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[code]
////if their name does not already exist in the relevant table, insert it////
$namesthere = mysql_fetch_array(mysql_query("SELCT * FROM `$compl` WHERE `Name`=$name"));
if (mysql_num_rows($namesthere)==0){
$query = mysql_query("INSERT INTO '$compl' ( `Name` , `Sail Number` , `Class` , `Score` ) VALUES ( '$name', '$snum', '$boattype', '$score')");
} [/code]

The code above is giving the "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource" error. Can anyone give me a hand please?!
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i'm sure ive got the table name correct but i now only get one error (before I got 2) saying:

[code]Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\Will\My Documents\xampp\htdocs\sailing\rescalc.php on line 91[/code]

with this code:

[code]////if their name does not already exist in the relevant table, insert it////
$namesthere = mysql_fetch_array(mysql_query("SELECT * FROM `$compl` WHERE `Name`='$name'"));
if (mysql_num_rows($namesthere)==0){
$query = mysql_query("INSERT INTO `$compl` ( `Name` , `Sail Number` , `Class` , `Score` ) VALUES ( '$name', '$snum', '$boattype', '$score')");
echo ("<pre>" . $query . "</pre>");
}

////////////
[/code]

the <pre> outputs:

[code]1[/code]

and thats it! any ideas?
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https://forums.phpfreaks.com/topic/5357-sql-num_rows-query-help/#findComment-19090
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That's because mysql_query is going to return the result set number.

You really need to stop nesting your functions so you can troubleshoot more easily:

[code]$query = "INSERT INTO `$compl` ( `Name` , `Sail Number` , `Class` , `Score` ) VALUES ( '$name', '$snum', '$boattype', '$score')";
$result = mysql_query($query) or die(mysql_error());
if($result && mysql_num_rows($result) > 0)
{
   $row = mysql_fetch_array($result);
}
[/code]

That's how it should be structured. Then you can echo the query quite easily.
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https://forums.phpfreaks.com/topic/5357-sql-num_rows-query-help/#findComment-19094
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You have too many compound statement to really determine what is causing the problem. Let's break them down to the individual statements with error checking along the way:
[code]<?php
////if their name does not already exist in the relevant table, insert it////
$q = "SELECT * FROM `$compl` WHERE `Name`='$name'";
$rs = mysql_query($q) or die('Problem with query: ' . $q . '<br>' . mysql_error());
$namesthere = mysql_fetch_assoc($rs);
if (mysql_num_rows($namesthere)==0){
    $q = "INSERT INTO `$compl` ( `Name` , `Sail Number` , `Class` , `Score` ) VALUES ( '$name', '$snum', '$boattype', '$score')";
    echo ("<pre>" . $q . "</pre>");
    $rs = mysql_query($q) or die('Problem with insert query: ' . $q . '<br>' . mysql_error());
}
?>[/code]

Ken
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https://forums.phpfreaks.com/topic/5357-sql-num_rows-query-help/#findComment-19097
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okay, ive used the improved code.

heres what i get when i input some dummy values

[code]Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\Will\My Documents\xampp\htdocs\sailing\rescalc.php on line 93

INSERT INTO `topper main points` ( `Name` , `Sail Number` , `Class` , `Score` ) VALUES ( 'Will Williams', '45052', 'Topper', '129')[/code]

i put the sql into phpmyadmin and it worked so what could be the problem?
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https://forums.phpfreaks.com/topic/5357-sql-num_rows-query-help/#findComment-19103
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if mysql_query() did not return anything then mysql_fetch_assoc() will fail. There is nothing there for it to fetch. You do no get num_rows from fetch_assoc...you get an associative array of one row from mysql_fetch_assoc() based on the query. Take the query and do mysql_num_rows() on it to get the number of rows or a SELECT count(*) with conditions...
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