INSERT INTO multiple times should be once

Jenny Dithe · 12-07-2010, 08:03 AM

Hi,

I know I have asked about multiple updates before but this seems to be due to a different issue.

My INSERT INTO is inserting three times in to my db, and I am not sure why?

My Code is:

PHP Code:


			
<?php



include("dbconnect.php");



function check_input($value){



if (get_magic_quotes_gpc()){

    $value=stripslashes($value);

    }

    

if (!is_numeric($value)){

    $value = mysql_real_escape_string($value);

    }

return $value;

}



foreach ($_POST as $key=>$value){

    $_POST[$key]= check_input($value);

    }





    $sql="UPDATE store SET country ='{$_POST['country']}' 

    WHERE PId={$_SESSION['Id']}";



echo $sql . '<br />';



    mysql_query($sql,$con) or die('Error: ' . mysql_error());





    $sql2="INSERT INTO purchases (Id, fullname, pphoto, category, subcategory, body, date)

        VALUES

        ('{$_SESSION['Id']}','{$_SESSION['fullname']}','{$_SESSION['pphoto']}','country','{$_POST['country']}',' ordered from ',NOW())";

echo $sql2 . '<br />';        

    mysql_query($sql2,$con) or die('Error: ' . mysql_error());



    if($_SESSION['country']=='online purchase'){

        $sql3="UPDATE countryepurchases SET {$_POST['ccountry']}=='{$_POST['ccountry']}'

        WHERE PId={$_SESSION['Id']}";

echo $sql3 . '<br />';



    mysql_query($sql3,$con) or die('Error: ' . mysql_error());



        }



?>

Now I worked out that there are three



	mysql_query($sql3,$con) or die('Error: ' . mysql_error());

which is probably the connection, I have changed my $sql, so they are $sql, $sql2, $sql3 so the same naming can't be a problem.

I originall had



	$result=mysql_query($sql) or die('Error: ' . mysql_error());

but I changed that to



	mysql_query($sql/2/3,$con) or die('Error: ' . mysql_error());

in case the three $results were creating the same issue.

I know the problem is probably very simple, but I can't work it out?

And everything is echoing out correctly, e.g. only once?

abduraooft · 12-07-2010, 08:09 AM

Quote:

My INSERT INTO is inserting three times in to my db, and I am not sure why?

Did you mean to the same table purchases?

Jenny Dithe · 12-07-2010, 09:36 AM

Yes.

This part inserts three times:



   $sql2="INSERT INTO purchases (Id, fullname, pphoto, category, subcategory, body, date)

        VALUES

        ('{$_SESSION['Id']}','{$_SESSION['fullname']}','{$_SESSION['pphoto']}','country','{$_POST['country']}',' ordered from ',NOW())";

echo $sql2 . '<br />';        

    mysql_query($sql2,$con) or die('Error: ' . mysql_error());

abduraooft · 12-07-2010, 10:34 AM

Quote:

function check_input($value){

Are you calling your function inside a loop or something like?

Jenny Dithe · 12-07-2010, 11:48 AM

No loop, the only thing on that page that I have omitted is the session information.

My code in my main page is:

Code:

<script type="text/javascript">
function loadXMLDoc(File,ID,Msg){
if (window.XMLHttpRequest)  {
  xmlhttp=new XMLHttpRequest();
  }
else {
	try{
  xmlhttp=new ActiveXObject("Msxml2.XMLHTTP");
  } catch (e) {
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }	
}

xmlhttp.onreadystatechange=function(){
  if (xmlhttp.readyState==4 && xmlhttp.status==200){
    document.getElementById(ID).innerHTML=xmlhttp.responseText;
    }
  }

var params=Msg;
xmlhttp.open("POST",File,true);

xmlhttp.setRequestHeader("Pragma", "Cache-Control:no-cache");
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp.setRequestHeader("Content-length", params.length);
xmlhttp.setRequestHeader("Connection", "close");
xmlhttp.send(params);
}

</script>
</head>
<body>
		<label class="two">Country:</label>
			<form action="" method="post">
			<select name="country"  class="input" required="required">
			<option value="please select"
			selected"selected">Please Select</option>
					<option value="AF|" >Afghanistan</option>
//country list
					<option value="ZW|" >Zimbabwe</option>

			</select>
			&nbsp; &nbsp; 
			<input type="button" value="Update" onclick="loadXMLDoc('update.php','txtHint','country='+encodeURI(document.getElementById('country').value))"; /> 
			</form>
			&nbsp; &nbsp;
				<span id="txtHint"></span>

(update.php is the code posted in my first post).

As you can see this has it's own separate form from any other information.

I do have other parts of the page refreshing but they are contained.

The only possible exception is that in my main page I have information in the session refreshing, but that doesn't include any of the information I use to submit. By refreshing I mean I use setInterval.

The other thing that might help, is that originally I made the mistake of having mysql_query($sql3,$con) or die('Error: ' . mysql_error()); out of the brackets, as below, and when I did that it was inserting the information four times.

PHP Code:


			
    $sql="UPDATE store SET country ='{$_POST['country']}' 
    WHERE PId={$_SESSION['Id']}";

echo $sql . '<br />';

    mysql_query($sql,$con) or die('Error: ' . mysql_error());


    $sql2="INSERT INTO purchases (Id, fullname, pphoto, category, subcategory, body, date)
        VALUES
        ('{$_SESSION['Id']}','{$_SESSION['fullname']}','{$_SESSION['pphoto']}','country','{$_POST['country']}',' ordered from ',NOW())";
echo $sql2 . '<br />';        
    mysql_query($sql2,$con) or die('Error: ' . mysql_error());

    if($_SESSION['country']=='online purchase'){
        $sql3="UPDATE countryepurchases SET {$_POST['ccountry']}=='{$_POST['ccountry']}'
        WHERE PId={$_SESSION['Id']}";
echo $sql3 . '<br />';
}
    mysql_query($sql3,$con) or die('Error: ' . mysql_error());