Update MySQL via PHP. [SOLVED] | DaniWeb

kunyomi 5 Newbie Poster

14 Years Ago

Hey guys, I've just started coding php, for a school project. I'm creating a member login system with functions to add/edit contacts.

Currently I'm stuck at updating the contacts. I'm able to pull up data into a form in a page called 'edit.php', but after updating and clicking submit, it just returns a blank page and the SQL database is not updated.

This is the code from 'edit.php'.

<?
include ("dbConfig.php");
require ("check.php");

if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit")
{
   if (!isset($_POST["submit"]))
   {
      $id2 = $_GET["id2"];
      $sql = "SELECT * FROM contacts WHERE id2=$id2";
      $result = mysql_query($sql);        
      $myrow = mysql_fetch_array($result);
      ?>
	  
      <form action="edit.php" method="post">
      <input type=hidden name="id2" value="<?php echo $myrow["id2"] ?>">
   
      Name:<INPUT TYPE="text" NAME="name" VALUE="<?php echo $myrow["name"] ?>" SIZE=30><br>
      Email:<INPUT TYPE="text" NAME="email" VALUE="<? echo $myrow["email"] ?>" SIZE=30><br>
      Who:<INPUT TYPE="text" NAME="age" VALUE="<?php echo $myrow["age"] ?>" SIZE=30><br>
      Birthday:<INPUT TYPE="text" NAME="birthday" VALUE="<? echo $myrow["birthday"] ?>" SIZE=30><br>
      Address:<TEXTAREA NAME="address" ROWS=10 COLS=30><? echo $myrow["address"] ?></TEXTAREA><br>
      Number:<INPUT TYPE="text" NAME="number" VALUE="<? echo $myrow["number"] ?>" SIZE=30><br>
   
      <input type="hidden" name="cmd" value="edit">
   
      <input type="submit" name="submit" value="submit">
   
      </form>
      
      
<? } ?>
<?
   if ($_POST["$submit"])
   {
   	  $id2 = $_POST["id2"];
      $name = $_POST["name"];
	  $email = $_POST["email"];
	  $age = $_POST["age"];
	  $birthday = $_POST["birthday"];
	  $address = $_POST["address"];
	  $number = $_POST["number"];
	  
	  $sql = "UPDATE contacts SET name='$name',email='$email',age='$age',birthday='$birthday',address='$address',number='$number' WHERE id2=$id2";
      
      $result = mysql_query($sql);
      echo "Thank you! Information updated.";
	}
}
?>

Please help, truly appreciated.

php

2 Contributors
17 Replies
424 Views
8 Hours Discussion Span
Latest Post 14 Years Ago Latest Post by kunyomi

cwarn23 commented: Good question +5

Recommended Answers

Answered by cwarn23 387 in a post from 14 Years Ago

Try replacing lines 33 to 50 with the following

<?php
   if (isset($_POST["$submit"]))
   {
   	  $id2 = mysql_real_escape_string(stripslashes($_POST["id2"]));
      $name = mysql_real_escape_string(stripslashes($_POST["name"]));
	  $email = mysql_real_escape_string(stripslashes($_POST["email"]));
	  $age = mysql_real_escape_string(stripslashes($_POST["age"]));
	  $birthday = mysql_real_escape_string(stripslashes($_POST["birthday"]));
	  $address = mysql_real_escape_string(stripslashes($_POST["address"]));
	  $number = mysql_real_escape_string(stripslashes($_POST["number"]));
 
	  $sql = "UPDATE contacts SET name='$name', email='$email', age='$age', birthday='$birthday', address='$address', number='$number' WHERE id2=$id2"; …

Answered by cwarn23 387 in a post from 14 Years Ago

Does column id2 equal the html form field named id2? That could be a problem if name=id2 doesn't exist in your html form and same with name=submit.

Answered by cwarn23 387 in a post from 14 Years Ago

Try the following.

<?
include ("dbConfig.php");
require ("check.php");

if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit")
{
   if (!isset($_POST["submit"]))
   {
      $id2 = $_GET["id2"];
      $sql = "SELECT * FROM contacts WHERE id2=$id2";
      $result = mysql_query($sql);        
      $myrow = mysql_fetch_array($result);
      ?>
	  
      <form action="<?php $v=explode('?',$_SERVER['PHP_SELF']); echo $v[0]; ?>" method="post">
      <input type=hidden name="id2" value="<?php echo $myrow["id2"]; ?>"> …

Answered by cwarn23 387 in a post from 14 Years Ago

This gives me an error. You took away the
<? } ?>
<?php
at line 32 to 33 right?
It renders the php code below useless. Hmm.
Oops, line 31 needs <?php

Answered by cwarn23 387 in a post from 14 Years Ago

Try the following.

