Control Output Within PHP While Loop

Question

I have a while loop that a query is in and the MySQL query works just fine. However when I try to output my text saying that the query is successful I get:

"Table Succesfully UpdatedTable Succesfully UpdatedTable Succesfully Updated"

I know this I happening becuase I am editing 3 objects, so I'm wondering is there a way to output the string without having it repeat however many objects I change?

Here is the loop:

while ($r3 = mysql_fetch_assoc($result3)) 
{
    $part_num = $r3['part_num'];
    $desc = $r3['desc'];
    $bom_qty = $r3['qty'];
    $need_qty = $qty * $bom_qty;
    //insert the part_num into the temp parts table and the qty
    $resut4 = mysql_query("INSERT INTO `temp_grabtec_parts`(`temp_id`,`part_num`,`desc`,`qty`) VALUES ('$temp_id','$part_num','$desc','$need_qty')");
    if ($resut4) {
        echo 'Table Succesfully Updated';
    } else {
        echo mysql_error();
    }
}

First of all... I think these updates should go in a transaction. Now if you don't want a transaction. Then just take a boolean and keep on ANDing it with each query success. Print according to this boolean. — sarveshseri
– sarveshseri, Commented Jan 28, 2015 at 6:27
About tansactions - tutorialspoint.com/mysql/mysql-transactions.htm,Transactions in php - mysqltutorial.org/php-mysql-transaction — sarveshseri
– sarveshseri, Commented Jan 28, 2015 at 6:28
Hi i have updated my answer and posted a more optimized way to achieve what you want — Veerendra
– Veerendra, Commented Jan 28, 2015 at 6:55

Realitätsverlust · Accepted Answer · 2015-01-28 06:27:24Z

3

while ($r3 = mysql_fetch_assoc($result3)) {
$part_num = $r3['part_num']; $desc = $r3['desc']; $bom_qty = $r3['qty']; $need_qty = $qty*$bom_qty;
   //insert the part_num into the temp parts table and the qty
$resut4 = mysql_query("INSERT INTO `temp_grabtec_parts`(`temp_id`,`part_num`,`desc`,`qty`) VALUES ('$temp_id','$part_num','$desc','$need_qty')");
    if ( $resut4 ) {
        $error = false;
    } else {
            $error = true;
            echo mysql_error();
    }
}
if(!$error) {
   echo "table successful updated"
}

Untested, but should work fine.

answered Jan 28, 2015 at 6:27

Realitätsverlust

3,9532 gold badges27 silver badges46 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Sougata Bose Over a year ago

You should use mysqli or PDO instead of mysql.

Realitätsverlust Over a year ago

@sgt i use AdoDB, but if he is using mysql, I'll give the sample code for mysql, everything else would be not helpful and offtopic. He asked for help because of the echo, not because of the mysql.

Realitätsverlust Over a year ago

@JoeyFeilmeier Glad that i could help! :) Feel free to mark this as accepted answer so other people might find it helpful too.

Realitätsverlust Over a year ago

@sgt you should tell him then, not me. Post the comment below the question, not below one of the answers.

Rajeev Ranjan Over a year ago

better to have this message on stackoverflow whenever its related to mqsql.

Sougata Bose · Accepted Answer · 2015-01-28 06:30:37Z

1

    $bool =false;
    while ($r3 = mysql_fetch_assoc($result3)) {
        $part_num = $r3['part_num']; $desc = $r3['desc']; $bom_qty = $r3['qty']; $need_qty = $qty*$bom_qty;
           //insert the part_num into the temp parts table and the qty
        $resut4 = mysql_query("INSERT INTO `temp_grabtec_parts`(`temp_id`,`part_num`,`desc`,`qty`) VALUES ('$temp_id','$part_num','$desc','$need_qty')");
            if ( $resut4 && $bool==false) {
                echo 'Table Succesfully Updated';
              $bool=true;
            } else if(!$result) {
                    echo mysql_error();
            }
    }

