PHP Forum

**gyrene2083** · 11-29-2009, 01:32 PM

Hello All,

I am a total noob, and I have been reading several threads, and I'm still stumped. I am hoping someone can help this 'old Marine, learn something new.

I am trying to create something simple, the Table has the following columns
ID, set to autonumber, Name, PicUrl, Description.

I would like to be able to update a record. For example I need to change the Description, or Name or even add a PicUrl. Below is what I have come up with so far.

Code:

<?php
$dbc = mysqli_connect("localhost","uname","pword","dbase");

$Name = $_POST['Name'];
$query = "UPDATE FROM table SET field1='$Name', field2='$PicUrl', field3-'$Description WHERE id='$_GET(Name)'";
$result = mysqli_query($dbc, $query);
echo 'Anime Updated: ';
mysqli_close ($dbc);
?>

I know I have to create an html form to point to the "Table", which I did, but not sure how to have it show the show the records before, and after the edit.

Any help, anyone could provide would be greatly appreciated.

**davidj** · 07:57 PM

where you have

PHP Code:


			
 WHERE id='$_GET(Name)'";

that is wrong

try

PHP Code:


			
$id= $_GET[id]; //<< your querystring needs to contain.... id={num}



$query = "UPDATE FROM table SET field1='$Name', field2='$PicUrl', field3-'$Description WHERE id='$id'";

i am asuming you are just passing an id and catching it in your GET so make sure you name it correctly

also

you are performing an UPDATE therefore there is nothing to return so you dont need..

PHP Code:


			
$result = mysqli_query($dbc, $query);

this will do

PHP Code:


			
mysqli_query($dbc, $query)

**gyrene2083** · 11-29-2009, 08:04 PM

davidj.

Thanks for the response. Seems I tried the suggestions but the table doesn't seem to be updating. I'm not getting an error msg., but I'm not getting any further either.

Could it be the HTML form?

**lizciz** · 11-30-2009, 02:02 AM

Your query has some errors. Try this

PHP Code:


			
"UPDATE table SET field1='$Name', field2='$PicUrl', field3='$Description' WHERE id=$id"

Also, make sure that all the variables exists, since we can't see them being created in your code above.

**gyrene2083** · 11-30-2009, 07:10 AM

Lizciz,

I tried what you said, and the same thing occurs. Here is what I have after, the modifications, also I'm going to post the HTML form I made. Nothing fancy, just something small.

Code:

<?php
$dbc = mysqli_connect("localhost","username","password","database");

$Name = $_POST['Name'];
$query = "UPDATE Downloads SET field1='$Name', field2='$PicUrl', field3='$Description' WHERE id=$id";  
mysqli_query($dbc, $query);
echo 'Anime Updated: ';
mysqli_close ($dbc);
?>

HTML

Code:

<html>
    <head>
        <title>Anime Update</title>
        <style type="text/css">
        body {
                background-color: #fffff;
                margin-left: 15%;
                margin-right: 15%;
                border: 1px solid gray;
                padding: 10px 10px 10px 10px;
                font-family: baar-sophia;
              }
    </style>
    </head>
    
<body>

    <p>Anime Update</p>
    <form method="post" action="112609Update.php">
    <label for="Name">Anime Name:</label>
    <input type="text" id="Name" name="Name" /><br />
    <label for="PicUrl">Picture Url:</label>
    <input type="text" id="Picture Url" name="PicUrl" /><br />
    <label for="Description">Description:</label>
    <input type="text" id="Description" name="Description" /><br />
    <input type="submit" value="Update" name="submit" />
    </form>
</body>
</html>

Not sure if you need this information, but here goes;
ID - int - autoincrement
Name - Varchar
PicUrl - Varchar
Description - Text

I sincerely appreciate all your help. I even went out and bought a book "Head First - PHP & MySql". I am reading threads, and I've been to W3Schools site as well, trying to understand the code.

**lizciz** · 11-30-2009, 09:49 AM

If that is all the code you use, it's no wonder it doesn't work, since the variables $PicUrl, $Description and $id doesn't exist. Then php will treat them as empty, so your query really says

PHP Code:


			
"UPDATE table SET field1='$Name', field2='', field3='' WHERE id="

You need to get the url and description variables in the same manner as the name, and the id would normally come from a GET-variable, session or another database query, depending on what you're doing (probably a GET-variable though).

**gyrene2083** · 11-30-2009, 04:31 PM

Ok I'm just stumped, and it's not for the sake of not trying. I looked up the GET variable, and changed the html from;

Code:

 <form method="post" action="112609Update.php">

to

Code:

 <form action="112609Update.php" method=GET>

I got that example from here . I also, tried to test out the way they did the .php file and I just kept getting errors. So, at this point, other than throwing my laptop against the wall, I'm hoping you guys can help me out. I appreciate all your patience.