PHP function to return SQL as html throwing error?

Question

Is this function correct? it keeps returning: Parse error: syntax error, unexpected ';', expecting T_FUNCTION in C:\Program Files (x86)\Apache Group\Apache2\htdocs\test\include\class.mysqltools.php on line 301

301 is the last line...any help is appreciated, thank you!

function DisplayA($query, $rowname1, $rowname2) {
    $result = mysql_query($query);
    $buffer = $buffer .="<table>";
    while($row = mysql_fetch_array($result)){
    $buffer = $buffer .="<tr><td>" . $row[$rowname1] . "</td><td>" . $row[$rowname2] . "</td></tr>";
    }
    $buffer = $buffer .="</table>";
    return $buffer;
}

http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php — Geoffrey Burdett, Nov 14 '13 at 17:23
I don't know if it's the issue, but `$buffer = $buffer .="";` isn't going to do what you want. You need either `$buffer = $buffer . "
";` or `$buffer .="
";` instead. — andrewsi, Nov 14 '13 at 17:26

score 1 · Accepted Answer · answered Nov 14 '13 at 17:25

Try:

function DisplayA($query, $rowname1, $rowname2) {
    $result = mysql_query($query);
    $buffer .= "<table>";
    while($row = mysql_fetch_array($result)){
        $buffer .= "<tr><td>" . $row[$rowname1] . "</td><td>" . $row[$rowname2] . "</td></tr>";
    }
    $buffer .= "</table>";
    return $buffer;
}

Not sure if your $buffer = $buffer .= may have been causing some unexpected results. Change all instances of this to just $buffer .=

PHP function to return SQL as html throwing error?

1 Answers1