Update Script Not Working

Ted_Striker · Sep 30, 2004

Hi, I'm hoping that someone can help me with my update script. I'm been playing around with php for a few months but I'm still a beginner so I apologize in advance if my code is hard to read, or fraught with errors.

I'm making an update script for my friend and it's just not working. I've created a private members area which works fine and now I'm trying to create a text box which has the old information already in it so that it's easy to modify.

Here are my problems:

1) The old info does not appear in the text box.

2) The script will not even post to the database.

It was actually posting before I made changes to it, but it would make an additional 2 blank duplicates and it would not edit the current entry - it would just make a new one.

Below is my code. I've been tweaking it for about 2 weeks now and I'm stuck.

<?php
require_once("connect.php");

// check session variable
  if (isset($_SESSION['valid_user'])) 
  {
      echo '<p>You are logged in as '.$_SESSION['valid_user'].'</p>';



// Gets the article_ID from link and sets it to a variable.  

$ArticleID = $_GET['article_ID'];
$Content = $_GET['articles_content'];


// Selects the fields that will be displayed
$query="SELECT articles_ID,articles_content FROM articles WHERE articles_ID=".$_GET['articles_ID']."";
$result=mysql_query($query,$connect);
?>



<html>
<head>
<title>Edit Article</title>
</head>

<body>

<table>
<form action="editarticles.php" method=GET>  

<font size=2>

<tr><td>Content</td><td><textarea name="articles_content" rows="3" cols="20"><?php  echo $Content ?></textarea></td></tr>
<tr><td><input type=submit name="submit" value="submit"></td>
<td><input type=Reset name="reset" value="Reset"></td></tr>
</font>
</form>
</table>


<?php


// This receives and handles the data generated from the form"    

  //$edit_content=$_GET['articles_content'];

  $query="UPDATE articles 
          SET articles_content='$articles_content' 
          WHERE articles_ID=".$_GET['articles_ID']."";
  $result=mysql_query($query,$connect);



 if ($_GET['submit'] == true)   //This means the form HAS been submitted:
 {

  if (mysql_query($query,$connect)) {
  	 echo "Article has been edited\n";
  	 echo "<a href='addarticles.php'>Click here to add another story</a>";


  }else {
  	echo "System error.  Article was not edited\n";
  }
 }

  mysql_close($connect);

  }


// Not logged in


else
  {
    echo '<p>You are not logged in.</p>';
    echo '<p>Only logged in members may see this page.</p>';
  }

echo '<a href="authmain.php">Back to main page</a>';

  ?>

A friend of mine told me that I need to put this line somewhere:

list($articles_content) = mysql_fetch_row($query);

Is this correct and if so why would I need it? I'm guessing that it would go right after my SELECT query.

Thanks.

Ted_Striker · Sep 30, 2004

I've been thinking about what could be wrong and I might have figured out part of what's wrong.

In the admin area it lists the articles but only as a link. So when I click on the "edit article" link, all it does is pass the article_ID, nothing else.

Do I need to make another script that lists the articles_ID and the articles_content and pass it through to the editarticles script or can I just attach the articles_content via the link?

I'm sure that I've made some coding errors but I think I'm missing a step as well.

gufmn · Sep 30, 2004

A friend of mine told me that I need to put this line somewhere:

list($articles_content) = mysql_fetch_row($query);

Is this correct and if so why would I need it? I'm guessing that it would go right after my SELECT query.

Thanks.

Kinda.

You should use something like:

$query="SELECT articles_ID,articles_content FROM articles WHERE articles_ID=".$_GET['articles_ID']."";
$result=mysql_query($query,$connect);
$row = mysql_fetch_array($result);

This will give you an array of collums per your query. So you could retrieve the contents by accessing:

$row['articles_ID'] (interchange for whatever collum you want)

I've been thinking about what could be wrong and I might have figured out part of what's wrong.

In the admin area it lists the articles but only as a link. So when I click on the "edit article" link, all it does is pass the article_ID, nothing else.

Do I need to make another script that lists the articles_ID and the articles_content and pass it through to the editarticles script or can I just attach the articles_content via the link?

I'm sure that I've made some coding errors but I think I'm missing a step as well.

You should be able to just pass article_ID through your link and use that to run your mysql query. Then just pull articles_content from the db.

HTH

cyberlew15 · Sep 30, 2004

$query="UPDATE articles

SET articles_content='$articles_content'
WHERE articles_ID=".$_GET['articles_ID']."";

// should read

$query="UPDATE articles

SET articles_content='$articles_oldcontent . $articles_content'
WHERE articles_ID=".$_GET['articles_ID'].";

This would input the old data before the new data or store the new data in a file by finding the old articles id maybe 64 so the new articles content would be 65

Have Fun:p

Ted_Striker · Oct 2, 2004

Thanks for replying guys. Seeing how it should be done makes me realize how flawed my logic is.

I'm going to modify the script with your corrections.

Any idea why it posts 3 times when I submit the form? I haven't been able to spot the problem.

Ted_Striker · Oct 2, 2004

I've tried to run the script after making the corrections you guys noted above and after changing a few other things I'm getting the following error:

error - Parse error: parse error, unexpected T_STRING in C:\apache\htdocs\members\editarticles.php on line 64 (might be different...I edited a few things afterwards but the error is the same)

This line is on line 64:

echo "Article has been edited \n";

Can you guys please take a look at the script again? I'm still not sure if I setup the form correctly and I don't know what's causing the error so I'm not able to test it out to see if it will update correctly.


