Newsboard script - I'm having trouble with the comments section

[Drebin]

I have created a very basic newsboard script. Right now I have managed to display news in seperate categories and each story appears as a link. When I click on the link, the full story appears on a seperate page.

I'm trying to add a comments feature where comments for each story would appear under each story but I don't know how to do this.

I created a comments table in my database with the following rows:

comments_ID
comments_Name
comments_Comment
comments_news_ID

I think that I have to connect the comment_news_ID row to the news_ID row so that the comment only appears under that story. I don't know how to input that into my database from my form.

I created a comments form and here is what I have so far:

<table>
<form action="showcomments.php" method=post>

<font size=2>

<tr><td>Name</td> <td><input type=text maxlength=20 Name="Array[comments_Name]" size=20></td></tr>
<tr><td>Comment</td><td><textarea name="Array[comments_Comment]" rows=3 cols=20></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>

This next part handles the form

<?php
// This receives and handles the data generated from the form"

$Array["comments_Name"] = trim($Array["comments_Name"]);

$Array["comments_Comment"] = trim($Array["comments_Comment"]);

$Array["comments_news_ID"] = trim($Array["comments_news_ID"]);

$DBName = "newsboard";
$TableName = "comments";

$query = "INSERT into $TableName values ('0', '$Array[comments_Name]', '$Array[comments_Comment]', '$Array[comments_news_ID]')";

// Selects the comments that will be displayed
$query="SELECT comments_Name,comments_Comment FROM $TableName";
$result=mysql_query($query,$connect);
list($comments_Name,$comments_Comment)=mysql_fetch_row($result);
echo "$comments_Name $comments_Comment'";

mysql_close($connect);

?>

Any suggestions on how I can get this to work?

[szym9341]

Is your $Array["comments_news_ID"] = trim($Array["comments_news_ID"]); given a value from the previous page?

[Drebin]

No, I'm not sure how I can pass that value. Here's my code:

This is the story.php page. This lists all of the categories and stories. No errors on this page

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

<?php
// Selects the fields that will be displayed
$query="SELECT news_ID,news_category_ID,news_title,news_story from news WHERE news_category_ID=1 ORDER BY news_ID DESC";
$result=mysql_query($query,$connect);

// This formats the table
 echo "<table width=200, border=1, align='center'>\n"; 
 echo "<tr><td bgcolor='red'>Sports</td></tr>";	

// This writes the data that has been selected
 for($i=0;$i<mysql_num_rows($result);$i++){
 list($news_ID,$news_category_ID,$news_title)=mysql_fetch_row($result);  

echo "<tr><td><a href='showstory.php?news_ID=$news_ID&news_category_ID=$news_category_ID'>$news_title</a></td>\n";
 } 



// This closes the table
echo "</tr><tr><td bgcolor='red'>&nbsp</td></tr>\n";
echo "</table>\n";





// Selects the fields that will be displayed
$query="SELECT news_ID,news_category_ID,news_title,news_story from news WHERE news_category_ID=2 ORDER BY news_ID DESC";
$result=mysql_query($query,$connect);

// This formats the table
 echo "<table width=200, border=1, align='center'>\n"; 
 echo "<tr><td bgcolor='red'>Entertainment</td></tr>";	

// This writes the data that has been selected
 for($i=0;$i<mysql_num_rows($result);$i++){
 list($news_ID,$news_category_ID,$news_title)=mysql_fetch_row($result);  

echo "<tr><td><a href='showstory.php?news_ID=$news_ID&news_category_ID=$news_category_ID'>$news_title</a></td>\n";
 } 



// This closes the table
echo "</tr><tr><td bgcolor='red'>&nbsp</td></tr></table>\n";






// Selects the fields that will be displayed
$query="SELECT news_ID,news_category_ID,news_title,news_story from news WHERE news_category_ID=3 ORDER BY news_ID DESC";
$result=mysql_query($query,$connect);

// This formats the table
 echo "<table width=200, border=1, align='center'>\n"; 
 echo "<tr><td bgcolor='red'>News</td></tr>";	

// This writes the data that has been selected
 for($i=0;$i<mysql_num_rows($result);$i++){
 list($news_ID,$news_category_ID,$news_title)=mysql_fetch_row($result);  

echo "<tr><td><a href='showstory.php?news_ID=$news_ID&news_category_ID=$news_category_ID'>$news_title</a></td>\n";
 } 



// This closes the table
echo "</tr><tr><td bgcolor='red'>&nbsp</td></tr>\n";
echo "</table>\n";







mysql_close($connect);
?>

<br>
<center><a href="addnews.php">Add a story</a></center>

This is my showstory.php page. It properly displays the correct story but when I try to enter a comment I get this error message.

