Cycling through POST array for Update

lerro

Can someone help me with this issue.

I have retrieved data from the database

Then i have displayed it on the screen within a form for the user to select an option

I read back the input via the $_POST array and then check what the selection is through each item that is in the form.

If the user selected a given option, then the specified field is updated with a given value, else, if it's another option, then a different value input during the update.

The problem i'm having is that when i try to carry out the UPDATE, it seems not to work as nothing is updated in the database.

This statement in my code seems to be flawed

$sql = "UPDATE users
SET option = 'Accept' WHERE id = '".$info['id']."' ";

Below is the code, can someone help?

<?php

include_once('db_connect.class.php');

$db = new DbConnect();

// get data from database:

$output = mysql_query("SELECT id, firstname, surname, option FROM users") or die(mysql_error());

// Display information
Print '<form action="' . $SERVER['PHP_SELF'] . '" method="POST">';
while($info = mysql_fetch_array($output))
{
Print "<table border cellpadding=1 >";
Print "<tr>";
Print "<td>".$info['id'] . "</td> ";
Print "<td>".$info['firstname'] . "</td> ";
Print "<td>".$info['surname'] . "</td> ";
Print "<td>".$info['users'] . "</td> ";
Print '<td><label><input type="radio" name="Choice'.$info['option'].'" value="1" />Yes</label>
<br />
<label><input type="radio" name="Choice_'.$info['option'].'" value="0" />No</label></td>';
Print "</tr>";
Print "</table>";
}
Print '<div align="right"> ';
Print '<input name="Send" type="submit" value="Submit"> ';
Print "</div>";
Print "</form>";

// Submit form with updated information, processed based on value of Choice radio button

while($info = mysql_fetch_array($output))
{
if($POST['Choice'.$info['option']] == 1){
//set app_status = approved, insert specified field values including app_no into customer table
$sql = "UPDATE users
SET option = 'Accept' WHERE id = '".$info['id']."' ";
mysql_query($sql);
} else {
// Refused
// set app_status = refused
}
}

?>

This is the problem i have

halojoy

The problem may be you loop mysql result 2 times.

Every loop will move the pointer to next row in result.
At the end of while loop the pointer is at the end.
If you try to loop again, then there is no more row.

The solution is to put the pointer at first row in result.
This row has got number=0
I think this is done in mysql using [man]mysql_data_seek[/man]

mysql_data_seek() moves the internal row pointer of the MySQL result associated with the specified result identifier to point to the specified row number. The next call to a MySQL fetch function, such as mysql_fetch_assoc(), would return that row.

I suggest you try to add this row

// Submit form with updated information, processed based on value of Choice radio button

mysql_data_seek($output, 0); // Reset result pointer
while($info = mysql_fetch_array($output))
{
if($_POST['Choice_'.$info['option']] == 1){

lerro

I tried running this code

// Submit form with updated information, processed based on value of Choice radio button

mysql_data_seek($output, 0); // Reset result pointer
while($info = mysql_fetch_array($output))
{
if($POST['Choice'.$info['option']] == 1){

But i get the following error

Notice: Undefined index: Choice_ in C:.....

Even when i try to remove 'Choice_', the error still happens, showing "Undefined index"

How do i resolve this?

halojoy

$_POST[Choice_xxxx] is defined AFTER you submit.
Before this has no value. Must post a value.

So, we can put in one test, to see if was $POSTed, Submitted.
Like this:

if(isset($_POST['Send'])){ //If form submitted

// Submit form with updated information, processed based on value of Choice radio button
mysql_data_seek($output, 0); // Reset result pointer 
while($info = mysql_fetch_array($output))
{
	if($_POST['Choice_'.$info['option']] == 1){
		//set app_status = approved, insert specified field values including app_no into customer table
		$sql = "UPDATE users
		SET option = 'Accept' WHERE id = '".$info['id']."' ";
		mysql_query($sql);
		echo 'Was UPDATED!';
	} else {
		// Refused
		// set app_status = refused
	}
}

}

lerro

I have tried implementing suggestions but still have problems. Please look at this code

There are two main problems
1 - i can't get the data from $POST to be processed into the database
2 - i identified what might be a flaw, if each field being printed out for radio button selection has

name="Choice_'.$info['app_status'].'"

as the field name, then it means the form does not have unique names for each form item. That means the radio button cannot be selected for each entry but only once on the form.