<?php
require_once("connect.php");

// check session variable
  if (isset($_SESSION['valid_user'])) 
  {
      echo '<p>You are logged in as '.$_SESSION['valid_user'].'</p>';


// Gets the article_ID from link and sets it to a variable.  

$ArticleID = $_GET['article_ID'];
$Content = $_GET['articles_content'];


// Selects the fields that will be displayed
$query="SELECT articles_ID,articles_content FROM articles WHERE articles_ID=".$_GET['articles_ID']."";
$result=mysql_query($query,$connect);
$row = mysql_fetch_array($result);


$OldContent = $_GET['articles_content'];
?>



<html>
<head>
<title>Edit Article</title>
</head>

<body>

<table>
<form action="editarticles.php" method=GET>  

<font size=2>

<tr><td>Content</td><td><textarea name="newarticles_content" rows="3" cols="20"><?php  echo $OldContent ?></textarea></td></tr>
<tr><td><input type=submit name="submit" value="submit"></td>
<td><input type=Reset name="reset" value="Reset"></td></tr>
</font>
</form>
</table>


<?php
//$NewContent = $_GET['newarticles_content'];

// This receives and handles the data generated from the form"       


 if ($_GET['submit'] == true)   //This means the form HAS been submitted:
 {

  $query="UPDATE articles 
          SET articles_content=".$_GET['newarticles_content']."
          WHERE articles_ID=".$_GET['articles_ID'].";
  $result=mysql_query($query,$connect);


  if (mysql_query($query,$connect)) {
  	 echo "Article has been edited \n";
  	 echo "<a href='addarticles.php'>Click here to add another story</a>";


  }else {
  	echo "System error.  Article was not edited\n";
  }
 }

  mysql_close($connect);

  }


// Not logged in


else
  {
    echo '<p>You are not logged in.</p>';
    echo '<p>Only logged in members may see this page.</p>';
  }

echo '<a href="authmain.php">Back to main page</a>';

  ?>

gufmn · Oct 2, 2004

Look at the syntax highlightling. You're not closing your string in you query.

$query="UPDATE articles 
          SET articles_content=".$_GET['newarticles_content']." 
          WHERE articles_ID=".$_GET['articles_ID'];

That should work

[edit]
I just noticed something else. You running the same query twice in a row.

$result=mysql_query($query,$connect); 


  if (mysql_query($query,$connect)) { 

//change to:

$result=mysql_query($query,$connect); 


if ($result) { doo stuff; }

[/edit]

Ted_Striker · Oct 2, 2004

Thanks. I really appreciate the help.

I made the changes you posted above but now I'm getting this error when I click submit :

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\apache\htdocs\members\editarticles.php on line 18

I'll just post the block of code that Line 18 is in.

<?php
require_once("connect.php");

// check session variable
  if (isset($_SESSION['valid_user'])) 
  {
      echo '<p>You are logged in as '.$_SESSION['valid_user'].'</p>';


// Gets the article_ID from link and sets it to a variable.  

//$ArticleID = $_GET['article_ID'];
//$Content = $_GET['articles_content'];


// Selects the fields that will be displayed
$query="SELECT articles_ID,articles_content FROM articles WHERE articles_ID=".$_GET['articles_ID']."";
$result=mysql_query($query,$connect);
list($articles_ID,$articles_content)=mysql_fetch_row($result);  // LINE 18


$OldContent = $_GET['articles_content'];
?>

The "list" line is causing the problem.  I referenced some other code I wrote in the past and I've used this same exact code on another little script I made and it's fine.  I double checked everything and the names match with what I have in the database.  

Any idea what's wrong?  Also, when I go to the edit script, the current info still doesn't show up in the form.  Is this related to the error I'm getting now?

gufmn · Oct 2, 2004

Ok, it is t hat list line that's giving you problems. So lets try assigning the vars one at a time.

$query="SELECT articles_ID,articles_content FROM articles WHERE articles_ID=".$_GET['articles_ID']."";
$result=mysql_query($query,$connect);
$row=mysql_fetch_row($result);
$articlesID = $row[0];
$artices_content=$row[1];

print_r($row);//this should print all the contents of the $row array for testing purposes.

And yes, it is most likely that this is why your form fields aren't being filled in.

Ted_Striker · Oct 3, 2004

When I go to the editarticles page the print statement returns the following:

Array ( [0] => 98 [1] => This is only a test )

But when I type something in the form and press submit I get this error message again:
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\apache\htdocs\members\editarticles.php on line 18

So it's reading the article ID and article content but something is getting messed up along the way when I submit the changes.

Is it possible that I have the form in the wrong place? Maybe I should have it at the bottom of the page after all of the code?

Ted_Striker · Oct 5, 2004

I haven't been able to figure this out yet. Would it be possible to get some further help with this?

Update Script Not Working

TTed_Striker

TTed_Striker

Ggufmn

Ccyberlew15

TTed_Striker

TTed_Striker

Ggufmn

TTed_Striker

Ggufmn

TTed_Striker

TTed_Striker