updating multiple rows in mysql using php

ary1080 · November 20, 2010

I am trying to update multiple rows in a mysql db through php but I am getting an error message in this line:

$size = count($_POST['uid']);

My db field names are uid (which autoincrements and does not need to be updated), fname, lname and email fields. Here is a link for the project I need to develophttp://users.cis.fiu.edu/~vagelis/classes/COP5725/project.htm

Here is the content for the first page:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head> 
<style type="text/css"> 

body { 
margin:50px 0px; padding:0px; 
text-align:center; 
font:13px Tahoma,Geneva,sans-serif 
} 

#content { 
width:1000px; 
margin:0px auto; 
text-align:left; 
padding:15px; 
border:1px dashed #333; 
background-color:#eee; 
} 
</style> 
<link rel="icon" href="http://www.mysite.com/favicon.ico" type="image/vnd.microsoft.icon" /> 
</head> 
<body> 
<div id='content'><h3><center>Update Information</center></h3> 
<?php

//create a connection
$connection = mysql_connect("localhost","root","r2d2c3po");

//test the connection
if (!$connection)
{
die("Database connection failed: " . mysql_error());
}
//select database to use
mysql_select_db("music_social_network", $connection);
//setup query
$sql = "SELECT * FROM `user` ORDER BY uid";
//get results from query or die
$result = mysql_query($sql, $connection) or die(mysql_error());
//fetch rows from query
$rows = mysql_fetch_assoc($result);
?>

<?php 

// find out how many records there are to update 
$size = count($_POST['uid']); 

// start a counter in order to number the input fields for each record 
$i = 0; 
print "<table width='100%' border='0' cellspacing='1' cellpadding='0'><tr><td>"; 
// open a form 
print "<form name='namestoupdate' method='post' action='update.php'> 
<table width='100%' border='0' cellspacing='1' cellpadding='1'><tr> 

<td align='center'><strong>User ID</strong></td> 
<td align='center'><strong>First Name</strong></td> 
<td align='center'><strong>Last Name</strong></td> 
<td align='center'><strong>Email</strong></td> 
</tr>\n"; 

// start a loop to print all of the users with their information 
// the mysql_fetch_array function puts each record into an array. each time it is called, it moves the array counter up until there are no more records left 
while ($Update = mysql_fetch_array($result)) { 
print "</tr>\n"; 
// assuming you have three important columns (the index (id), the course name (course), and the book info (bookinfo)) 
// start displaying the info; the most important part is to make the name an array (notice bookinfo[$i]) 
print "<td align='center'><p>{$Update['uid']}</p></td>\n"; 
print "<td align='center'><input type='text' fname='fname[$i]' value='{$Update['fname']}' /></td>"; 
print "<td align='center'><input type='text' size='40' lname='lname[$i]' value='{$Update['lname']}' /></td>\n"; 
print "<td align='center'><input type='text' size='40' email='email[$i]' value='{$Update['email']}' /></td>\n"; 
print "</tr>\n"; 
// add 1 to the count, close the loop, close the form, and the mysql connection 
++$i; 
} 
print "<tr> 
<td colspan='4' align='center'><input type='submit' value='submit' />"; 
print "</td> 
</tr> 
</table> 
</td> 
</tr> 
</form> 
</table>"; 
mysql_close($connection); 
?><br /><br /><div></body> 
</html>

here is the content for the update.php page:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head> 
<style type="text/css"> 

body { 
margin:50px 0px; padding:0px; 
text-align:center; 
font:13px Tahoma,Geneva,sans-serif 
} 

#content { 
width:1000px; 
margin:0px auto; 
text-align:left; 
padding:15px; 
border:1px dashed #333; 
background-color:#eee; 
} 
</style> 
<link rel="icon" href="http://www.mysite.com/favicon.ico" type="image/vnd.microsoft.icon" /> 
</head> 
<body> 
<div id='content'><h3><center>Success! </center></h3> 
<table width='100%' border='0' cellspacing='1' cellpadding='0'><tr><td> 


