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How to display current selection in a select box

I have the following php form which is linked to a mysql database

<?php

$sql="SELECT `fldTransactionId` , DATE_FORMAT( `fldDate` , '%d %b %Y' ) , `fldWithdrawingAccount` , `fldSalesPerson` ,`fldInvoice` , `fldAccount` FROM `tbTransactions` LEFT JOIN `tbAccounts` ON `fldAccountID` = `fldWithdrawingAccount` WHERE `tbTransactions`.`fldtype` =2";

 $j=$_GET['nextrow'];

switch ($_GET['recnav']){

  case "Prev":
      $j=$j-1;
      break;

 case "Next":

  $j=$j+1;
  break;

  case null:
   $j=$j+1;
   break;

  }
  $result=mysql_query($sql);

  switch($j){

      case 0:
      case 1:

            $msg= "This is the first record. You can't go back any further";
            $j=1;
            break;

      case mysql_num_rows($result):

            $msg= "This is the last record";
            break;

      case mysql_num_rows($result)+1:
            $j=mysql_num_rows($result);
            $msg= "This is the last record";
            break;

  //Main form starts here

  echo "<form class='mainform' method='GET' action='../pages/pgMain.php'>";

      echo "<input type='hidden' name='nextrow' value='".$j."'>".$msg."<br/>";
      echo "<input type='hidden' name='navbutton' value='Expenses'>";

      echo "<div>Date:</div><input value='".mysql_result($result, $j-1,1)."' ><br/>";
       echo "<div>Withdrawing Account:</div><input value='".mysql_result($result, $j-1,5)."' ><br/>";

      include "../queries/qryAccounts.php";

      $WithdrawingAccountssql="Select `tbTransactions`.`fldWithdrawingAccount`,`tbAccounts`.`fldAccount` From `tbAccounts` LEFT JOIN `tbTransactions`  ON `tbAccounts`.`fldAccountID`=`tbTransactions`.`fldWithdrawingAccount`";

      $Accounts=mysql_query($WithdrawingAccountssql);

      while ($AccountsRow = mysql_fetch_array($Accounts)){

      $AccountsOption=$AccountsOption."<option VALUE=".$AccountsRow[0].">". $AccountsRow[1]."</option>";

      }

      echo "<div>Withdrawing Account:</div><select>".$AccountsOption."</select><br/>";
      echo "<div>Sales Person:</div><input value='".mysql_result($result, $j-1,3)."' ><br/>";

      echo "<div class='navbuttons'>"
      echo "<input type='submit' name='recnav' value='Prev' />"

      echo "<input type='submit' name='recnav' value='Next' />"

</div>
  echo "</form>

?>

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I want the select box on line 66 to show the data in the table when they are available and allow me to edit that data. At present it only shows the first option and does not seem to be linked to the data.

So the question is how do I make the select box show the data for that record and allow me to update.

Thanks

SOLUTION

sivagnanam chandrakanth

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Sheils

ASKER

The form moves through a number of records. So the selected cannot be static. Are you suggesting that I use some more code to move the selected attribute to the relevant option each time I move to a new record?

ASKER CERTIFIED SOLUTION

sivagnanam chandrakanth

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Sheils

ASKER

This code works when there is data in the field. But all the option disappears when there is no data in the field.

while ($AccountsRow = mysql_fetch_array($Accounts)){
      if($AccountsRow[1]=mysql_result($result, $j-1,5)){

      $selected=" selected ";

      }


      $AccountsOption=$AccountsOption."<option".$selected."VALUE=".$AccountsRow[0].">". $AccountsRow[1]."</option>";

      }

      echo "<div>Withdrawing Account:</div><select>".$AccountsOption."</select><br/>";

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I believe that it has to do with spacing but I just can't get it right.

Sheils

ASKER

I have tried the following if..else statement to no avail

if($AccountsRow[1]=mysql_result($result, $j-1,5)){

$selected=" selected ";

}

else{
$selected=" ";
}

sivagnanam chandrakanth

add $selected=""; in else

while ($AccountsRow = mysql_fetch_array($Accounts)){
if($AccountsRow[1]=mysql_result($result, $j-1,5)){

$selected=" selected ";

}else{
$selected="";
}

sivagnanam chandrakanth

What do you mean by this //But all the option disappears when there is no data in the field.//

Can you explain more

Sheils

ASKER

Chaged as follows still no luck

while ($AccountsRow = mysql_fetch_array($Accounts)){
$selected="";
if($AccountsRow[1]=mysql_result($result, $j-1,5)){

$selected=" selected ";

}
else{
$selected="";
}

$AccountsOption=$AccountsOption."<option ".$selected." VALUE=".$AccountsRow[0].">". $AccountsRow[1]."</option>";

}

see output on: http://www.wowislandcharter.com/wowaccounts/pages/pgMain.php?nextrow=4&navbutton=Expenses&recnav=Prev

Sheils

ASKER

It looks like the while loop is not running if the if statement is false. I also notice that when is does work the 4 options are the same.

SOLUTION

sivagnanam chandrakanth

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sivagnanam chandrakanth

Did you check if condition?

Sheils

ASKER

ok that fixed it. Thanks a lot mate