[RESOLVED] Updating past entries

[SpecialFX]

I am working on a content management system. So I need to be able to edit past entries. I am having trouble loading the content into the form fields so that it may be edited.

<?php
require ('dbconnect.php');
require ('storyfns.php');
if (isset($_REQUEST['blogid']))
{
$blogid = get_story_record($_REQUEST['blogid']);
}
?>

<form action="edited.php" method="post" enctype="multipart/form-data">
<table>
<tr>
 <td><font face="Tahoma" size="2">Title</font><td>
</tr>
<tr>
<td><input size="53" name="title"
		value="<?php echo $row['title'];?>"></td>
</tr>
<tr>
 <td><font face="Tahoma" size="2">Body</font></td>
</tr>
<tr>
<td><textarea cols="40" rows="7" name="blog"
	wrap="virtual"><?php echo $row['blog'];?></textarea>
</td>
</tr>
<tr>
 <td align="center"><input type="image" src="http://www.fallenskydesigns.com/blog/publish.gif" name="publish"></td>
</tr>
</table>
</form>

get_story_record() is stored in this file

<?php

function get_story_record(){
$query = "SELECT * FROM blognews WHERE blogid = '$blogid'";   

$myresult = mysql_query($query) or die(mysql_error());
return($myresult->fetch_assoc());
}

?>

[cgraz]

$blogid is not defined in your function (it's not global), but you're passing it as a parameter when you call your function. You need to change your function definition to this:

function get_story_record($blogid){
$query ...

I'm also not sure how this line will work

return($myresult->fetch_assoc()); 

so after making that first change, if you're still not seeing any results, then try replacing it with:

$row = mysql_fetch_assoc($query) or die(mysql_error());
return $row;

[SpecialFX]

I changed somethings and got it so that it the text shows up in the text fields and when you publish it echoes that it was updated however when you check it, It isn't updated here is the php.

 <?php
$blogid = $_REQUEST['blogid'];
require ('dbconnect.php');
$query = "SELECT title, blog FROM blognews WHERE blogid='$blogid'";   

$myresult = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($myresult)
?>

 <form name="shows"action="edited.php" method="post" enctype="multipart/form-data">
<table>
<tr>
 <td><font face="Tahoma" size="2">Title</font><td>
</tr>
<tr>
<td><input size="53" name="title"
		value="<?php echo $row['title'];?>"></td>
</tr>
<tr>
 <td><font face="Tahoma" size="2">Body</font></td>
</tr>
<tr>
<td><textarea cols="40" rows="7" name="blog"
	wrap="virtual"><?php echo $row['blog'];?></textarea>
</td>
</tr>
<tr>
 <td align="center"><input type="image" src="http://www.fallenskydesigns.com/blog/publish.gif" name="publish"></td>
</tr>
</table>
</form>

The script is then sent here "edited.php".

<?php
require ('dbconnect.php');

$title = $_POST['title'];
$blog = $_POST['blog'];
$blogid = $_REQUEST['blogid'];

mysql_query("UPDATE blognews SET title='$title', blog='$blog' WHERE blogid='$blogid'")or die(mysql_error());  {
echo '<font face="Tahoma" size="2">Your entry <b>'.$title.'</b> has been updated.</font>';
}
//and
mysql_close(); 
?>

[cgraz]

Are you sure $blogid is set? I'm not seeing where you are grabbing it from...?

In the first php code you posted, try something like this

$query = "SELECT title, blog FROM blognews WHERE blogid='$blogid'";
print $query . "<br />\n";
$myresult = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($myresult);

I only posted what I changed, so you'll need to add in the additional code, but what this does is prints the string you are passing to SQL so you can see what you are sending to mysql_query() and catch any problems with your statement.

Do the same with your UPDATE query like this

$strSQL = "UPDATE blognews SET title='".$title."', blog='".$blog."' WHERE blogid='".$blogid."'");
print $strSQL . "<br />\n";

$result = mysql_query($strSQL) or die(mysql_error());

That should help you determine the cause of the failure to update.

[SpecialFX]

I did what you said and in the edited.php I don't think it can find the blogid. But I don't know how to make it so it can.

[cgraz]

well i would assume it would have to come from your form, since that is what is POSTing values to edited.php.

My guess from what you have is to add the following to your form

<input type="hidden" name="blogid" value="<?php echo $row["blogid"]; ?>" />

Change $REQUEST["blogid"] to $POST["blogid"]

Then, before you submit the form, go to View > Source and make sure that the value is there. If it is, submit, still printing the SQL query string and using mysql_error()). It should upate at that point.

[SpecialFX]

Perfect thank you, Resolved.