Updating a database using PHP/MySQL

[iorrasnet]

I have a database which I want to update the individual records using PHP / My SQL.

I have got as far as the following :

Listing all the records with an edit link after each record.

Listing the specific record, using the zid, in a form which allows the individual items of the records to be updated.

I can get the edited information into a SQL statement to UPDATE the record in the database. I cannot seem to get the zid to go into the SQL statement

So the SQL reads

$sqlQuery = "UPDATE table SET fieldInInfo = '$cleanInfo' where zid = '$cleanZID'";

When I run this, echoing the sql I get :

UPDATE table SET fieldInfo = 'Brian Cowen' where zid = ''

and it doesnt update the database.

Any thoughts ?

I am a newbie so it is probably something very simple that i just cannot see.

Thanks

 

[louie]

put the id into a hidden field or pass into the querystring.

 

[garycocs]

What louie said,

<input type="hidden" name="zid" value="<?php echo $cleanZID; ?>" />

 

[iorrasnet]

Thanks for your replies.

I had the ZID in a hidden field on the form. The problem I have is getting it from that hidden field into the SQL statement.

In the hidden field of the form displaying the records it is ZID.

It becomes $cleanZID in the function after it is posted from the form and passed through htmlspecialchars.

The coding I used is the same as the coding I am using for the other variables which are getting into the SQL statement correctly, but because the ZID is not being carried, the SQL statement does not know which record to update.

Am I missing something very obvious ?

Thanks again

 

[garycocs]

Could you put up the part of the code where you are using _GET or _POST?

 

[iorrasnet]

$cleanZID= htmlspecialchars ($_POST['recordID']);
$cleanCanditate= htmlspecialchars ($_POST ['electionCanditate']);

recordID is the name of the hidden field.

 

[garycocs]

Cool, you sure recordID is the name of the hidden field??

Just run an echo $cleanZID; after the _POST to see if it actually comes through.

 

[iorrasnet]

Actually it doesnt !??

Now I may have an overly complicated way of doing this so I will walk you through what i am doing.

There is a file called viewrecords.php which list the records and has an edit link.

When you click on the edit link you bring up a file called editrecord.php This file contains the following code :

if (isset($_GET['id'])) {

displayEditRecords (trim($_GET['id']));

}

displayEditRecords () outputs the form allowing you to edit the fields. When you click the submit button it activates the edit.php file. It is this file which has the function that makes the MySql statement and which starts off with the code that I posted earlier.

But echo the $cleanZID is not outputting anything.

 

[garycocs]

Hi,

I don't think the edit form is working right. Could you send us on the code in the form?

Thanks
Gary

 

[iorrasnet]

function output_form_edit ($zid) {
echo "<form action='edit.php' method='post' onSubmit=return validateForm();>";
echo "<input type ='hidden' name='recordID' value = '$strContents'>";

This function is called from within a function called displayEditRecords

 

[garycocs]

Right think that might be it.

function output_form_edit ($zid) {
echo "<form action='edit.php' method='post' onSubmit=return validateForm();>";
echo "<input type ='hidden' name='recordID' value = '$zid'>";


You have to use zid rather than $strContents as the value??

 

[iorrasnet]

Yahoo !!! It worked !!!

Thanks very much !!!!!

Gosh basic error but sure I have to make them to learn.

Thanks again

 

[garycocs]

Glad to hear, best of luck with it anyways, it seems like a long enough way to go about it but I'm sure you're learning as you go along.

 

[iorrasnet]

Yeah, when I wrote it out for you it struck me that it was a bit long winded. Now that i have it working I might tidy it up a bit.

Thanks again