<?php
include ("dbConfig.php");
require ("check.php");

if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit")
{
   if (!isset($_POST["submit"]))
   {
      $id2 = $_GET["id2"];
      $sql = "SELECT * FROM contacts WHERE id2=$id2";
      $result = mysql_query($sql);        
      $myrow = mysql_fetch_array($result);
      ?>
	  
      <form action="<?php $v=explode('?',$_SERVER['PHP_SELF']); echo $v[0]; ?>" method="post">
      <input type=hidden name="id2" value="<?php echo $myrow["id2"]; ?>"> …

All 17 Replies

cwarn23 387 Occupation: Genius

14 Years Ago

Try replacing lines 33 to 50 with the following

<?php
   if (isset($_POST["$submit"]))
   {
   	  $id2 = mysql_real_escape_string(stripslashes($_POST["id2"]));
      $name = mysql_real_escape_string(stripslashes($_POST["name"]));
	  $email = mysql_real_escape_string(stripslashes($_POST["email"]));
	  $age = mysql_real_escape_string(stripslashes($_POST["age"]));
	  $birthday = mysql_real_escape_string(stripslashes($_POST["birthday"]));
	  $address = mysql_real_escape_string(stripslashes($_POST["address"]));
	  $number = mysql_real_escape_string(stripslashes($_POST["number"]));
 
	  $sql = "UPDATE contacts SET name='$name', email='$email', age='$age', birthday='$birthday', address='$address', number='$number' WHERE id2=$id2";
 
      $result = mysql_query($sql) or die(mysql_error());
      echo "Thank you! Information updated.";
	}
}
?>

kunyomi 5 Newbie Poster

14 Years Ago

Hi cwarn23,

I tried doing as you said, but I'm still redirected to a blank page and the database is still not updated.

I've been at this for 2 days but I still can't find what went wrong.

cwarn23 387 Occupation: Genius

14 Years Ago

Does column id2 equal the html form field named id2? That could be a problem if name=id2 doesn't exist in your html form and same with name=submit.

kunyomi 5 Newbie Poster

14 Years Ago

In my idea, the user goes to 'contacts.php', and in this page the user sees all the contacts that he/she have created in the DB table called 'contacts'.

Here is my code for 'contacts.php'. The field for id2 is same throughout, even in the MySQL column.

I am able to pull information so that it displays the contacts the user have added, after clicking on 'Edit', goes to 'edit.php', but after updating the data and clicking submit, it redirects me to a blank page.

<?php
include ("dbConfig.php");
require ("check.php");

 echo "<div align =\"center\">";
 echo "Welcome " . $_SESSION["valid_user"];echo ". Not you? <a href=\"http://localhost/cas/logout.php\">Click here.</a>";
 echo "<br />";
 echo "<h1>Contacts</h1>";
 echo "</div>";
 echo "<table align=\"center\" width=\"1100\" border=\"0\">";
 echo "<tr>";
 echo "<td width=\"100\"><b>Name</b></td>";
 echo "<td width=\"200\"><b>Email</b></td>";
 echo "<td width=\"100\"><b>Age</b></td>";
 echo "<td width=\"100\"><b>Birthday</b></td>";
 echo "<td width=\"300\"><b>Address</b></td>";
 echo "<td width=\"200\"><b>Number</b></td>";
 echo "<td><b>Photo</b></td>";
 echo "</tr>"; 
 echo "</table>";
 
//selectdatabase
//mysql_select_db("cas"); 

//store the current user into a variable
$username  = $_SESSION["valid_user"];

//select table
$result = mysql_query("SELECT id2, name, email, age, birthday, address, number FROM contacts WHERE username='$username'");

//grab content
while($r=mysql_fetch_array($result))
{	
  
   $id2=$r["id2"];
   $name=$r["name"];
   $email=$r["email"];
   $age=$r["age"];
   $birthday=$r["birthday"];
   $address=$r["address"];
   $number=$r["number"];
   $photo=$r["photo"];
   
