PHP / MySQL中的多词搜索

Question

我正在努力创建一个搜索多个单词的搜索。我的第一次尝试没有任何结果，结果如下：

  require_once('database_conn.php');
  if($_POST){
  $explodedSearch = explode (" ", $_POST['quickSearch']);


  foreach($explodedSearch as $search){
  $query = "SELECT * 
            FROM jobseeker 
            WHERE forename like '%$search%' or surname like '%$search%' 
            ORDER BY userID 
            LIMIT 5";
  $result = mysql_query($query);
}

while($userData=mysql_fetch_array($result)){
    $forename=$userData['forename'];
    $surname=$userData['surname'];
    $profPic=$userData['profilePicture'];
    $location=$userData['location'];

    echo "<div class=\"result\">
    <img class=\"quickImage\" src=\"" . $profPic. "\" width=\"45\" height=\"45\"/>
    <p class=\"quickName\">" . $forename . " " . $surname . "</p>
    <p class=\"quickLocation\"> " . $location . "</p>
    </div>";

}
}

我还尝试了以下方法，这些方法产生了结果，但是正如您可以想象的那样，我输入的每个单词都得到了重复的结果：

if($_POST){
$explodedSearch = explode (" ", $_POST['quickSearch']);


foreach($explodedSearch as $search){
$query = "SELECT * 
          FROM jobseeker 
          WHERE forename like '%$search%' or surname like '%$search%' 
          ORDER BY userID 
          LIMIT 5";
$result .= mysql_query($query);


while($userData=mysql_fetch_array($result)){
    $forename=$userData['forename'];
    $surname=$userData['surname'];
    $profPic=$userData['profilePicture'];
    $location=$userData['location'];

    echo "<div class=\"result\">
    <img class=\"quickImage\" src=\"" . $profPic. "\" width=\"45\" height=\"45\"/>
    <p class=\"quickName\">" . $forename . " " . $surname . "</p>
    <p class=\"quickLocation\"> " . $location . "</p>
    </div>";
}
}
}

我对如何进行此操作非常迷惑，不胜感激。

编辑：

if($_POST){
$quickSearch = $_POST['quickSearch'];
$explodedSearch = explode (" ", trim($quickSearch));


$queryArray = array();

foreach($explodedSearch as $search){
$term = mysql_real_escape_string($search);
$queryArray[] = "forename like '%" . $term .  "%' surname like '%" . $term . "%'";
}

$implodedSearch = implode(' or ', $queryArray);

$query="SELECT *
        FROM jobseeker
        WHERE ($implodedSearch)
        ORDER BY userID
        LIMIT 5";

$result = mysql_query($query);

while($userData=mysql_fetch_array($result, MYSQL_ASSOC)){
    $forename=$userData['forename'];
    $surname=$userData['surname'];
    $profPic=$userData['profilePicture'];
    $location=$userData['location'];


    echo "<div class=\"result\">
    <img class=\"quickImage\" src=\"" . $profPic. "\" width=\"45\" height=\"45\"/>
    <p class=\"quickName\">" . $forename . " " . $surname . "</p>
    <p class=\"quickLocation\"> " . $location . "</p>
    </div>";

}
}

Answer 1

我已经在同一个主题（搜索关键字）上工作了一段时间，这是我的工作方式：

$words = $_POST['keywords'];
if(empty($words)){
    //redirect somewhere else!
}
$parts = explode(" ",trim($words));
$clauses=array();
foreach ($parts as $part){
    //function_description in my case ,  replace it with whatever u want in ur table
    $clauses[]="function_description LIKE '%" . mysql_real_escape_string($part) . "%'";
}
$clause=implode(' OR ' ,$clauses);
//select your condition and add "AND ($clauses)" .
$sql="SELECT * 
      FROM functions 
      WHERE
      user_name='{$user_name}'
      AND ($clause) ";
$results=mysql_query($sql,$connection);
 if(!$results){
    redirect("errors/error_db.html");
 }
 else if($results){
 $rows = array();
<?php 
 while($rows = mysql_fetch_array($results, MYSQL_ASSOC))
{
   // echo whatever u want !
}
?>

-现在这就是我尝试使用FULLTEXT搜索运行它的样子：但是您应该将表类型设置为“ MyISAM”

<?php
$words = mysql_real_escape_string($_POST['function_keywords']);
if(empty($words)){
    redirect("welcome.php?error=search_empty");
}
//if the columns(results)>1/2(columns) => it will return nothing!(use "NATURAL LANGUAGE"="BOOLEAN")
$sql="SELECT * FROM functions
     WHERE MATCH (function_description)
     AGAINST ('{$words}' IN NATURAL LANGUAGE MODE)";
$results=mysql_query($sql,$connection);
 if(!$results){
    redirect("errors/error_db.html");
 }
 else if($results){
$rows = array();
while($rows = mysql_fetch_array($results, MYSQL_ASSOC))
{
     // echo 
}
}
?>