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Update Mysql Data with PHP Code not working
Posted on 2011-03-08
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> </td>
<td colspan="3"><strong>Update data in mysql</strong> </td>
</tr>
<tr>
<td align="center"> </td>
<td align="center"> </td>
<td align="center"> </td>
<td align="center"> </td>
</tr>
<tr>
<td align="center"> </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> </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> </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> </td>
</tr>
</table>
</td>
</form>
</tr>
</table>
<?
// close connection
mysql_close();
?>
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3
2
6 Comments
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.
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.
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
http://www.sitepoint.com/books/phpmysql4/
HTH, ~Ray
Active today
Your query is not returning anything. Try this on line 19
$row_num = mysql_num_rows($result);
echo "number of rows returned $row_num";
$row_num = mysql_num_rows($result);
echo "number of rows returned $row_num";
@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
Man page here (see the return values):
http://us.php.net/manual/en/function.mysql-query.php
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
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
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