  //display the row
  echo "<table align=\"center\" width=\"1100\" border=\"0\">";
 
  
  echo "<tr>";
  echo "<td width=\"100\">$name <a href=\"edit.php?cmd=edit&id2=$id2\">(Edit)</a></td>";
  echo "<td width=\"200\">$email</td>";
  echo "<td width=\"100\">$age</td>";
  echo "<td width=\"100\">$birthday</td>";
  echo "<td width=\"300\">$address</td>";
  echo "<td width=\"200\">$number</td>";
  echo "<td>$photo</td>";
  echo "</tr>";
  echo "</table>";
  echo "<br />";

}
  echo "<br />";
  echo "<br />";
  echo "<div align =\"center\">";
  echo "<a href=\"http://localhost/cas/members.php\">Member Home</a>";
  echo "</div>";

?>

Edited 14 Years Ago by kunyomi because: n/a

cwarn23 387 Occupation: Genius

14 Years Ago

Try the following.

<?
include ("dbConfig.php");
require ("check.php");

if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit")
{
   if (!isset($_POST["submit"]))
   {
      $id2 = $_GET["id2"];
      $sql = "SELECT * FROM contacts WHERE id2=$id2";
      $result = mysql_query($sql);        
      $myrow = mysql_fetch_array($result);
      ?>
	  
      <form action="<?php $v=explode('?',$_SERVER['PHP_SELF']); echo $v[0]; ?>" method="post">
      <input type=hidden name="id2" value="<?php echo $myrow["id2"]; ?>">
   
      Name:<INPUT TYPE="text" NAME="name" VALUE="<?php echo $myrow["name"]; ?>" SIZE=30><br>
      Email:<INPUT TYPE="text" NAME="email" VALUE="<?php echo $myrow["email"]; ?>" SIZE=30><br>
      Who:<INPUT TYPE="text" NAME="age" VALUE="<?php echo $myrow["age"]; ?>" SIZE=30><br>
      Birthday:<INPUT TYPE="text" NAME="birthday" VALUE="<?php echo $myrow["birthday"]; ?>" SIZE=30><br>
      Address:<TEXTAREA NAME="address" ROWS=10 COLS=30><?php echo $myrow["address"]; ?></TEXTAREA><br>
      Number:<INPUT TYPE="text" NAME="number" VALUE="<?php echo $myrow["number"]; ?>" SIZE=30><br>
   
      <input type="hidden" name="cmd" value="edit">
   
      <input type="submit" name="submit" value="submit">
   
      </form>
      
      
   }
}
   if (isset($_POST['submit'])) {
   	  $id2 = mysql_real_escape_string(stripslashes($_POST["id2"]));
      $name = mysql_real_escape_string(stripslashes($_POST["name"]));
	  $email = mysql_real_escape_string(stripslashes($_POST["email"]));
	  $age = mysql_real_escape_string(stripslashes($_POST["age"]));
	  $birthday = mysql_real_escape_string(stripslashes($_POST["birthday"]));
	  $address = mysql_real_escape_string(stripslashes($_POST["address"]));
	  $number = mysql_real_escape_string(stripslashes($_POST["number"]));
 
	  $sql = "UPDATE contacts SET name='$name', email='$email', age='$age', birthday='$birthday', address='$address', number='$number' WHERE id2=$id2";
 
      $result = mysql_query($sql) or die(mysql_error());
      echo "Thank you! Information updated.";
	}
?>

kunyomi 5 Newbie Poster

14 Years Ago

This gives me an error. You took away the

<? } ?>
<?php

at line 32 to 33 right?

It renders the php code below useless. Hmm.

cwarn23 387 Occupation: Genius

14 Years Ago

This gives me an error. You took away the
<? } ?>
<?php
at line 32 to 33 right?
It renders the php code below useless. Hmm.

Oops, line 31 needs <?php

kunyomi 5 Newbie Poster

14 Years Ago

Can you tell me what you changed from my previous code? except for the <?php?

cwarn23 387 Occupation: Genius

14 Years Ago

I changed to position of some brackets and fixed the php short tags. Also fixed some echo bugs for future versions of php. I also prevented sql injections. Fixed the additional slashes from being inserted into the mysql database. Also make parts of the code look pretty. And made the form to post to the the same php file the user is currently viewing.