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\apache\htdocs\newsboard\showstory.php on line 9

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

<?php
// Selects the fields that will be displayed
$query1="SELECT news_title,news_story FROM news WHERE news_ID=$news_ID";
$result1=mysql_query($query1,$connect);
list($news_title,$news_story)=mysql_fetch_row($result1);

echo "$news_title <br> $news_story";
echo "<br><br>";
echo "Visitors comments <br><br>";

// Selects the comments that will be displayed
$query2="SELECT comments_Name,comments_Comment FROM comments";
$result2=mysql_query($query2,$connect);
list($comments_Name,$comments_Comment)=mysql_fetch_row($result2);

echo "$comments_Name <br> $comments_Comment";


//mysql_close($connect);
?>

<table>
<form action="showstory.php" method=post>  

<font size=2>

<tr><td>Name</td>  <td><input type=text maxlength=20 Name="Array[comments_Name]" size=20></td></tr>
<tr><td>Comment</td><td><textarea name="Array[comments_Comment]" rows=3 cols=20></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"

  $Array["comments_Name"] = trim($Array["comments_Name"]);    

  $Array["comments_Comment"] = trim($Array["comments_Comment"]);   

  $Array["comments_news_ID"] = trim($Array["comments_news_ID"]);


   $DBName = "newsboard";
   $TableName = "comments";

  $query = "INSERT into comments values ('0', '$Array[comments_Name]', '$Array[comments_Comment]', '$Array[comments_news_ID]')";  

mysql_close($connect);

 ?>

[szym9341]

I'm not too sure about this..

add this in your form?? I'm curious if this works.. lemme know

<input name="Array[comments_news_ID]" value="<?=$news_ID?>"

[Drebin]

I'll try your code, but I think there's something wrong with one of my queries too.

[Drebin]

Your code works. It added a cell that shows the ID of the story. Thanks.

However, I'm still getting that same error message and it won't let me post a comment.

[szym9341]

Sheesh.. I dunno...

I looked at the mysql_fetch_row and kinda shook my head. Sorry. What if you didn't pass the $connect into the $result1 query... would that do anything..?? good luck

[Drebin]

I tried taking out the $connect part and I got the same error. I can't figure out what's wrong. I messed around with the close mysql line too but it's not that.

[Drebin]

I've changed a few things in my script. Here's what it looks like now:

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

<?php
// Selects the fields that will be displayed
$query1="SELECT news_title,news_story FROM news WHERE news_ID=$news_ID";
$result1=mysql_query($query1,$connect);
list($news_title,$news_story)=mysql_fetch_row($result1);


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

  $Array["comments_Name"] = trim($Array["comments_Name"]);    

  $Array["comments_Comment"] = trim($Array["comments_Comment"]);   

  $Array["comments_news_ID"] = trim($Array["comments_news_ID"]);


   $DBName = "newsboard";
   $TableName = "comments";


   if($_POST['submit']==true) {

   $query = "INSERT into comments values ('0', '$Array[comments_Name]', '$Array[comments_Comment]', '$Array[comments_news_ID]')";  

 }
//mysql_close($connect);



// Selects the comments that will be displayed
$query2="SELECT comments_Name,comments_Comment FROM comments WHERE comments_news_ID=$news_ID";
$result2=mysql_query($query2,$connect);
list($comments_Name,$comments_Comment)=mysql_fetch_row($result2);



echo "$news_title <br> $news_story";
echo "<br><br>";
echo "Visitors comments <br><br>";

echo "$comments_Name <br> $comments_Comment";


 mysql_close($connect);
?>

<table>
<form action="showstory.php" method=post>

<font size=2>

<tr><td>Name</td> <td><input type=text maxlength=20 Name="Array[comments_Name]" size=20></td></tr>
<tr><td>Comment</td><td><textarea name="Array[comments_Comment]" rows=3 cols=20></textarea></td></tr>
<tr><td><input type="hidden" name="Array[comments_news_ID]" value="<?=$news_ID?>"</td>
<tr><td><input type=submit name="submit" value="Submit"></td>
<td><input type=Reset name="reset" value="Reset"></td></tr>
</font>
</form>
</table>

The news displays correctly and if I add a comment (with a comment ID) it will display correctly as well under the right news story. However, when I try to post a comment I get an error on both List lines.

Does anyone know what I'm doing wrong?

[leatherback]

Hi,

Have you echo-ed the queries? To your mysql_query(..) add:
or die(mistake inquery1: mysql_error());

That way you will get an error which describe which error occurs. That shoul dhelp tackle the problem.

J.

[Drebin]

Hi, do I add that line after the list line? I've searched some threads to see if I can find where to put this error checking line of code but I couldn't find it.

**edit: I did a search and found where to put this line. I'll try it out now.

[Drebin]

I added or die(mistake inquery1: mysql_error()); to both of my queries and the stories wouldn't even appear.

I took out the 'or die' lines and when I post I get these error messages;

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\apache\htdocs\newsboard\showstory.php on line 9

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\apache\htdocs\newsboard\showstory.php on line 34

I'm going to try to print the query lines. I double checked the spelling on the entries in my database and they all match.

[leatherback]

That is the whole idea! Instead of processing any further, it will stop at the first error in your queries.

You get an error on the query. So if you add the or die(...) part, you will get a description of the error / it will tell you exactly what is wrong with the query.

I have the feeling that the $news_ID is no passed though..

[Drebin]

I see what you mean. The error message that was generated with the 'or die' line was exactly the same as the one I posted.

I think that $news_ID is being passed because when I click on a story, only that one is displayed.

Thanks for your help, I have a feeling this will take me a while to figure out.