PHP函数返回SQL为html抛出错误?

时间:2013-11-14 17:21:55

标签: php html mysql

这个功能是否正确?它不断返回: 解析错误:语法错误,意外';',期待第301行的C:\ Program Files(x86)\ Apache Group \ Apache2 \ htdocs \ test \ include \ class.mysqltools.php中的T_FUNCTION

301是最后一行...感谢任何帮助,谢谢!

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;
}

1 个答案:

答案 0 :(得分:1)

尝试:

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;
}

不确定您的$buffer = $buffer .=是否可能导致一些意外结果。将此所有实例更改为$buffer .=