So does that all work or are there still errors?

Edited 14 Years Ago by cwarn23 because: added info

kunyomi 5 Newbie Poster

14 Years Ago

Did what you said and it gave me this error:

Parse error: parse error, unexpected $end in C:\Program Files\xampp\htdocs\cas\edit.php on line 46

cwarn23 387 Occupation: Genius

14 Years Ago

Try the following.

<?php
include ("dbConfig.php");
require ("check.php");

if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit")
{
   if (!isset($_POST["submit"]))
   {
      $id2 = $_GET["id2"];
      $sql = "SELECT * FROM contacts WHERE id2=$id2";
      $result = mysql_query($sql);        
      $myrow = mysql_fetch_array($result);
      ?>
	  
      <form action="<?php $v=explode('?',$_SERVER['PHP_SELF']); echo $v[0]; ?>" method="post">
      <input type=hidden name="id2" value="<?php echo $myrow["id2"]; ?>">
   
      Name:<INPUT TYPE="text" NAME="name" VALUE="<?php echo $myrow["name"]; ?>" SIZE=30><br>
      Email:<INPUT TYPE="text" NAME="email" VALUE="<?php echo $myrow["email"]; ?>" SIZE=30><br>
      Who:<INPUT TYPE="text" NAME="age" VALUE="<?php echo $myrow["age"]; ?>" SIZE=30><br>
      Birthday:<INPUT TYPE="text" NAME="birthday" VALUE="<?php echo $myrow["birthday"]; ?>" SIZE=30><br>
      Address:<TEXTAREA NAME="address" ROWS=10 COLS=30><?php echo $myrow["address"]; ?></TEXTAREA><br>
      Number:<INPUT TYPE="text" NAME="number" VALUE="<?php echo $myrow["number"]; ?>" SIZE=30><br>
   
      <input type="hidden" name="cmd" value="edit">
   
      <input type="submit" name="submit" value="submit">
   
      </form>
      
<?php      
   }
}
   if (isset($_POST['submit'])) {
   	  $id2 = mysql_real_escape_string(stripslashes($_POST["id2"]));
      $name = mysql_real_escape_string(stripslashes($_POST["name"]));
	  $email = mysql_real_escape_string(stripslashes($_POST["email"]));
	  $age = mysql_real_escape_string(stripslashes($_POST["age"]));
	  $birthday = mysql_real_escape_string(stripslashes($_POST["birthday"]));
	  $address = mysql_real_escape_string(stripslashes($_POST["address"]));
	  $number = mysql_real_escape_string(stripslashes($_POST["number"]));
 
	  $sql = "UPDATE contacts SET name='$name', email='$email', age='$age', birthday='$birthday', address='$address', number='$number' WHERE id2=$id2";
 
      $result = mysql_query($sql) or die(mysql_error());
      echo "Thank you! Information updated.";
	}
?>

kunyomi commented: Patient and just simply genius! +1

kunyomi 5 Newbie Poster

14 Years Ago

Ok,

and I still get this error that tells me it's an unexpected end:

Parse error: parse error, unexpected $end in C:\Program Files\xampp\htdocs\cas\edit.php on line 46

I wonder why?

-UPDATE BELOW-

Edited 14 Years Ago by kunyomi because: n/a

kunyomi 5 Newbie Poster

14 Years Ago

No wait, I edited the wrong file. Goodness! It's working! Thank you so much!

Can you tell me what I did wrong? Were it just syntax errors?

cwarn23 387 Occupation: Genius

14 Years Ago

That error normally indicates that there is a missing curly backet { or } but according to the code I posted it looks like it's all there. Perhaps when you copied and paste you missed the last two lines? Is that true?

[Edit]
I guess in your mistake your forgot to place one of the brackets.
[/Edit]

Edited 14 Years Ago by cwarn23 because: n/a

kunyomi 5 Newbie Poster

14 Years Ago

It's working now! You won't imagine my excitement. Thank you.

I'm now working on uploading and displaying contact image. I hope I wouldn't need to add another thread on this, but really, thank you.

cwarn23 387 Occupation: Genius

14 Years Ago

Good to hear. Could you please mark this thread as solved by clicking the solved link as it gives others on the topic credit and shows the status as solved in the forum.

kunyomi 5 Newbie Poster

14 Years Ago

Done!

Thanks once again! God bless.

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