Duplicates posting to database

raustin

The following script posts form data to a mysql database, but every time the page is refreshed, a blank entry is entered in the db. why is that and how do i fix it?

<H1>Add an Entry</H1>
<form method="POST" action="adminhouse5.php">
<strong>Area:</strong><br>
<span class="box"> </span>
<select name="area" id="area">
<option>Area 1</option>
<option>Area 2</option>
<option>Area 3</option>
<option>Area 4</option>
<option>Area 5</option>
<option>Area 6</option>
<option>Area 7</option>
<option>Area 8</option>
<option>Area 9</option>
<option>Area 10</option>
<option>Area 11</option>
<option>Area 12</option>
<option>Area 13</option>
<option>Area 14</option>
<option>Area 15</option>
<option>Area 16</option>
<option>Area 17</option>
<option>Area 18</option>
<option>Area 19</option>
<option>Other Area</option>
</select>
<br>
<br>
<b>Address:</b><br>
<input name="address" type="text" id="address" size="100">
<br>
<br>
<b>Details:</b><br>
<input name="details" type="text" id="details" size="100">

<br>
<input type="submit" name="submit" value="Submit">
</form>

<?php
mysql_connect("db.xxx.com","xxxn","xxx");
mysql_select_db("xxxx");
$string = "INSERT INTO openhouse (area, address, details) VALUES('$area', '$address', '$details')";
$query = mysql_query($string);
?>
<br>
<table width="100%" border="0" cellpadding="5">
<tr>
<td><h2>Current Listings:
</h2>
<br>
<?

// Delete Row
if ($GET['delete'] > 0) {
$sql = mysql_query("DELETE FROM openhouse WHERE id = '$GET[delete]' ") or die ( mysql_error());
echo "Row Deleted";
}
// Start Table
echo ("<table width=100% border=1 cellpadding=1 cellspacing=1 >
<tr>
<td width=2%>id</td>
<td width=5%>area</td>
<td width=10%>address</td>
<td width=10%>details</td>
<td width=7%>Modify</td>
<td width=5%>Delete</td>
</tr>");

$sql = mysql_query("SELECT * FROM openhouse") or die ( mysql_error() );
while ($row = mysql_fetch_array($sql)) {
echo "<TR>"."<TD>".$row['id']."</TD>"."<TD>".$row['area']."</TD><TD>".$row['address']
."</TD><TD>".$row['details']."</TD>"; // Display Fields
// Display Commands
// Modify
echo "<TD>"."<A HREF=modify.php?id=".$row['id'].">"."Modify"."</a> </TD>";
// Delete
echo "<TD>"."<A HREF=?delete=".$row['id'].">"."Delete"."</a> </TD>";
echo " </TR>";
}
echo "</TABLE>"; // end table
?>
</td>
</tr>
</table>

eoghain

Your problem is here:

<?php
mysql_connect("db.xxx.com","xxxn","xxx");
mysql_select_db("xxxx");
$string = "INSERT INTO openhouse (area, address, details) VALUES('$area', '$address', '$details')";
$query = mysql_query($string);
?>

Everytime your page is loaded the above code gets executed, and it atempts to enter the form information into your database. You probably wanted to do something like this:

<?php
mysql_connect("db.xxx.com","xxxn","xxx");
mysql_select_db("xxxx");
if (isset($_POST['submit'])) { 
   $string = "INSERT INTO openhouse (area, address, details) VALUES('$area', '$address', '$details')";
   $query = mysql_query($string);
}
?>

So that the code to add an entry is only processed if the user hit the submit button.

raustin

Thanks a ton...

I have a modify page set-up to make changes to existing records in the db, but for some reason, it does not change anything; it just re-loads the existing data:

<?
$sql = mysql_query("SELECT * FROM openhouse WHERE id = ('$GET[id]')");
while ($row = mysql_fetch_array($sql))
{
$field_one = $row['area'];
$field_two = $row['address'];
$field_three = $row['details'];
}
?>
<table width="714" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="291"><H1>Modify an Entry</H1></td>
<td width="423"><a href="adminhouse5.php">Add New Listing</a> </td>
</tr>
</table>
<br>
<form name="form1" method="post" action="">
<table width="709" border="0" cellspacing="3" cellpadding="3">
<tr>
<td width="86">Area: </td>
<td width="602"><select name="field_one" size="1" id="field_one">
<option value="<? echo $field_one; ?>"><? echo $field_one; ?></option>
<option>Area 1</option>
<option>Area 2</option>
<option>Area 3</option>
<option>Area 4</option>
<option>Area 5</option>
<option>Area 6</option>
<option>Area 7</option>
<option>Area 8</option>
<option>Area 9</option>
<option>Area 10</option>
<option>Area 11</option>
<option>Area 12</option>
<option>Area 13</option>
<option>Area 14</option>
<option>Area 15</option>
<option>Area 16</option>
<option>Area 17</option>
<option>Area 18</option>
<option>Area 19</option>
<option>Other Area</option>
</select></td>
</tr>
<tr>
<td>Address: </td>
<td><input name="field_two" type="text" id="field_two" value="<? echo $field_two; ?>
" size="100"></td>
</tr>
<tr>
<td>Details: </td>
<td><input name="textfield" type="text" value="<? echo $field_three; ?>" size="100"></td>
</tr>
<tr>
<td colspan="2"><div align="right">
</div>
<div align="right">
<input type="submit" name="Submit" value="Modify Data">
</div></td>
</tr>
</table>
</form>
<?
if ($GET['update'] > 0)
{
mysql_query("UPDATE openhouse SET field_one = ('$POST[field_one]') WHERE id = '$GET[update]'") or die ( mysql_error() );
mysql_query("UPDATE openhouse SET field_two = ('$POST[field_two]') WHERE id = '$GET[update]'") or die ( mysql_error() );
mysql_query("UPDATE openhouse SET field_three = ('$POST[field_three]') WHERE id = '$GET[update]'") or die ( mysql_error() );
mysql_query("UPDATE openhouse SET field_five = now() WHERE id = '$_GET[update]'") or die ( mysql_error() );

