Multiple row insert to table if check box is selected

Question

I am trying to insert multiple rows to a database table if check box is selected. But in my code when I am trying to insert, new rows are inserting based on check box selection. But no data is passing. I need some advice on below code to modify:

<?php
  $db=mysql_connect("localhost","root","");
  mysql_select_db("kkk",$db);
  $qry="select * from  pi";
  $result=mysql_query($qry);
?>

<form action="check.php"   method="post">
  <table>
    <tr>
      <th>A</th>
      <th>B</th>
      <th>C</th>
   </tr>

<?php
while($row=mysql_fetch_array($result))
{
  echo "<tr><td><input type=checkbox name=name[] value='".$row['id']."'>".$row['PI_NO']."</td><td>".$row['CUSTOMER_NAME']."</td><td>".$row['PI_ADDRESS']."</td></tr>";
}
?>

<input type="submit" value="save" id="submit">

<?php
  $db=mysql_connect("localhost","root","");
  mysql_select_db("kkk",$db);

  $name=$_POST['name'];
  foreach($_POST['name'] as $x)
  {
    $qry="INSERT INTO pi (PI_NO, CUSTOMER_NAME, PI_ADDRESS)VALUES ('$PI_NO','$CUSTOMER_NAME','$PI_ADDRESS')";
    mysql_query($qry);
  }     
?>

all these `'$PI_NO','$CUSTOMER_NAME','$PI_ADDRESS'` are undefined. I also don't understand why you're selecting and wanting to insert from the same table. — Funk Forty Niner, May 12 '15 at 05:08
also. prepared statements. use them. `mysql_*`, do not use them. — pala_, May 12 '15 at 05:09

score 1 · Answer 1 · edited May 23 '17 at 11:44

Notes:

You forgot to bind the name of your checkbox using a single tick (')
You used variables in your query which you didn't defined and assigned value with yet
You only passed on the value of name, and did not include the Pi Address and Customer name. I'll be passing them by hidden input using <input type="hidden">.
I'll change the way you check your passed on form by looping them and check them using for() and if()
Use mysql_real_escape_string() before using them in your queries to prevent some of the SQL injections. But better if you consider using mysqli prepared statement rather than the deprecated mysql_*.
Is your post a single file? If it is, you must enclose your query using an isset() to prevent error upon loading the page.
You didn't close your <form>

Here's your corrected while loop:

<?php
while($row=mysql_fetch_array($result))
{
  ?>
    <tr>
      <td>
        <input type="checkbox" name="name[]" value="<?php echo $row['id']; ?>">
        <?php echo $row["PI_NO"]; ?>
        <!-- HERE IS THE START OF YOUR TWO HIDDEN INPUT -->
        <input type="hidden" name="piaddress[]" value="<?php echo $row["PI_ADDRESS"]; ?>">
        <input type="hidden" name="customer[]" value="<?php echo $row["CUSTOMER_NAME"]; ?>">
      </td>
      <td><?php echo $row['CUSTOMER_NAME']; ?></td>
      <td><?php echo $row['PI_ADDRESS']; ?></td>
    </tr>
  <?php
} /* END OF WHILE LOOP */
?>

<input type="submit" value="save" id="submit">
</form> <!-- YOU DID NOT CLOSE YOUR FORM IN YOUR POST -->

And your query:

<?php

  $db=mysql_connect("localhost","root","");
  mysql_select_db("kkk",$db);

  $counter = count($_POST["name"]); /* COUNT THE PASSED ON NAME */

  for($x=0; $x<=$counter; $x++){

    if(!empty($_POST["name"][$x])){
      $PI_NO = mysql_real_escape_string($_POST["name"][$x]);
      $CUSTOMER_NAME = mysql_real_escape_string($_POST["customer"][$x]);
      $PI_ADDRESS = mysql_real_escape_string($_POST["piaddress"][$x]);

      $qry="INSERT INTO pi (PI_NO, CUSTOMER_NAME, PI_ADDRESS) VALUES ('$PI_NO','$CUSTOMER_NAME','$PI_ADDRESS')";
      mysql_query($qry);

    } /* END OF CHECKING THE CHECKBOX IF SELECTED */

  } /* END OF FOR LOOP */

?>

Thanks to all for your valuable advice. I have corrected the problem and understood the error in my coding. is there any way i can display value in text box, change and then send it to table to add. — Ripon, May 12 '15 at 05:41
I have able to do text box insert into database. thanks to all. — Ripon, May 12 '15 at 05:54
I am unable to save the rows i am selecting. it is taking rows 1,2,3..... from the table when save. — Ripon, May 12 '15 at 10:55

score 0 · Answer 2 · answered May 12 '15 at 05:23

Lots of little problems. And some big ones.

as $x){ .. $x is not being used so I assume you just loop for the number of checked boxes.

These have no values: '$PI_NO','$CUSTOMER_NAME','$PI_ADDRESS'

Missing </form>

Not being used: $name=$_POST['name'];

<?php
echo '<form action="check.php"   method="post"><table><tr><th>A</th><th>B</th><th>C</th></tr>';

$db=mysql_connect("localhost","root","");
mysql_select_db("kkk",$db);
$sql = "select `id`,`PI_NO`, `CUSTOMER_NAME` ,`PI_ADDRESS` from  `pi`";
$result=mysql_query($sql);
while($row=mysql_fetch_array($result)){
  echo "<tr><td><input type=\"checkbox\" name=\"name[]\" value=/"$row[0]/"'>$row[1]</td><td>$row[2]</td><td>$row[3]</td></tr>";
}

echo '<input type="submit" value="save" id="submit"></form>';
foreach($_POST['name'] as $x){
  $sql="INSERT INTO pi (`PI_NO`, `CUSTOMER_NAME`, `PI_ADDRESS`)VALUES ('$PI_NO','$CUSTOMER_NAME','$PI_ADDRESS')";
  mysql_query($sql);
}
?>

Multiple row insert to table if check box is selected

2 Answers2

Notes:

Linked