loop?

[petesmc]

Here is script:

<?php

$result = mysql_query("SELECT category.ID, category.Cat, subcategory.ID, subcategory.CID, " .
"subcategory.SubCat, authors.ID, authors.Name FROM category, subcategory, authors");
if (!$result) {
echo("<P>Error retrieving from database!<BR>".
"Error: " . mysql_error());
exit();
}

$result = mysql_fetch_array($result);
$listcatid = $result[category.ID];
$listcat = $result["Cat"];
$listsubid = $result[subcategory.ID];
$listsubcid = $result["CID"];
$listsubcat = $result["SubCat"];
$listauthorid = $result[authors.ID];
$listname = $result["Name"];
?>

<table width="100%"><tr>
<td><b>Category/b><br><? echo("$listcatid. $listcat <br>"); ?></td>
<td><b>SubCategory/b><br><? echo("$listsubcid - $listsubid. $listsubcat <br>"); ?></td>
<td><b>Authors/b><br><? echo("$listauthorid. $listname <br>"); ?></td>
<br><br>

Firstly it doesn't display: $listcatid, $listsubid, $listauthorid

Secondly, the results in table only dispaly one result, is there a way, i can display every record?

[PeterW]

"Firstly it doesn't display: $listcatid, $listsubid, $listauthorid".. make sure you select all the fields you plan to use, else use *.

I usually use the following code for loops and retrieving mysql data..
(cannot use $result = mysq_fetch.. otherwise it will not work for more than one loop).

if($myrow = mysql_fetch_array($result))
do {

echo ("results here");

} while($myrow = mysql_fetch_array($result));

else
{
echo "No records!";
}

[petesmc]

okay, now i have this:

-------

<?php
$result = mysql_query("SELECT category.ID, category.Cat, subcategory.ID, subcategory.CID, " .
"subcategory.SubCat, authors.ID, authors.Name FROM category, subcategory, authors");
if (!$result) {
echo("<P>Error retrieving from database!<BR>".
"Error: " . mysql_error());
exit();
}
?>

# there is form and table here
<select name="authorid">
<?

while ($result = mysql_fetch_array($result)) {
$listauthorid = $result[authors.ID];
$listname = $result["Name"];
echo("<option value='$listauthorid'>$listname</option> <br>");
} ?></select>

</b></td>
<td width="114"><b>SubcategoryID: </b></td>
<td width="150"> <b>
<select name="subid">
<?
while ($result = mysql_fetch_array($result)) {
$listsubid = $result[subcategory.ID];
$listsubcat = $result["SubCat"];
echo("<option value='$listsubid'>$listsubcat</option> <br>");
} ?></select>
</b></td>
</tr>
<tr>
<td width="71"><b>Title: </b></td>
<td width="151"> <b>
<input type=TEXT name="title" size=25 maxlength=100>
</b></td>
<td width="114"><b>Category/b></td>
<td width="150"> <b>
<select name="catid">
<?
while ($result = mysql_fetch_array($result)) {
$listcatid = $result[category.ID];
$listcat = $result["Cat"];
echo("<option value='$listcatid'>$listcat</option> <br>");
} ?></select>

--------

and only the first select box displays data, if i remove that select box, the next one along the line is the only to display. How can i make al display.

[PeterW]

!!$result = mysql_fetch_array($result))!!

If you over write the value in $result, mysql is not provided with a valid array descriptor to fetch!

pretend I had code with a command in..

$command = "Peter, go and get that data and put the value in $command";

so Peter did that..
but next time, $command said something like "two oranges, three bananas", Peter wouldn't know what to do.

so, you must do something like

$somethingdifferenttoresult = mysql_fetch_array($result))

I also notice that you are trying to use the $result over and over again for each set of list boxes. This is incorrect, you would need to create a copy of $result and then copy it in to $result each time you wanted to start from the first record again.
<Edited by PeterW on 01-06-2001 at 11:00 AM>

[petesmc]

Got it...

Thanx

It was what you said
<Edited by petesmc on 01-06-2001 at 11:02 AM>

[PeterW]

<edit>Okay </edit>

I dont think you understand what I mean..
<?

while ($THISMUSTBEDIFFERENT = mysql_fetch_array($TOTHIS))
{
//...
}
?>

[petesmc]

Oh my god....i h8 PHP

Now, it doesn't add the results to database

Only the results from all of the select boxes.....

any idea?

[PeterW]

We would probably need to see the code to tell what is wrong

[petesmc]

<!-- newarticle.php -->
<HTML>
<HEAD>
<TITLE> Add New article </title><meta http-equiv="expires" content="0">
<meta http-equiv="pragma" content="no-cache">
<meta http-equiv="cache-control" content="no-cache"> </head>

<BODY>
<?php
include("articles.inc");
mysql_select_db("petesmc_articles");
if ($submit): // A new article has been entered
// using the form below.

