使用PHP獲取點擊圖像的ID並從MYSQL數據庫加載它的信息

Question

我需要幫助爲加載點擊圖像的特定信息。信息存儲在MYSQL數據庫中。

1. 我需要determin單擊的圖像，並獲得圖像 ID
2. 然後從數據庫
3. 獲取關於該圖像的信息和回聲出有關圖像

信息

鏈接到圖像數據庫結構：https://www.dropbox.com/sh/8xrifn64psmc9qh/AAA5j_eous8sBpg4WP0dgrmya?dl=0

謝謝！這是我用來從數據庫加載圖像的代碼行。

$sql="SELECT * FROM klubbar ORDER BY namn ASC"; 
$res = mysql_query($sql); 

if (!$res) { 
    echo "Could not successfully run query ($sql) from DB: " . mysql_error(); 
    exit; 
} 

while ($row = mysql_fetch_assoc($res)) { 
    echo '<div class="klubbar"><a href=""><img  src="'.$row['bild'].'"alt="'.$row['namn'].'"title="'.$row['namn'].'"/></a></div>'; 
} 

Answer 1

PHP：

$sel_club = $_GET['klub']; 
     $sql="SELECT * FROM klubbar ORDER BY namn ASC"; 
     $res = mysql_query($sql); 
     if (!$res) { 
     echo "Could not successfully run query ($sql) from DB: " . mysql_error(); 
     exit; 
     } 

     if($sel_club) 
     { 
     $sqlC= "SELECT * FROM klubbar WHERE namn ='$sel_club'"; 
     $resC= mysql_query($sqlC); 

     if($rowC = mysql_fetch_assoc($resC)) 
     { 
      $id = $rowC['id']; 
      $namn = $rowC['namn']; 
      $bild = $rowC['bild']; 
      $lank = $rowC['lank']; 
      $alder = $rowC['alder']; 
      $dresscode = $rowC['dresscode']; 

      echo 'Klub: '.$namn.'<br>Webbsida: '.$lank.'<br>Åldersgräns: '.$alder.'+ <br> Dresskod: '.$dresscode; 
      echo 
      '<form method="post" action="anmalan.php"> 
      <label name="namn">Namn:</label> 
      <input type="text" name="namn" placeholder="Förnamn Efternamn"/><br> 
      <label name="antalGuess">Antal gäster</label> 
      <select> 
       <option>+0</option> 
       <option>+1</option> 
       <option>+3</option> 
       <option>+4</option> 
       <option>+5</option> 
       <option>+6</option> 
      </select><br> 
      <label name="mejl">E-post:</label> 
      <input type="text" name="mejl" placeholder="example@ex.com"/><br> 
      <label name="telefon">Mobilnr:</label> 
      <input type="text" name="telefon" placeholder="07x xxx xx xx"/><br> 
      <input type="submit" name="submit" value="Skicka" /> 
      </form>'; 
     } 
     else 
     { 
      echo "Could not successfully run query ($sql) from DB: " . mysql_error(); 
     } 

     } 
     else 
     { 
     while ($row = mysql_fetch_assoc($res)) 
     { 
      $id = $row['id']; 
      $namn = $row['namn']; 
      $bild = $row['bild']; 
      $lank = $row['lank']; 
      $alder = $row['alder']; 
      $dresscode = $row['dresscode']; 
      echo '<div class="klubbar" ><a href="?klub='.$namn.'" id="'.$id.'"><img src="'.$bild.'"alt="'.$namn.'"title="'.$namn.'"/></a></div>'; 
     } 
     }