1

我正在尝试使用多个 WHERE 子句运行查询。如果我进行多项搜索，它会从单个条件返回一条记录。我需要这个查询来返回一个包含所有条件而不仅仅是一个条件的结果。

你可以在这里看到它。

另外，我提供了代码： 

if (isset ( $_POST ["btnSearch"] )) {
echo "<br>Selected Options are :<br>";
$checked = $_POST ["criteria"];

$criteria = "";
$separator = ", ";
for($i = 0; $i < count ( $checked ); $i ++) {
    echo "  " . $checked [$i] . "<br/>";

    if ($i == count ( $checked ) - 1) {
        $separator = "";
    }

    $criteria = $criteria . "'" . $checked [$i] . "'" . $separator;
}
echo "<br><br>";

echo $criteria . "<br><br>";
include "config.php";

mysql_select_db ( "MyHead", $con );
//$DM = implode(',',$criteria);
$mysqlQuery = "SELECT tblRestaurants.RestName, tblLocDet.LocationID, tblLocDet.DetailID, tblDetails.DetailName, tblRestaurants.RestName
FROM tblRestaurants INNER JOIN (tblLocations INNER JOIN (tblLocDet INNER JOIN tblDetails ON  tblLocDet.DetailID = tblDetails.DetailID) ON tblLocations.LocationID = tblLocDet.LocationID) ON  tblRestaurants.RestID = tblLocations.RestID
GROUP BY tblRestaurants.RestName, tblLocDet.LocationID, tblLocDet.DetailID, tblDetails.DetailName
 HAVING tblDetails.DetailName IN (" . $criteria . ");";

if (! $rs = mysql_query ( $mysqlQuery )) {
    echo "Cannot parse query";
} elseif (mysql_num_rows ( $rs ) == 0) {
    echo "No records found";
} else {
    echo "<table id=\"myTable\" table width=\"710\" class=\"beautifuldata\" align=\"Left\" cellspacing=\"0\">\n";
    echo "<thead>\n<tr>";
    echo "<th>PLACE</th>";
    echo "<th>ADDRESS</th>";
    echo "<th>PHONE</th>";
    echo "<th>PRICE</th>";
    echo "<th>RATING</th>";
    echo "</tr>\n</thead>\n";
    while ( $row = mysql_fetch_array ( $rs ) ) {
        echo "<tr><td><strong><a href='" . $row [RestPage] . "'>" . $row ['RestName'] . "</a></strong></td>";
        echo "<td>" . $row ['DetailName'] . "</td>";
        echo "<td>" . $row ['Phone'] . "</td>";
        echo "<td>" . $row ['Price'] . "</td>";
        echo "<td>" . $row ['Rating'] . "</td>";
        echo "</tr>";
    }
}
echo "</table><br />\n";

mysql_close ( $con );
}
?>

表：

tblRestaurants (RestID, RestName)
tblLocations (LocationID, CityID, AreaID, CuisineID)
tblLocDet (DetailID, LocationID)
tblDetails (DetailID, DetailName, DetailType)
4

3 回答 3

2

为确保所有选定的行都具有 中的所有项目$criteria，一种方法是计算此条件变量中的这些项目，然后具有HAVING COUNT(DISTINCT DetailName) = $n，以便任何选定的行都应具有所有项目，例如：

SELECT 
  r.RestName, 
  ld.LocationID, 
  ld.DetailID, 
  d.DetailName
FROM tblRestaurants     AS r
INNER JOIN tblLocations AS l  ON r.RestID     = l.RestID
INNER JOIN tblLocDet    AS ld ON l.LocationID = ld.LocationID
INNER JOIN
(
  SELECT l.Locationid
  FROM tblLocDet l
  INNER JOIN tbldetails d ON l.detailid = d.detailid
  WHERE d.detailname IN ('det1', 'det2', 'det3')
  GROUP BY l.locationid
  HAVING COUNT(DISTINCT DetailName) = $n
)                       AS ld2 ON ld.locationid = ld2.locationid
INNER JOIN tblDetails   AS d   ON ld.DetailID   = d.DetailID   
GROUP BY r.RestName, 
         ld.LocationID, 
         ld.DetailID, 
         d.DetailName;