echo "Fields Updated";
}
?>
<br>

Any thoughts?

eoghain

This is because you are selecting from the database and displaying your form before you update the record.

If you were to click submit a second time you would see the updated information in your form. But by doing that you would have just updated the database with the old information overwriting the new information you'd entered the time before.

You probably want to move your update code above your select code.

raustin

when i move the code up, then nothing is loaded in the form...

eoghain

You code should probably look like this:

<?
if ($_GET['update'] > 0)
{
mysql_query("UPDATE openhouse SET field_one = ('$_POST[field_one]'),
                                  field_two = ('$_POST[field_two]'),
                                  field_three = ('$_POST[field_three]'),
                                  field_five = now()
              WHERE id = '$_GET[update]'") or die ( mysql_error() );

echo "Fields Updated";
}

$sql = mysql_query("SELECT * FROM openhouse WHERE id = ('$_GET[id]')");
while ($row = mysql_fetch_array($sql))
{
$field_one = $row['area'];
$field_two = $row['address'];
$field_three = $row['details'];
}
?>
<table width="714" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="291"><H1>Modify an Entry</H1></td>
<td width="423"><a href="adminhouse5.php">Add New Listing</a> </td>
</tr>
</table>
<br>
<form name="form1" method="post" action="">
<table width="709" border="0" cellspacing="3" cellpadding="3">
<tr>
<td width="86">Area: </td>
<td width="602"><select name="field_one" size="1" id="field_one">
<option value="<? echo $field_one; ?>"><? echo $field_one; ?></option>
<option>Area 1</option>
<option>Area 2</option>
<option>Area 3</option>
<option>Area 4</option>
<option>Area 5</option>
<option>Area 6</option>
<option>Area 7</option>
<option>Area 8</option>
<option>Area 9</option>
<option>Area 10</option>
<option>Area 11</option>
<option>Area 12</option>
<option>Area 13</option>
<option>Area 14</option>
<option>Area 15</option>
<option>Area 16</option>
<option>Area 17</option>
<option>Area 18</option>
<option>Area 19</option>
<option>Other Area</option>
</select></td>
</tr>
<tr>
<td>Address: </td>
<td><input name="field_two" type="text" id="field_two" value="<? echo $field_two; ?>
" size="100"></td>
</tr>
<tr>
<td>Details: </td>
<td><input name="textfield" type="text" value="<? echo $field_three; ?>" size="100"></td>
</tr>
<tr>
<td colspan="2"><div align="right">
</div>
<div align="right">
<input type="submit" name="Submit" value="Modify Data">
</div></td>
</tr>
</table>
</form>
<br>

raustin

i tried the code you posted, and the form is populated with the listing to be modified, but after i hit 'submit', the form reverts to the original data and does not change anything. I'm stumped...

eoghain

Sorry I wasn't trying to debug your code so I missed the problem.

You are looking for the $GET['update'] variable to tell you that your form was submitted but you are submiting your form via the POST method (<form name="form1" method="post" action="">). Besides the different methods you also don't seem to have a variable in your form called update. So what you probably want is this:

if ($_POST['Submit'] > 0)
{
mysql_query("UPDATE openhouse SET field_one = ('$_POST[field_one]'),
                                  field_two = ('$_POST[field_two]'),
                                  field_three = ('$_POST[field_three]'),
                                  field_five = now()
              WHERE id = '$_GET[update]'") or die ( mysql_error() );

echo "Fields Updated";
}

raustin

same result... any other ideas?

eoghain

Well I'll point out the other mistake I missed, but you should really learn how to spot and fix these things on your own.

The other problem is in your queries where clause. You say: WHERE id = '$GET[update]' $GET variables are the ones passed in the URL (i.e. http://www.mysite.com/submitform.php?update=1). You don't have any GET variables in your URL after you click on your submit button do you?

To fix this problem you can either add in the GET variable to your FORM tag:

<form name="form1" method="post" action="<?php echo "?update=$_GET[update]"; ?>">

Or you can add the update variable as a hidden element in your form and access it in your where clause using $_POST[update] (this is a safer way)"

<form name="form1" method="post" action=""> 
<input type="hidden" name="update" value="<?php echo $_GET[update]; ?>">

When dealing with database modification scripts such as this it is always better to use POSTed variables.