出乎意料 但我不明白为什么

Question

Parse error: syntax error, unexpected ';' in /home/realcas/public_html/eshop/ecms/system/classes/database.php on line 29

This is the code on line 29

return empty($resultArray) ? "Error in Query " ? json_encode($resultArray);

this is the section of code that is the issue

public function select($table,$options,$where,$orderby)
{
    $options = empty($options) ? "*"   : $options;
    $where  =  empty($where)   ? "1=1" : $where;
    $orderby = empty($orderby) ? ""    : $orderby;

    $qry = "SELECT $options FROM $table WHERE $where $orderby ";
    $result = mysql_query($qry) or die(json_encode(array("error",mysql_error())));

    while(($resultArray[] = mysql_fetch_assoc($result)));
    return empty($resultArray) ? "Error in Query " ? json_encode($resultArray);

    return json_encode($resultArray);
}

Answer 1

You want probably this one

return empty($resultArray) ? "Error in Query " : json_encode($resultArray);

because its

($condition) ? "condition is true" : "condition is false";

I see you're using it already, so I assume it was just a typo

Also remove this line

    return json_encode($resultArray);

because it's unnecessary and never will happen. Additionally I'm not sure that your while loop is correct.

Result

public function select($table,$options,$where,$orderby)
{
    $options = empty($options) ? "*"   : $options;
    $where  =  empty($where)   ? "1=1" : $where;
    $orderby = empty($orderby) ? ""    : $orderby;

    $qry = "SELECT $options FROM $table WHERE $where $orderby ";
    $result = mysql_query($qry) or die(json_encode(array("error",mysql_error())));

    while(($row = mysql_fetch_assoc($result))){ $resultArray[] = $row; }
    return count($resultArray) < 1 ? "Error in Query " : json_encode($resultArray);
}