<table width='100%' border='0' cellspacing='1' cellpadding='1'> 
<tr> 

<td align='center'><strong>User ID</strong></td> 
<td align='center'><strong>First Name</strong></td> 
<td align='center'><strong>Last Name</strong></td> 
<td align='center'><strong>Email</strong></td> 
</tr> 
<?php 

//create a connection
$connection = mysql_connect("localhost","root","r2d2c3po");

//test the connection
if (!$connection)
{
die("Database connection failed: " . mysql_error());
}
//select database to use
mysql_select_db("music_social_network", $connection);
//setup query
$sql = "SELECT * FROM `user` ORDER BY uid";
//get results from query or die
$result = mysql_query($sql, $connection) or die(mysql_error());
//fetch rows from query
$rows = mysql_fetch_assoc($result);




// find out how many records there are to update 
$size = count($_POST['uid']); 

// start a loop in order to update each record 
$i = 0; 
while ($i < $size) { 
// define each variable 
$uid = $_POST['uid'][$i]; 
$fname = $_POST['fname'][$i]; 
$lname = $_POST['lname'][$i]; 
$email = $_POST['email'][$i]; 

// do the update and print out some info just to provide some visual feedback 

$query = "UPDATE `user` SET `fname` = '$fname', `lname` = '$lname', `email` = '$email' WHERE `uid` = '$uid' "; 
mysql_query($sql) or die ("Error in query: $query"); 
print " 

</tr> 
<td align='left'><p>$uid</p></td> 
<td align='left'>$fname</td> 
<td align='left'>$fname</td> 
<td align='left'>$lname</td> 
</tr> 
"; 
++$i; 
} 
mysql_close($connection); 
?> 
<tr> 
<td colspan='4' align='center'> 
</td> 
</tr> 
</table> 
</td> 
</tr> 

</table></div></body></html>

Mod edit:

 . . .

tags added.

Pikachu2000 · November 20, 2010

You say you're getting an error message, but you leave out the most important part: the error message.

I've enclosed your code in the BBCode

 . . .

tags for you. Please use them in the future when posting code.

ary1080 · November 20, 2010

Thanks! The error I'm getting is:

Notice: Undefined index: uid in C:\wamp\www\one.php on line 46

this is line 46 in my code: $size = count($_POST['uid']);

Pikachu2000 · November 20, 2010

That means that $_POST['uid'] is undefined, thus has no value. Where is that value supposed to come from?

ary1080 · November 20, 2010

The uid value is set to autoincrement in the db. I don't want it to be a value that users input nor update. I thought I could use it to count the number of records by the userid(uid) and then use the $size variable for the while loop that followed.

Pikachu2000 · November 20, 2010

A $_POST var, by definition, comes from form data. I suspect this is what you meant to do . . .

$sql = "SELECT COUNT(`uid`) as num FROM `user`";

//get results from query or die
$result = mysql_query($sql, $connection) or die(mysql_error());

//fetch rows from query
$rows = mysql_fetch_assoc($result);

// find out how many records there are to update 
$size = $rows['num'];

ary1080 · November 20, 2010

Ok my error went away, but my while loop does not seem to be functioning. What I am trying to accomplish is for users to see the current data that is stored in the database in text boxes and then allow users to update the information they see in the boxes.

// start a loop in order to update each record 
$i = 0; 
while ($i < $size) { 
// define each variable 
$uid = $_POST['uid'][$i]; 
$fname = $_POST['fname'][$i]; 
$lname = $_POST['lname'][$i]; 
$email = $_POST['email'][$i]; 


// do the update and print out some info just to provide some visual feedback 

$query = "UPDATE `user` SET `fname` = '$fname', `lname` = '$lname', `email` = '$email' WHERE `uid` = '$uid' "; 
mysql_query($sql) or die ("Error in query: $query"); 
print " 

</tr> 
<td align='left'><p>$uid</p></td> 
<td align='left'>$fname</td> 
<td align='left'>$fname</td> 
<td align='left'>$lname</td> 
</tr> 
"; 
++$i; 
}

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updating multiple rows in mysql using php

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