tensionx
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Posts posted by tensionx
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Well. Took out the extra {
Now its saying
[b]"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/wekillmo/public_html/tablemaker.php on line 20"
[/b]
lol...
my select is really simple
select * from XXXX
I just need a damn excel style table...
$result=mysql_query ("SELECT * FROM XXX"); -
I started doing PHP yesterday
Excuse the syntax :)
I come from ASP and am not tooo comfortable in this world yet. -
I broke this down and tested each section.
Somehow i am missing something in the last line of my code.
I could use some helping eyes.
Here is my error
[b]"Parse error: syntax error, unexpected $end in /home/wekillmo/public_html/tablemaker.php on line 37"[/b]
[code]
<html>
<table border="0" CELLSPACING="0" CELLPADDING="10" align="center">
<?php
//db connect and select
$db = @mysql_connect("localhost", "XXX", "XXX");
if( ! ($db = @mysql_connect("localhost", "name", "password")) ) {
} else {
mysql_select_db("XXX",$db) or die("Select DB Error: ".mysql_error());
}
$result=mysql_query ("SELECT * FROM XXX");
$count = 1;
$column = 1;
//loop statement
while($myrow = mysql_fetch_array($result)) {
{
// first column display
if ($column == 1)
{
//field is the column in your table
printf("<tr><td>%s</td>",$myrow["name"]);
}
else{
//second column display
printf("<td>%s</td></tr>",$myrow["class"]);
}
$count += 1;
$column = $count % 2;
}
?>
</table>
</html>
[/code] -
How many comments do you want to see per user? Do you want to see the last ten?
SELECT top 10,* from (table)
WHERE blah blah
What exactly do you want to see? -
[table]
[tr][td]<b>Name</b>[/td][/tr][/table]
Bob
Steve
Mark
Sara
John
[/table]
7 columns
the primary key is the userid
Sorts tied to the primary key.
Side by side like an excel spreadsheet hehe
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I am trying to build my DKP tracking tool. I have my beta up right now. My biggest conundrum is displaying my MySql table in an organized fashion with sort functions. my current out put is:
[code]$sql = "SELECT * FROM UFDKP";
$result = mysql_query($sql);
if ($myrow = mysql_fetch_array($result)) {
do
{
$USERID=$myrow["USERID"];
$Name=$myrow["Name"];
$Class=$myrow["Class"];
$SWR=$myrow["SWR"];
$RP=$myrow["RP"];
$Tier1=$myrow["Tier1"];
$Tier2=$myrow["Tier2"];
$Tier3=$myrow["Tier3"];
echo "<BR>USERID: $USERID";
echo "<BR>NAME: $Name";
echo "<BR>CLASS: $Class";
echo "<BR>SWR: $SWR";
echo "<BR>RP: $RP";
echo "<BR>TIER 1: $Tier1";
echo "<BR>TIER 2: $Tier2";
echo "<BR>TIER 3: $Tier3";
}
while ($myrow = mysql_fetch_array($result));
}
?>
[/code]
http://wekillmonsters.com/show.php
Its very dissapointing. lol
I tried a table Function
[code]
function GenerateTable($sqlResult, $Headers) {
$row = mysql_fetch_assoc($sqlResult);
$FieldList = Array_Keys($row);
echo "<table Border=2>";
Echo "<tr>";
For ($Cnt=0; $Cnt < Count($FieldList); $Cnt++) {
echo "<th>" . $Headers[$FieldList[$Cnt]] . "</th>";
}
echo "</tr>";
While ($row) {
$RowColor = ($RowCnt % 2) ? "row1":"row2";
Echo "<tr>";
For ($Cnt=0; $Cnt < Count($FieldList); $Cnt++) {
echo "<td class='$RowColor'>" . $row[$FieldList[$Cnt]] . "</td>";
}
echo "</tr>";
$RowCnt++ ;
$row = mysql_fetch_assoc($sqlResult);
}
echo "</table>"; }
[/code]
I could not get it working. I am a beginner at this. The only thing I am confident in is my queries because thats my work.
If someone could point me in the write direction so I can figure out how to bring these 2 things together It would be much appreciated.
Thanks.
Strange
in PHP Coding Help
Posted
when i throw a {
at the end
It stops giving me that error.
Then it sends me on my merry way to line 32. The last line... "</html>"
and says
Parse error: syntax error, unexpected $end in /home/wekillmo/public_html/tablemaker.php on line 32