this javascript alert to test the id s of textboxes and their values
*/

}
</script>
</head>

<body>

<form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post">

<!-- These fields are dynamically created -->

<input type="text" id="'.$var_id.'" name="'.$var_id.'" value="'.$row['qid'].'" readonly size="2 px"/>

<input type="text" id="'.$var_question.'" name="'.$var_question.'" value="'.$row['question'].'" style="size:auto"/>

<input type="button" id="'.$var_question.'" name="'.$var_question.'" value="Change" onclick="getvalue(this.name)"/> 

<!-- These fields are dynamically created -->

</form>

</body>
</html>

<?php

    $msg= "";

    function display($qid,$question)
    {
           require('uni_db_conn.php'); // this is my db connection          
           $qid = $_POST[$qid];       
           $question= $_POST[$question];
           $query = "UPDATE question SET question='.$question.' WHERE qid='.$qid.'";
           $result = mysql_query($query);

           if(!$result)
           {
            $msg= 'Cant Insert Values to the Table !'.mysql_error();
           }
           else
           {
            $msg = 'Successfully Added to the Table !';
           }

           echo '<label>'.$msg.'</label>';
    }


    function set_id($qid,$question)
    {
        if(isset($_POST[$question]))
        {
           display($qid,$question);
        } 
    }

?>

Thank You ! Sorry If there was any mistake.

share|improve this question
    
You can put your PHP code in a separate file, and use AJAX to send data to it and it will return you the values, without your page being refreshed. – I Can Has Kittenz Mar 18 '14 at 6:09
up vote 1 down vote accepted

Try this code

<?php
if(isset($_POST['submit'])){
$QID = $_POST["qid"];
$QUE = $_POST["question"];
echo $QID;
echo $QUE;
}
?>

<html>
<head>
<script language="javascript">


function getvalue()
{
var valid= true;
var id = document.getElementById("ID").value;
var ques = document.getElementById("ques").value;
return valid;
}
</script>
</head>
<body>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" onSubmit=" return getvalue()" >

<input type="text" id="ID" name="qid"/>
<input type="text" id="ques" name="question"/>
<input type="submit" name="submit" value="Change"/> 

</form>
</body>
</html>
share|improve this answer

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