Question

我的工具提示有问题。

我在表格中有一些行

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

现在我的问题是，我可以用这样的

进行MySQL查询的动态工具提示

$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'].'";
}

并在工具提示中获取此结果，在我的表中的第二行$zmiennacal

我正在尝试一些脚本，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

提前谢谢，请帮助

Answer 1

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>';

带有MySQL查询的工具提示

1 个答案: