使用LIKE和空格的MySQL查詢

Question

如果搜索輸入中有空格，我將無法查詢顯示結果。 數據庫中有一個包含兩個字符串的列。 問題是，如果字符串中有空格，我不知道如何使用LIKE來過濾信息。 我不確定是否需要MATCH AGAINST，因為我對表中的數據的兩個獨立列不感興趣，我對由兩個單詞隔開一個空格的列感興趣，該列為fullNameNormal。

編輯：我剛剛意識到，我忘記了搜索功能實際上是$ search的輸入。 最初將其記錄為$ letter，但這僅用於字母搜索。

$search = '';
        if(isset($_REQUEST['search'])){
            $search = substr($this->encode($_REQUEST['search']), 0, 50);
        }
$letter = '';
if (isset($_REQUEST['letter'])) {
    $letter = substr($_REQUEST['letter'], 0, 1);
}
// Get total amount of records for current search...for paging
$total = 0;
$sql = "SELECT COUNT(*) FROM " . $wpdb->prefix . "example_directory";
$clean_where = " WHERE dirListing = 'public' AND active = 1 AND (((employeeType = 'faculty') AND (shortPositionCode NOT IN ('47', '49', '58', '59')) AND (status IN ('fullTime', 'partTime', 'proRata'))) OR ((employeeType = 'staff') AND (shortPositionCode NOT IN ('47', '49', '58', '59')) AND (status IN ('fullTime', 'partTime'))))";
$order = " ORDER BY lastName, firstName, department";
$limit_query = $wpdb->prepare(" LIMIT %d, %d", $start, $limit);
$where = "";
$args = array();
if ($search != '') {
    $where = " AND (lastName LIKE %s OR firstName LIKE %s OR department LIKE %s OR fullName LIKE %s OR nickName LIKE %s OR jobTitle LIKE %s OR phoneExt LIKE %s OR phone LIKE %s OR email LIKE %s OR locationBuilding LIKE %s OR locationBuildingAbbrev LIKE %s OR locationRoom LIKE %s or fullNameNormal LIKE %s)";
    $arg = '%' . $search . '%';
    $args = array($arg, $arg, $arg, $arg, $arg, $arg, $arg, $arg, $arg, $arg, $arg, $arg, $arg);
} elseif ($letter != '') {
    $where = " AND lastName LIKE %s";
    $arg = $letter . '%';
    $args = array($arg);
} else {
    $where = "";
}
    $where = $wpdb->prepare($where, $args);
    $total = $wpdb->get_var($sql . $clean_where . $where . $order);

如前所述，我只對fullNameNormal感興趣，例如在單元格中以“ John Smith”為例，但似乎沒有辦法使用LIKE並將其限制為僅輸入到我變成$ search的內容中。 我還嘗試過像這樣添加MATCH和AGAINST-

$where = " AND (lastName LIKE %s OR firstName LIKE %s OR department LIKE %s OR fullName LIKE %s OR nickName LIKE %s OR jobTitle LIKE %s OR phoneExt LIKE %s OR phone LIKE %s OR email LIKE %s OR locationBuilding LIKE %s OR locationBuildingAbbrev LIKE %s OR locationRoom LIKE %s OR MATCH(fullNameNormal) AGAINST (".$search."))";

編輯：也嘗試了此-

$where = " AND (lastName LIKE %s OR firstName LIKE %s OR department LIKE %s OR fullName LIKE %s OR nickName LIKE %s OR jobTitle LIKE %s OR phoneExt LIKE %s OR phone LIKE %s OR email LIKE %s OR locationBuilding LIKE %s OR locationBuildingAbbrev LIKE %s OR locationRoom LIKE %s OR (fullNameNormal LIKE 'something%' AND fullNameNormal LIKE '% something%')";

盡管我輸入了搜索輸入，但最終還是查詢了所有內容。

我不確定在這里需要做什么。

Answer 1

您實際上應該采用自然語言搜索方法來解決此問題。 使用這種方法的查詢如下所示：

SELECT COUNT(*) FROM *example_directory
WHERE MATCH (
  lastName,
  firstName,
  department,
  fullName,
  nickName,
  jobTitle,
  phoneExt,
  phone,
  email,
  locationBuilding,
  locationBuildingAbbrev,
  locationRoom,
  fullNameNormal
) AGAINST (? IN NATURAL LANGUAGE MODE)
AND  ... /* other filter conditions */

在? 是您的搜索詞組。

您將需要在搜索的列上創建FULLTEXT索引。

參見MySQL文檔 。

旁注：我不知道為什么您要在COUNT(*)的上下文中在查詢中應用排序或限制，而在選擇中沒有任何其他字段的情況下不能返回多行。