$_POST-ing data to table - Problem on Last Post

j3rmain3 · November 15, 2006

I am trying to display numbers which have been sent from a form but i get the error message:

Parse error: parse error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' in C:\Apache\Apache2\PHPFiles\brainstorm\btable.php on line 42

This is the code:

while ($row=mysql_fetch_assoc($result)){

$num = $row;
for ($i=1;$i<=$row;$i++) {
$post = $row['filename']."tech";
$$post = $_POST[$post];
}
[b]42:[/b] echo "<td>".${$post}"</td>";

Can anyone see what is wrong IN LINE 42? Isn't that the way i display something using variable variable

Thanks
J3RMAIN3

marcus · November 15, 2006

You can't have a double $.

You could always just replace it with

$post .= $_POST[$post];

or

$post1 = $_POST[$post];

j3rmain3 · November 15, 2006

I have nearly finished creating a rating system but i am stuck on one last thing

I am about to use UPDATE to add data to my databases but i have a problem trying to write the query here is what i want to do:

UPDATE doc_rate SET tech_rating = $_POST[$post] WHERE $fileID = $ID

[b]my problem is the $ID is part of another table(doc_data).[/b]

How do i get $fileID from the doc_rate table look for a matching ID in the doc_data table.

J3RMAIN3

j3rmain3 · November 15, 2006

[quote author=mgallforever link=topic=115058.msg468310#msg468310 date=1163591899]
You can't have a double $.

You could always just replace it with

$post .= $_POST[$post];

or

$post1 = $_POST[$post];
[/quote]

mgallforever: I thought that is how i use variable variable? with double $

And the error was mentioning the row under it, so i think it accepted the $$ ok.

j3rmain3 · November 15, 2006

I have created a form which should allow me to add values into a database. I am using dropdown menus to input the values. I am having trouble moving the values into the database. This is supposed to be a rating system.

Here is the coding for the form

[code]<html>
<head>
<title>Brainstorm Table</title>
</head>
<body>
<?php
if (!isset($_POST['submit'])){
?>
<?php

$host = "myserver";
$user = "myself";
$pass = "password";
$db = "help";

$connection = mysql_connect($host,$user,$pass) or die ("ERROR:Unable To Connect".mysql_error());

mysql_select_db($db) or die ("ERROR:Unable To Connect To DB".mysql_error());

$query = "SELECT * FROM doc_data ORDER BY session ";

$result = mysql_query($query) or die ("ERROR:Unable To Run Query".mysql_error());

echo "<h3>Department: Technical</h3>";

if (mysql_num_rows($result) > 0) {
echo "<form name=ratings method=POST action=btable.php>";
echo "<table border=1 cellpadding=3 cellspacing=3>";
echo "<tr>";
echo "<td>Session</td>";
echo "<td>Title of Brainstorm Idea</td>";
echo "<td>Inventor Name</td>";
echo "<td>Rating</td>";
echo "</tr>";

while ($row=mysql_fetch_assoc($result)){
echo "<tr>";
echo "<td>".$row['session']."</center></td>";
echo "<td><a href=http://www.mysite.com/$row[url]>".$row['idea']. "</td>";
echo "<td>".$row['inventor_name']."</td>";
echo "<td><select name=rates> <option value= selected> <option value=1 name=$row[idea]tech>1 - Poor <option value=2 name=$row[idea]tech>2 - Fair <option value=3 name=$row[filename]tech>3 - Average <option value=4 name=$row[filename]tech>4 - Good <option value=5 name=$row[filename]tech>5 - Excellent</select></td>";
echo "</tr>";
}
echo "</table>";
} else {
echo "No Rows Found";
}
echo " ";
echo "<input type=submit name=submit value=Submit> ";
echo "<a href=http://localhost/brainstorm/index.php>Back to Form</a>";
?>

<?php
} else {

$host = "myserver";
$user = "myself";
$pass = "password";
$db = "help";

