[SOLVED] changing data in an autopopulating form

[matthewst]

I have a form which is autopopulated. I need my users to be able to make changes to it when necessary. The problem is my users can change whatever needs to be changed but when they hit submit it doesn't change anything in the database.

<form name="FormName" action="<?=$PHP_SELF;?>" enctype="multipart/form-data" method="post">

$query="SELECT * FROM abc_tables WHERE table_id=$table_id";
$result=mysql_query($query);
while ($row = mysql_fetch_assoc($result))
{
$rest_name = $row['rest_name'];

<input type="text" class="formTextbox" name="rest_name" size="24" value="<?="$rest_name"?>">

<input type="submit" name="submit" class="formTextbox" value="Submit">

[pocobueno1388]

Well...you don't have any code telling it to update the DB...you need something like this at the top of your script:

 

<?php
if ($_POST['submit']){
   $rest_name = $_POST['rest_name'];
   mysql_query("UPDATE tbl SET rest_name='$rest_name' WHERE condition");
   
    echo 'Updated Successfully';
}
?>

 

also as your form action you have:

$PHP_SELF

 

It should be:

$_SERVER['PHP_SELF']

 

You also need to end your while loop, and use echo to display the HTML in between, or close the PHP tags.

[matthewst]

I've got all the quotes and everything that was just a snippet so you would have an idea of what I was talking about.

Heres what I have now:

At the top:

function update_db()
{
if ($_POST['submit']){
$contact_fname = $_POST['contact_fname'];
mysql_query("UPDATE abc_tables SET contact_fname='$contact_fname'");
}}

 

Then:

<input type="text" class="formTextbox" name="contact_fname" size="24" value="<?="$contact_fname"?>">

 

At the bottom:

<input type="submit" name="submit" class="formTextbox" value="Submit" onsubmit="<?=update_db();?>">

 

But it still dosen't update the database.

[pocobueno1388]

On this line:

mysql_query("UPDATE abc_tables SET contact_fname='$contact_fname'");

 

You have no condition, so it is going to update every records contact_fname to that variable, which I am sure you don't want.

 

Why are you putting the code into a function? Take it out of the function and try it.

[matthewst]

here is the current query:

mysql_query("UPDATE abc_tables SET contact_fname='$contact_fname' WHERE table_id=$table_id");

 

 

I have it in a function so it dosen't update when the page loads.

example:

 

a user gets a legal name change and needs to update his info

 

he would load this page (all fields autopopulate)

 

change his name

 

then click submit (onsubmit="<?=update_db();?>")

[pocobueno1388]

Oh, your trying to mix JavaScript and PHP together....I can't help you there.

[matthewst]

we can do it without java, i don't care as long as it works

[penguin0]

did you try an if statement:

<?
if (update_db) {
"code you want to happen";
}
?>
<input type="submit" name="update_db" value="Update">

[matthewst]

Still no good. Here is all the relevent code:

 

 

<?
if (submit) {
"UPDATE abc_tables SET contact_fname=$contact_fname WHERE tabel_id=$table_id";
}
?>
<form name="FormName" action="<?=$_server['PHP_SELF'];?>" enctype="multipart/form-data" method="post">
<?php
$query="SELECT * FROM abc_tables WHERE table_id=$table_id";
$result=mysql_query($query);
while ($row = mysql_fetch_assoc($result))
{
$contact_fname = $row['contact_fname'];
}
?>
<table>
<tr>
<td>Contact First Name</td>
</tr>
<tr>
<td>
<input type="text" class="formTextbox" name="contact_fname" size="24" value="<?="$contact_fname"?>"></td>
</tr>
//////////////////////
<tr>
<td>
<input type="submit" name="submit" class="formTextbox" value="Submit">
</td>
</tr>
</table>
<?
mysql_close();
?>

[pocobueno1388]

<?php

if ($_POST['submit']) {

$table_id = "?"; //You need to give this variable the value of whatever you want it to be that fits the query

mysql_query("UPDATE abc_tables SET contact_fname=$contact_fname WHERE tabel_id='$table_id'");
}

?>

<form name="FormName" action="<?=$_server['PHP_SELF'];?>" enctype="multipart/form-data" method="post">
<?php
$query="SELECT * FROM abc_tables WHERE table_id=$table_id";
$result=mysql_query($query);
while ($row = mysql_fetch_assoc($result))
{
$contact_fname = $row['contact_fname'];
}
?>
<table>
<tr>
<td>Contact First Name</td>
</tr>
<tr>
<td>
<input type="text" class="formTextbox" name="contact_fname" size="24" value="<?="$contact_fname"?>"></td>
</tr>

<tr>
<td>
<input type="submit" name="submit" class="formTextbox" value="Submit">
</td>
</tr>
</table>

[matthewst]

OK so I changed this:

<form name="FormName" action="<?=$_server['PHP_SELF'];?>" enctype="multipart/form-data" method="post">

to this:

<form name="FormName" action="table_order_handler.php" enctype="multipart/form-data" method="post">

and added this:

table_order_handler.php

<?php
error_reporting(0); 
include('include/user_check.php'); 
include('include/db_con.php');

$query="SELECT * FROM abc_tables WHERE table_id=$table_id";
$result=mysql_query($query);
while ($row = mysql_fetch_assoc($result))
{
$rest_name = $row['rest_name'];
}

$query_update="UPDATE abc_tables SET contact_fname=$contact_fname WHERE tabel_id=$table_id";
mysql_query($query_update);

echo "The Table Order Form for $rest_name has been updated.";
?>

 

but still nothing

[matthewst]

As it turns out I needed to add a hidden field to my form

<input type="hidden" name="table_id" value="<?="$table_id"?>">

that way $table_id gets posted to the handler page

 

thanks for all your help!