PHP函数将SQL作为html返回并抛出错误？

Question

这个函数正确吗？它不断返回:解析错误:语法错误，意外的“;”，期望 C:\Program Files (x86)\Apache Group\Apache2\htdocs\test\include\class.mysqltools.php 第 301 行出现 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;
}

Answer 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。将其所有实例更改为 $buffer .=