工具提示与MySQL查询

Question

我有工具提示问题。

我在表中的某些行

echo '<td><center>'.$dana['material'].'</td>'; 
echo '<td><center>'.$zmiennacal.'</td>'; 
$mat=$dana['material']; 

，现在它我的问题，我可以做一个动态的提示与MySQL查询像这样

$connect = mysqli_connect(CONNECTION QUERY); 
$zapytanie = "SELECT iloscsurmaterialu, Produkt FROM receptaprodukt WHERE surowiec='$mat'" 
or die(mysql_error()); 
$result = mysqli_query($connect, $zapytanie); 

while($row = mysqli_fetch_assoc($result)) 
{ 
    echo "'.$row['iloscsurmaterialu'].' - '.$row['Produkt'].'"; 
} 

，并以此结果提示，在第二在我的表中排在哪里$zmiennacal

我想一些脚本，但没有人工作:(（我知道它必须在ajax中，但我不知道PHP的结构：P

预先感谢您，请帮助

Answer 1

要使用ajax。你需要有一个触发ajax的'事件'。也许悬停会起作用？

<?php 
//ajax php file 
$mat = $_POST['mat']; 
$sql = 'SELECT iloscsurmaterialu, Produkt FROM receptaprodukt WHERE surowiec = :mat'; 
// to see where $dbh comes from: http://php.net/manual/en/pdo.construct.php 
$sth = $dbh->prepare($sql, array(PDO::ATTR_CURSOR => PDO::CURSOR_FWDONLY)); 
$sth->execute(array(':mat' => $mat)); 
$result = $sth->fetchAll(); 
$title = ''; 
if(!$result){ 
    echo json_encode(array('success' => false)); 
    exit; 
} 
foreach($result as $row) 
{ 
    $title .= "'".$row['iloscsurmaterialu'].' - '.$row['Produkt']."'"; 
} 
echo json_encode(array(
    'success' => true, 
    'title' => $title 
)); 
exit; 



//jQuery post: https://api.jquery.com/jquery.post/ 
$('body').on('hover', '#mytable td', function(){ 
    $td = $(this); 
    $td.attr('title', 'Loading...'); 
    var mat = $(this).data('mat'); 
    $.post("ajax/get_title.php", {"mat" : mat}, function(data) { 
     if(data && data.success){ 
      var title = data.title; 
      $td.attr('title', title); //note that td is not specific, it will add title to all the tds. You need to select the correct one. 
     } 
    }, "json"); 
}); 

//Html 
$mat = $dana['material']; 
echo '<td data-mat = "'.$mat.'"><center>'.$mat.'</td>'; 
echo '<td><center>'.$zmiennacal.'</td>';