$conn = mysql_connect($host,$user,$pass) or die ("ERROR:Unable to Connect".mysql_error());

mysql_select_db($db) or die ("ERROR:Unable to Connect to DB");

$query = "SELECT * FROM doc_data ORDER BY session";

$result = mysql_query($query) or die ("ERROR:Unable to Run Query".mysql_query());

while ($row=mysql_fetch_assoc($result)){
#$num = $row;
#for ($i=1;$i<=$row;$i++) {
$posttech = $row['filename']."tech";
# $postfin = $row['filename']."fin";
$postt = $_POST[$posttech];
# $postf = $_post[$postfin];
$query1 = "UPDATE doc_rate SET tech_rating=$_POST[$postt] WHERE $rate_id = $doc";

$result = mysql_query($result) or die ("ERROR:Unable To Run Query".mysql_query());

}

#$row['filename'] = empty($_POST[$postt])
}
?>
</body>
</html>[/code]

Here is the coding for the table which the results are supposed to be displayed in:

[code]<html>
<head>
<title>Brainstorm Table</title>
</head>
<body>
<?php

$host = "myserver";
$user = "myself";
$pass = "password";
$db = "help";

$connection = mysql_connect($host,$user,$pass) or die ("ERROR:Unable To Connect".mysql_error());

mysql_select_db($db) or die ("ERROR:Unable To Connect To DB".mysql_error());

$query = "SELECT * FROM doc_data ORDER BY session ";

$result = mysql_query($query) or die ("ERROR:Unable To Run Query".mysql_error());

if (mysql_num_rows($result) > 0) {
echo "<table border=1 cellpadding=3 cellspacing=3>";
echo "<tr>";
echo "<td>Session</td>";
echo "<td>Title of Brainstorm Idea</td>";
echo "<td>Inventor Name</td>";
echo "<td>Technical Rating</td>";
echo "<td>Finance Rating</td>";
echo "</tr>";

while ($row=mysql_fetch_assoc($result)){
echo "<tr>";
echo "<td>".$row['session']."</center></td>";
echo "<td><a href=http://www.mysite.com/$row[url]>".$row['idea']."</td>";
echo "<td>".$row['inventor_name']."</td>";

#$num = $row;
#for ($i=1;$i<=$row;$i++) {
# $posttech = $row['filename']."tech";
# $postfin = $row['filename']."fin";
# $postt = $_post[$posttech];
# $postf = $_post[$postfin];
#}

#echo "<td>".${$postt}"</td>";
#echo "<td>".${$postf}"</td>";
#echo "<td>"$row['filename']."fin</td>";
echo "</tr>";
}
echo "</table>";
} else {
echo "No Rows Found";
}
echo " ";
echo "<a href=http://localhost/brainstorm/index.php>Back to Form</a>";

?>
</body>
</html>[/code]

Can anyone help me work out how to place the data into the table.

Here are the problems i have encountered:

1. I have placed $_POST in a loop but that didnt seem to work. This is the way i done the loop:

[code]
while ($row=mysql_fetch_assoc($result)){
for ($i=1;$i<=$row;$i++) {
$posttech = $row['filename']."tech";
$postfin = $row['filename']."fin";
$postt = $_post[$posttech];
$postf = $_post[$postfin];
}
[/code]

Should this work?

2. Do i record the ratings by creating another table in MySQL. I thought i had to, but then i created a foreign key so that i could match the $rate_id from the rating table to the $id in the doc table. Is this right?
I stopped this because i was confused on how to populate the FK with the $id from the doc table.

3. I was using UPDATE so that the rating did not have to be placed in numerically. It could be placed in anywhere in the database. This is the coding i was using:

[code]$update = "UPDATE doc_rate SET tech_rating = $postf WHERE $rate_id = $id [/code]

$id is supposed to come from another table which is doc_data, thats is why i added a FK in doc_rate. But like i said above i do not know how to populate the FK with data from another table.

Someone help me plz ???

J3RMAIN3

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