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我试图通过过滤为用户选择的选项从数据库中获取结果。用户可以根据酒名或酒厂或两者搜索酒店。我编写了以下命令，它应该根据用户的选择附加特定的 sql 命令。但它给了我服务器错误。我认为我的阵列或任何东西有问题。请有人帮我解决这个问题。 

<?php
      $searchResult = "";
      $WineName = preg_replace('#[^a-z]#i','',$_POST['WineName']);
      $winery = preg_replace('#[^a-z]#i','',$_POST['winery']);
      $arr = array();
      if(!empty($_POST['WineName']))
      $arr[] = "wine_name LIKE '%$WineName'";
      if(!empty($_POST['winery']))
      $arr[] = "winery_name LIKE '%$winery";

      $str = impode("and", $arr);
      if(!empty($str)) $str = "and ".$str;

      include_once("connect.php");//requesting to open the database
      $query = mysql_query("select *from WineSearchView where 1 $str") or die(mysql_error());
      $count = mysql_num_rows($query);
      if($count > 1){
        $searchResult .= "$count results for $WineName";
        while($row = mysql_fetch_array($query)){
          $wName = $row["wine_name"];
          $variety = $row["variety"];
          $year = $row["year"];
          $wineryName = $row["winery_name"];
          $region = $row["region_name"];
          $cost = $row["cost"];
          $stock = $row["on_hand"];
          $ordered = $row["qty"];

          $searchResult .= "Wine Name: $wName, V: $variety, Y: $year, WN: $winery_name, R: $region, C: $cost, S: $stock, O: $ordered<br />";
        }//close while
      }
     else { $searchResults = "0 results for $WineName $sqlCommand";}
?>
4

1 回答 1

1

改变

$str = impode("and", $arr);

至

$str = implode("and", $arr);
于 2013-08-21T05:27:12.393 回答