Home/ Questions/Q 7404235
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-29T05:13:38+00:00

ive been trying to make a form with drop downs which are populated from

  • 0

ive been trying to make a form with drop downs which are populated from different tables and then upload that data into another table.

ive cobbled together the following code but when submitting the form the data in the table shows as ‘Array’ and not the selected item from the dropdown, the ‘Qty’ at the end submits fine.

<?php
include ('../connect.php');

if(isset($_POST['submit']))
{

$qty = $_POST['qty'];
$mob_name = $_POST['mob_name'];
$item_img = $_POST['item_img'];
$item_name = $_POST['item_name'];


//save the name of image in table
$query = mysql_query("INSERT INTO tbl_drop VALUES('','$mob_name','$item_img','$item_name','$qty')") or die(mysql_error());


}

$query_Recordset1 = "SELECT mob_name FROM tbl_img ORDER BY mob_name ASC";
$Recordset1 = mysql_query($query_Recordset1) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);

$query_Recordset2 = "SELECT item_name FROM tbl_itm ORDER BY item_name ASC";
$Recordset2 = mysql_query($query_Recordset2) or die(mysql_error());
$row_Recordset2 = mysql_fetch_assoc($Recordset2);
$totalRows_Recordset2 = mysql_num_rows($Recordset2);

$query_Recordset3 = "SELECT item_img FROM tbl_itm ORDER BY item_name ASC";
$Recordset3 = mysql_query($query_Recordset3) or die(mysql_error());
$row_Recordset3 = mysql_fetch_assoc($Recordset3);
$totalRows_Recordset3 = mysql_num_rows($Recordset3);

?>

<html>
<head>
<title></title>
</head>

<body>
<form id="form1" name="form1" enctype="multipart/form-data" method="post" action="new_drop.php">
<label>Mob Name:
<select name="mob_name" id="mob_name">
<?php
do {
?>
<option value="<?php echo $row_Recordset1?>"><?php echo $row_Recordset1['mob_name']?></option>
<?php
} while ($row_Recordset1 = mysql_fetch_assoc($Recordset1));
$rows = mysql_num_rows($Recordset1);
if($rows > 0) {
mysql_data_seek($Recordset1, 0);
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
}
?>
</select>
</label>
<p>
<label>Item Name:
<select name="item_name" id="item_name">
<?php
do {
?>
<option value="<?php echo $row_Recordset2?>"><?php echo $row_Recordset2['item_name']?></option>
<?php
} while ($row_Recordset2 = mysql_fetch_assoc($Recordset2));
$rows = mysql_num_rows($Recordset2);
if($rows > 0) {
mysql_data_seek($Recordset2, 0);
$row_Recordset2 = mysql_fetch_assoc($Recordset2);
}
?>
</select>
</label>
<p>
<label>Item Image:
<select name="item_img" id="item_img">
<?php
do {
?>
<option value="<?php echo $row_Recordset3?>"><?php echo $row_Recordset3['item_img']?></option>
<?php
} while ($row_Recordset3 = mysql_fetch_assoc($Recordset3));
$rows = mysql_num_rows($Recordset3);
if($rows > 0) {
mysql_data_seek($Recordset3, 0);
$row_Recordset3 = mysql_fetch_assoc($Recordset3);
}
?>
</select>
</label>
<p>
<label>Quantity:
<input type='text' name='qty' />
</label>
<p>
<label>
<input type="submit" name="submit" id="submit" value="Submit" />
</label>
</p>
</form>
</body>
</html>
<?php
mysql_free_result($Recordset1);
mysql_free_result($Recordset2);
mysql_free_result($Recordset3);
?>
Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    There is a mistake in your code:

    <option value="<?php echo $row_Recordset3?>"><?php echo $row_Recordset3['item_img']?></option>

    Should be 

    <option value="<?php echo $row_Recordset3['item_img']?>"><?php echo $row_Recordset3['item_img']?></option>

    $row_Recordset3 is an array containing values. So if you are echoing the value of an array ,it will show as array and the value of option block become array.

    so you need to change $row_Recordset3 in option value to $row_Recordset3['item_img']

    Hope this helps 🙂

    • 0