Created a new variable by pulling a value from a MySQL table

I am using a page where a variable $submissionid is being posted to it. I would like to use this variable and pull the field subcheck from a MySQL table called submission. I would like the value of t开发者_运维技巧he field subcheck to simply be a new variable $subcheck. I'm not sure how to do this. I have a query below, but how to I convert the result of the query below into a variable called $subcheck? (For any given submissionid, there will only be one $subcheck.)

Thanks in advance,

John

$querysub = mysql_query("SELECT subcheck FROM submission WHERE submissionid = '$submissionid' ");
mysql_query($querysub) or die(mysql_error());

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You can try:

$querysub = mysql_query("SELECT subcheck FROM submission WHERE submissionid = ".
                                   mysql_real_escape_string($submissionid));
$result = mysql_query($querysub);
if (!$result) {
    die 'Could not run query: ' . mysql_error();
}

$subcheck = mysql_result($result, 0);

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This is more of a 'php' question, than it is for mysql.

Look up the 'extract' keyword for PHP Link. Effectively 'extract' takes the contents of an associative array and creates php variables (symbol table entries) using the names of keys. Each php variable will then contain the associated value.

You should be able to just:

   $result = mysql_query("SELECT * FROM table");
   $row    = mysql_fetch_array( $result, MYSQL_ASSOC );
   extract( $row );  // Create php variables, named after each column in the table.

   $row["field"] == $field;  // Will be a true statement after 'extract()'

Enjoy, you now have the ability to have your code dynamic adjust to a DB schema that could be changed.

-- J Jorgenson --

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This should work:

$querysub = mysql_query("SELECT subcheck FROM submission WHERE submissionid = '" . $submissionid ."' ");
$result = mysql_query($querysub) or die(mysql_error());
$row = mysql_fetch_assoc( $result );
if ($row ) {
   $subcheck = $row['subcheck'];
} else {
    echo "Subcheck not found"; 
}

Be careful with the escape characters around $submissionid in your query string. In your sample, they are probably letting the name of the variable go into the string you send to the mysql server.