Go Premium for a chance to win a PS4. Enter to Win

x
?
Solved

Update Mysql Data with PHP Code not working

Posted on 2011-03-08
6
Medium Priority
?
348 Views
Last Modified: 2012-08-13
This code is supposed to allow me to view data in a mysql database & then allow me to edit it & submit it updating the database. I'm getting an error "attached in the photo"  when I go to the page in the browser.
<?php
$host="localhost"; // Host name 
$username="uname"; // Mysql username 
$password="password"; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="test_mysql"; // Table name 

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

// get value of id that sent from address bar
$id=$_GET['id'];


// Retrieve data from database 
$sql="SELECT * FROM $tbl_name WHERE id='$id'";
$result=mysql_query($sql);

$rows=mysql_fetch_array($result);
?>
<table width="400" border="0" cellspacing="1" cellpadding="0">
<tr>
<form name="form1" method="post" action="update_ac.php">
<td>
<table width="100%" border="0" cellspacing="1" cellpadding="0">
<tr>
<td>&nbsp;</td>
<td colspan="3"><strong>Update data in mysql</strong> </td>
</tr>
<tr>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
</tr>
<tr>
<td align="center">&nbsp;</td>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>Lastname</strong></td>
<td align="center"><strong>Email</strong></td>
</tr>
<tr>
<td>&nbsp;</td>
<td align="center"><input name="name" type="text" id="name" value="<? echo $rows['name']; ?>"></td>
<td align="center"><input name="lastname" type="text" id="lastname" value="<? echo $rows['lastname']; ?>" size="15"></td>
<td><input name="email" type="text" id="email" value="<? echo $rows['email']; ?>" size="15"></td>
</tr>
<tr>
<td>&nbsp;</td>
<td><input name="id" type="hidden" id="id" value="<? echo $rows['id']; ?>"></td>
<td align="center"><input type="submit" name="Submit" value="Submit"></td>
<td>&nbsp;</td>
</tr>
</table>
</td>
</form>
</tr>
</table>

<?

// close connection 
mysql_close();

?>

Open in new window

mysql-error.JPG
0
Comment
Question by:wantabe2
  • 3
  • 2
6 Comments
 
LVL 111

Accepted Solution

by:
Ray Paseur earned 2000 total points
ID: 35068992
Same issues as the other script.  You need to use data visualization to see the error messages!

Change this code from this:
$result=mysql_query($sql);

... to this:
$result=mysql_query($sql) or die( mysql_error() );

Add the error_reporting(E_ALL) and "display errors" settings to the script and run it again.
0
 
LVL 111

Expert Comment

by:Ray Paseur
ID: 35069009
Also, please let me suggest a really good book to help you get some foundation in how PHP and MySQL work together.  It will not make you a pro, but it has great examples and is very readable.  Also comes with a downloadable code library that you can copy and use for your own apps.
http://www.sitepoint.com/books/phpmysql4/

HTH, ~Ray
0
 
LVL 17

Expert Comment

by:Chris Harte
ID: 35069013
Your query is not returning anything. Try this on line 19

$row_num = mysql_num_rows($result);
echo "number of rows returned $row_num";
0
Get your Disaster Recovery as a Service basics

Disaster Recovery as a Service is one go-to solution that revolutionizes DR planning. Implementing DRaaS could be an efficient process, easily accessible to non-DR experts. Learn about monitoring, testing, executing failovers and failbacks to ensure a "healthy" DR environment.

 
LVL 111

Expert Comment

by:Ray Paseur
ID: 35069154
@Munterman:  Actually the query is returning an error value.  The warning message is pretty clear.  MySQL_Query() returns either a results resource on success or FALSE on failure.  The message says, "... expects parameter 1 to be a resource, boolean given..."  In this case, the boolean is FALSE.

Man page here (see the return values):
http://us.php.net/manual/en/function.mysql-query.php
0
 
LVL 27

Expert Comment

by:Lukasz Chmielewski
ID: 35069197
There are 3 possibilities of this error:
1. The name of the table is for non-existing one
2. There is no ?id=yournumber in the GET variable from the browser addressbar
3. Well, I forgot the 3rd one

Turn on error reporting as Ray told, it is essential to "debug" mysql/php errors
0
 
LVL 27

Expert Comment

by:Lukasz Chmielewski
ID: 35069205
What was the cause ?
0

Featured Post

New feature and membership benefit!

New feature! Upgrade and increase expert visibility of your issues with Priority Questions.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Q&A with Course Creator, Mark Lassoff, on the importance of HTML5 in the career of a modern-day developer.
By, Vadim Tkachenko. In this article we’ll look at ClickHouse on its one year anniversary.
The viewer will learn how to count occurrences of each item in an array.
The viewer will learn how to create a basic form using some HTML5 and PHP for later processing. Set up your basic HTML file. Open your form tag and set the method and action attributes.: (CODE) Set up your first few inputs one for the name and …
Suggested Courses

879 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question