edited Jan 28, 2015 at 6:30

Sougata Bose

31.8k8 gold badges52 silver badges89 bronze badges

answered Jan 28, 2015 at 6:26

habib ul haq

8246 silver badges13 bronze badges

Comments

Ian Thompson · Accepted Answer · 2015-01-28 06:36:31Z

1

$check=true;
while ($r3 = mysql_fetch_assoc($result3)) {
$part_num = $r3['part_num']; $desc = $r3['desc']; $bom_qty = $r3['qty']; $need_qty = $qty*$bom_qty;   
$resut4 = mysql_query("INSERT INTO `temp_grabtec_parts`(`temp_id`,`part_num`,`desc`,`qty`) VALUES ('$temp_id','$part_num','$desc','$need_qty')");
    if (!$resut4 ) {
      $check=false;
      echo mysql_error();
    }
}
if($check)
{
echo "table successful updated";
}

answered Jan 28, 2015 at 6:36

Ian Thompson

1871 silver badge11 bronze badges

Comments

Veerendra · Accepted Answer · 2015-01-28 06:52:28Z

The more optimized way to do this is create the query in the loop and fire it once and not inside the loop again and again.

This way you can reduce the load and the execution will be fast

<?php
$update_count = 0;
$query_part = "INSERT INTO `temp_grabtec_parts`(`temp_id`,`part_num`,`desc`,`qty`) VALUES ";
while ($r3 = mysql_fetch_assoc($result3)) 
{
    $part_num = $r3['part_num'];
    $desc = $r3['desc'];
    $bom_qty = $r3['qty'];
    $need_qty = $qty * $bom_qty;
    //create the query part here in the loop 
    $query_part .= "('$temp_id','$part_num','$desc','$need_qty'),";
}

$query_final = substr($query_part, 0,-1);//remove the extra comma at the end
$resut4 = mysql_query($query_final);
if ($resut4) 
{
    echo 'Table(s) Succesfully Updated';
} 
else
{
    echo mysql_error();
}
?>

Below is the answer with your approach

You can print the message one time at the end with the exact count with the below code

<?php
$update_count = 0;
while ($r3 = mysql_fetch_assoc($result3)) 
{
    $part_num = $r3['part_num'];
    $desc = $r3['desc'];
    $bom_qty = $r3['qty'];
    $need_qty = $qty * $bom_qty;
    //insert the part_num into the temp parts table and the qty
    $resut4 = mysql_query("INSERT INTO `temp_grabtec_parts`(`temp_id`,`part_num`,`desc`,`qty`) VALUES ('$temp_id','$part_num','$desc','$need_qty')");
    if ($resut4) 
    {
        $update_count++;
    } 
    else
    {
        echo mysql_error();
    }
}
echo $update_count.' Table(s) Succesfully Updated';
?>

Kavita · Accepted Answer · 2015-01-28 07:00:30Z

-2

Put if-else part outside the while loop.

  while ($r3 = mysql_fetch_assoc($result3)) {
      $part_num = $r3['part_num']; $desc = $r3['desc']; $bom_qty = $r3['qty']; $need_qty = $qty*$bom_qty;
      //insert the part_num into the temp parts table and the qty
      $resut4 = mysql_query("INSERT INTO `temp_grabtec_parts`(`temp_id`,`part_num`,`desc`,`qty`) VALUES ('$temp_id','$part_num','$desc','$need_qty')") or die(mysql_error());
       }
      if ( $resut4 ) {
      echo 'Table Succesfully Updated';
      } else {
      echo 'Table Update Fail';
      }

edited Jan 28, 2015 at 7:00

answered Jan 28, 2015 at 6:29

Kavita

797 bronze badges

3 Comments

Sougata Bose Over a year ago

What if one of them fails??

Veerendra Over a year ago

If he will put the condition outside the loop the condition will only check the result for the last query in the loop as the variable $result4 will be over written on each loop iteration and by this way he will not get the perfect result

Veerendra Over a year ago

If you are posting an answer you should post it completely and should not post a part in the comment

Collectives™ on Stack Overflow

Control Output Within PHP While Loop

5 Answers 5

5 Comments

Comments

Comments

Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

5 Comments

Comments

Comments

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related