SQL 小提琴演示。

这会给你类似的东西：

| RESTNAME | LOCATIONID | DETAILID | DETAILNAME |
-------------------------------------------------
|     res1 |          1 |        1 |       det1 |
|     res1 |          1 |        2 |       det2 |
|     res1 |          1 |        3 |       det3 |

但是，您可以缩短此查询；例如，如果您从子句中删除detailid,并使用来在一行中选择它们，如下所示：detailnameGROUP BYGROUP_CONCAT,

SELECT 
  r.RestName, 
  ld.LocationID, 
  GROUP_CONCAT(DISTINCT d.DetailName separator ',') Details
FROM tblRestaurants     AS r
INNER JOIN tblLocations AS l  ON r.RestID     = l.RestID
INNER JOIN tblLocDet    AS ld ON l.LocationID = ld.LocationId
INNER JOIN tblDetails   AS d  ON ld.DetailID  = d.DetailID   
WHERE d.detailname IN ('det1', 'det2', 'det3')
GROUP BY r.RestName, 
         ld.LocationID
HAVING COUNT(DISTINCT d.DetailName) = 3;

更新了 SQL Fiddle 演示。

这会给你类似的东西：

| RESTNAME | LOCATIONID |        DETAILS |
------------------------------------------
|     res1 |          1 | det3,det2,det1 |

请注意：HAVING COUNT(DISTINCT d.DetailName) = 3如果您想获取至少 将其更改为>=.

于 2013-02-10T02:30:00.347 回答
0

这是一个解决方案：

<?php
if (isset ( $_POST ["btnSearch"] )) {
$checked = (array) $_POST ["criteria"];
sort($checked);
$criteria = implode(",", $checked);
echo "<br>Selected Options are :<br>" . $criteria . "<br><br>";

include "config.php";

mysql_select_db ( "MyHead", $con );

$criteria = mysql_real_escape_string($criteria);

$mysqlQuery = "
SELECT r.RestName, ld.LocationID, ld.DetailID, d.DetailName, r.RestName
FROM tblRestaurants AS r
INNER JOIN tblLocations AS l 
  ON r.RestID = l.RestID
INNER JOIN tblLocDet AS ld 
  ON l.LocationID = ld.LocationID 
INNER JOIN tblDetails AS d 
  ON ld.DetailID = d.DetailID
INNER JOIN (
  SELECT DetailID, GROUP_CONCAT(DetailName ORDER BY DetailName) AS DetailList 
  FROM tblDetails GROUP BY DetailID) AS dx
  ON ld.DetailID = dx.DetailID
WHERE dx.DetailList = '$criteria'";

if (! $rs = mysql_query ( $mysqlQuery )) {
  . . .

其余的将与您的代码相同。

于 2013-02-10T02:39:43.010 回答
0

你需要这个查询：

SELECT tblRestaurants.RestName, ...
FROM tblRestaurants 
INNER JOIN tblLocations ON tblLocations.RestID = tblRestaurants.RESTID
INNER JOIN tblLocDet ON tblLocDet.LocationID = tblLocations.LocationID
INNER JOIN tblDetails ON tblLocDet.DetailID  = tblDetails.DetailID
WHERE tblDetails.DetailName IN (" . $criteria . ");";
ORDER BY...

您不需要GROUP BY，因为您没有聚合函数（MIN、MAX 等）。用于ORDER BY对列表进行排序。

转移到 mysqli_*（或 PDO）函数并使用准备好的语句来避免 SQL 注入也是一个好主意

于 2013-02-10T02:25:29.053 回答