搜索结果显示表中的所有信息

时间:2014-02-24 16:42:48

标签: php html mysql

我正在尝试从数据库中搜索某些数据并显示它。但是,每当我点击“搜索”时,都会显示表格中的所有结果。我能用任何方式只显示用户正在搜索的信息吗? 这是表单的html代码。

<h2> Search </h2>
<form action = "search.php" method = "post" >
  Search for: <input type = "text" name ="find" placeholder="Animal Type/Date"><span   class="required"</span> in
  <select NAME = "field">
    <Option VALUE = "Animal Type"> Animal Type</option>
    <Option VALUE = "dateseen"> Date Required</option>

  </Select>
  <input type= "hidden" name = "searching" value ="yes"/>
  <input type= "submit" name = "search" value ="Search"/>
</form>

这是我正在使用的PHP代码。

$link=mysqli_connect("localhost","root","");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$db_selected = mysqli_select_db($link,"AnimalTracker1");

if (!$db_selected)
  {
  die ("Can\'t use test_db : " . mysqli_error($link));
  }

//$find = strtoupper($find);
//$find = strip_tags($find);
//$find = trim($find);

$sql=mysqli_query($link, "Select * FROM Locations  ");

if ($sql == FALSE)
{
  die($sql." Error on query: ".mysqli_error($link)); 
}

while($result = mysqli_fetch_array($sql))
{
echo  $result ['Animal Type'];
echo "<br>";
echo $result ['Latitude'];
echo "<br> ";
echo $result ['Longitude'];
echo " <br>";
echo $result ['Seen'];
echo " <br> ";
echo $result ['Time'];
echo "<br> ";
echo "<br> ";
}
//}
?>

1 个答案:

答案 0 :(得分:1)

为此,您需要首先抓住文本框的find字段..

$searchKeyword = $_POST['find']; // Sanitize this value first !!

接下来,您需要将其传递给您的查询...更改您的yourcolumn以适合您的列名称。

$sql=mysqli_query($link, "Select * FROM Locations WHERE `yourcolumn` LIKE '%$searchKeyword%' ");

<强> EDIT :

你可以抓住两个字段并进行检查..

if(!empty($_POST['find']) && !empty($_POST['field']))
    {
        //do your query like..
        $searchKeyword = $_POST['find']; // Sanitize this value first !!
        $searchKeyword2 = $_POST['field']; // Sanitize this value first !!

        $sql=mysqli_query($link, "Select * FROM Locations WHERE `yourcolumn` LIKE '%$searchKeyword%' AND `yourcolumn2` LIKE '%$searchKeyword2%' ");
    }
    else
    {
        echo "The search criteria cannot be empty !";
    }