$sql = "INSERT INTO articles SET " .
"AuthorID='$aauthorid', " .
"CatID='$ccatid', " .
"SubID='$ssubid', " .
"DateAdded='CURDATE()', " .
"Description='$description', " .
"Title='$title', " .
"Status='$status', " .
"Text='$text'";
if (mysql_query($sql)) {
echo("<P>New article added</P>");
} else {
echo("<P>Error adding new article: " .
mysql_error() . "</P>");
}

?>

<P><A HREF="<?php echo($PHP_SELF); ?>">Add another article</A></P>
<P><A HREF="articles.php">Return to article list</A></P>

<?php
else: // Allow the user to enter a new article

$result = mysql_query("SELECT authors.ID, authors.Name FROM authors");

$result2 = mysql_query("SELECT category.ID, category.Cat FROM category");

$result3 = mysql_query("SELECT subcategory.ID, subcategory.SubCat FROM subcategory");

?>
<FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=GET>
</tr></table><br><br>
<P><font size="6" color="#0066FF">Enter the new Article/font><BR>
</P>


<blockquote><table width="536" border="1" cellspacing="0" cellpadding="5" align="left">
<tr>
<td width="71"><b>AuthorID/b></td>
<td width="151"> <b>
<select name="authorid" size="1">
<?
while ($resulta = mysql_fetch_array($result)) {
$listauthorid = $resulta[authors.ID];
$listname = $resulta["Name"];
echo("<option value=$listauthorid>$listname</option>");
} ?></select>

</b></td>
<td width="114"><b>SubcategoryID: </b></td>
<td width="150"> <b>
<select name=subid>
<?
while ($result3a = mysql_fetch_array($result3)) {
$listsubid = $result3a[subcategory.ID];
$listsubcat = $result3a["SubCat"];
echo("<option value=$listsubid>$listsubcat</option>");
} ?></select>
</b></td>
</tr>
<tr>
<td width="71"><b>Title: </b></td>
<td width="151"> <b>
<input type=TEXT name="title" size=25 maxlength=100>
</b></td>
<td width="114"><b>Category/b></td>
<td width="150"> <b>
<select name=catid>
<?
while ($result2a = mysql_fetch_array($result2)) {
$listcatid = $result2a[category.ID];
$listcat = $result2a["Cat"];
echo("<option value=$listcatid>$listcat</option>");
} ?></select>
</b></td>
</tr>
<tr>
<td width="71"><b>Status/b></td>
<td width="151"> <b>
<select name="status">
<option>Select Status type .....</option>
<option value="active">Active</option>
<option value="featured">Featured</option>
<option value="hidden">Hidden</option>
</select>
</b></td>
<td colspan="2"><b></b><b></b></td>
</tr>
<tr>
<td colspan="4" height="26"><b> Description: </b></td>
</tr>
<tr>
<td colspan="4"> <b>
<textarea name="description" rows=6 cols=60 WRAP></textarea>
</b></td>
</tr>
<tr>
<td colspan="4"><b>Article Text/b></td>
</tr>
<tr>
<td colspan="4"> <b>
<textarea name="text" rows=25 cols=100 WRAP></textarea>
</b></td>
</tr>
<tr>
<td colspan="4">
<input type=SUBMIT name="submit" value="SUBMIT ARTICLE">
<input type="reset" name="Reset" value="RESET">
</td>
</tr>
</table></blockquote>
<P> <BR>
</P>
</FORM>



<?php endif; ?>


</BODY>
</HTML>

[PeterW]

$sql = "INSERT INTO articles SET " .
"AuthorID='$aauthorid', " .
"CatID='$ccatid', " .
"SubID='$ssubid', " .
"DateAdded='CURDATE()', " .
"Description='$description', " .
"Title='$title', " .
"Status='$status', " .
"Text='$text'";

By looking at your subsequent sql queries it seems you have multiple tables, so trying to insert data into fields that dont exist will not work.

[petesmc]

I am only inserting to one table in this script.

Could you try and create a simple script, where you insert using a select box?

<Edited by petesmc on 01-06-2001 at 11:49 AM>

[petesmc]

i did it. For some reason you can't do tis:

$listcatid = $result2a["category.ID"];

I got rid of category, and made there different queries to table:


$listcatid = $result2a["ID"];
<Edited by petesmc on 01-06-2001 at 12:11 PM>

[petesmc]

Hmmm...

Can you do this:

<select name="authorid" size="1">
<? $result = mysql_query("SELECT ID, Name FROM authors");
while ($resulta = mysql_fetch_array($result)) {
$listauthorid = $resulta["ID"];
$listname = $resulta["Name"];
echo("<option value=$listauthorid); ?><? if ($listauthorid==$authorid) {
echo "selected"; } ?> <? echo("$listname</option>");
} ?></select>


authorid is specified above the above text. I am gettign a parse error where it say echo "selected";

What i am trying to do is, pull data from table, put into select obx, then with an if statemnt, i am saying if the id is equal to the id in the other table ($authorid), then that is selected.

[petesmc]

Don't worry, i downloaded the files from kevin Yanks tutorials:

<SELECT NAME="authorid" SIZE=1>
<?php
$authors = mysql_query("SELECT ID, Name FROM authors");
while ($author = mysql_fetch_array($authors)) {
$listauthorid = $author["ID"];
$listaname = $author["Name"];
if ($authorid == $listauthorid) {
echo("<OPTION SELECTED VALUE='$listauthorid'>$listaname\n");
} else {
echo("<OPTION VALUE='$listauthorid'>$listaname\n");
}
}
?>
</SELECT>


That is how you do it.