How do i resolve these two problems so that
- Data is retrieved
- Data is displayed on the screen with choice for user to select an option for each data record
- Data is read back
- Data is processed based on user selection and then submitted to the database to update the specified field for each record as defined in the sql query

include_once('../../classes/db_connection.class.php');

</head>

<body>

$db = new DatabaseConnection();

// get the row's primary key from the database so as to update that specific row:

$output = mysql_query("SELECT app_no, app_date, app_firstname, app_status FROM application") or die(mysql_error());


// Display information
Print '<form action="' . $_SERVER['PHP_SELF'] . '" method="POST">'; 
while($info = mysql_fetch_array($output))
{
Print "<table border cellpadding=5 >";

Print "<tr>";		
Print "<th>APP NO:</th>"; 
Print "<th>APP DATE:</th>"; 
Print "<th>FIRSTNAME:</th>"; 
Print "<th>PROCESS:</th>"; 
Print "</tr>";


Print "<tr>";
Print "<td>".$info['app_no'] . "</td> ";
Print "<td>".$info['app_date'] . "</td> ";
Print "<td>".$info['app_firstname'] . "</td> ";
Print '<td><label><input type="radio" name="Choice_'.$info['app_status'].'" value="0" />Approved</label>
<br />
<label><input type="radio" name="Choice_'.$info['app_status'].'" value="0" />Declined</label></td>';
Print "</tr>";
Print "</table>"; 
}
Print '<div align="right"> ';
Print '<input name="Send" type="submit" value="Submit"> '; 
Print "</div>";
Print "</form>";


if(isset($_POST['Submit'])){ //If form submitted 

// Submit form with updated information, processed based on value of Choice radio button 
mysql_data_seek($output, 0); // Reset result pointer 
while($info = mysql_fetch_array($output)) 
{ 
    if($_POST['Choice_'.$info['app_status']] == 1){ 
        //set app_status = approved, insert specified field values including app_no into customer table 
        $sql = "UPDATE application 
        SET app_status = 'Approved' WHERE app_no = '".$info['app_no']."' "; 
        mysql_query($sql); 
        echo 'Was UPDATED!'; 
    } else { 
        // Refused 
        // set app_status = refused 
    } 
} 

}

</body>
</html>

halojoy

Isnt $info['app_no'] unique for each row?

So you could use something like this:

// NOTICE!!! you have value=0 for both butttons .......

Print '<td><label><input type="radio" name="Choice_'.$info['app_no'].'" value="0" />Approved</label>
<br />
<label><input type="radio" name="Choice_'.$info['app_no'].'" value="0" />Declined</label></td>';
Print "</tr>";

lerro

It's unique for each row, i tried but it still fails to work. There seems to be a flaw in the logic.

every steps seems fine, but still data is not input into the database

Data is retrieved from the database as an array - that's okay
It's displayed on the screen - that's okay

The issue begins with the radio buttons, each now has a unique name using the appended app_no which is unique

Then person submits

A check is made to see if submit is clicked
Then the $output array pointer is reset to zero
Then a looop is made through the array
if $_POST field name == 1
then sql code is run to set the specified field to "Approved" based on "app_no" field value in database being equal to "app_no" value in the array being currently looped through

But still, it doesn't work. There is no database update and even the echo "WAS UPDATED" message doesn't show up, which means the "if" part doesn't run so the failure is before the "if" statement

Might this difference in quotation marks in the form

name="Choice_'.$info['app_no'].'"

and this difference in the if $_POST clause

['Choice_'.$info['app_no']]

be a contributing factor?

lerro

instead of submitting to database, i tried outputting to screen to monitor what happens

It seems that the radio button option issn't working correctly.

When i set to check whether == 1
Then what is echoed out is "Was not updated"

When i se to check whether == 0
The what is echoed out is "Was updated"

And that is regardless whether i have selected the radio button approved or Declined. It ignores that

Are you able to figure out what causes this flaw

$db = new DatabaseConnection();

// get the row's primary key from the database so as to update that specific row:

$output = mysql_query("SELECT app_no, app_date, app_firstname, app_status FROM application") or die(mysql_error());


// Display information
//Print "<form  action="..........."  method="POST" >";
Print '<form action="' . $_SERVER['PHP_SELF'] . '" method="POST">'; 
while($info = mysql_fetch_array($output))
{
Print "<table border cellpadding=5 >";

Print "<tr>";		
Print "<th>APP NO:</th>"; 
Print "<th>APP DATE:</th>"; 
Print "<th>FIRSTNAME:</th>"; 
Print "<th>PROCESS:</th>"; 
Print "</tr>";


Print "<tr>";
Print "<td>".$info['app_no'] . "</td> ";
Print "<td>".$info['app_date'] . "</td> ";
Print "<td>".$info['app_firstname'] . "</td> ";
Print '<td><label><input type="radio" name='.$info['app_no'].' value="0" />Approved</label>
<br />
<label><input type="radio" name='.$info['app_no'].' value="0" />Declined</label></td>';

Print "</tr>";
Print "</table>"; 
}
Print '<div align="right"> ';
Print '<input name="Send" type="submit" value="Submit"> '; 
Print "</div>";
Print "</form>";


if(isset($_POST['Send'])){ //If form submitted 

// Submit form with updated information, processed based on value of Choice radio button 
mysql_data_seek($output, 0); // Reset result pointer 
while($info = mysql_fetch_array($output)) 
{
       if($_POST[$info['app_no']] == 0){
        //set app_status = approved, insert specified field values including app_no into customer table 
        //$sql = "UPDATE application SET app_status = 'Approved' WHERE app_no = '".$info['app_no']."' "; 
        //mysql_query($sql); 
        echo 'Was UPDATED!'; 
    } else { 
        // Refused 
        // set app_status = refused
         echo 'Was not updated TEST'; 
    } 




 }


}