Question

我有一个简单的HTML文件。在这个HTML文件中，我有一个包含行和列的表。在另一个PHP文件中。我有一个代码，可以获取我的银行信息MYSQL信息现在我尝试从第一个HTML文件发送到PHP。每个标签都有一个属性。家和客人。将2 VAR发送到PHP，然后使用我发送的信息加载PHP页面。现在我问你。我的方法是否正确？还有其他办法吗？非常感谢所有阅读所有代码的人。并试图帮助我。

HTML

＆＃13;

$("th").click(function() {
  var home = $(this).attr("home");
  var guest = $(this).attr("guest");
  $.post("getBetsTeam.php", {
      home: home,
      guest: guest
    })
    .done(function(response) {
      $("#controler").html(response);
    });
})

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<html>

<table>
  <tbody>
    <tr>
      <th class="without" home="Liverpool" guest="Bolton">Liverpool-Bolton</th>
      <td>3.3</td>
      <td>2.5</td>
      <td>2.8</td>
    </tr>
    <tr>
      <th class="without" home="manchster united" guest="newcasel">manchster united-newcasel</th>
      <td>2</td>
      <td>3</td>
      <td>4</td>
    </tr>
  </tbody>
</table>

＆＃13;

PHP

<?php $link = mysqli_connect('localhost', 'root', '');
mysqli_select_db($link,'bet369');

if(!link){
    //echo "connection to db feild!";
}else{
    //echo "conection succesful!!! </br>";
}

if(mysqli_select_db($link,'bet369')){
    //echo "db selected! <br>";
} ?>
<?php
    $fr = 'Fulltime Result';
    $home = $_POST['home'];
    $guest = $_POST['guest'];
    $res = mysqli_query($link,"select * from bets where line='$fr' and home='$home' and guest='$guest'");
    $rows =mysqli_fetch_row($res);
    if ($rows[0] == 0) {
    die;
}
else{
    $res = mysqli_query($link,"select * from bets where line='$fr'");
   //if ($rows[0] == 1) {
    echo '
    <div class="headBets" style="color:white; with:100%; text-align: center; background-color: #0c15d2;">Fulltime Result</div>
    <table>';

        while($row=mysqli_fetch_array($res))
        {

echo '<tr>';
//echo '<th class="without">'; echo $row[home] . '-' . $row[guest]; echo '</th>';
echo '<td class="with" cheak="0" idbet="'.$row[idbet].'"numid="'. $row[idbet].'AAbet" a="0" b="0" c="0" d="0" e="0" f="0" z="1" value="'.$row[bethome].'"'.'team="'.$row[home].'  v  '.$row[guest].'"choice="'.$row[home].'"liga="'.$row[liga].'"line="'.$row[line].'">'.$row[home]. "   ".$row[bethome].'</td>';
echo '<td class="with" cheak="0" idbet="'.$row[idbet].'"numid="'. $row[idbet].'XXbet" a="0" b="0" c="0" d="0" e="0" f="0" z="x" value="'.$row[betdraw].'"'.'team="'.$row[home].'  v  '.$row[guest].'"choice="'.$row[draw].'"liga="'.$row[liga].'"line="'.$row[line].'">'.$row[draw]. "   ".$row[betdraw].'</td>';
echo '<td class="with" cheak="0" idbet="'.$row[idbet].'"numid="'. $row[idbet].'BBbet" a="0" b="0" c="0" d="0" e="0" f="0" z="2" value="'.$row[betguest].'"'.'team="'.$row[home].'  v  '.$row[guest].'"choice="'.$row[guest].'"liga="'.$row[liga].'"line="'.$row[line].'">'.$row[guest]. "   ".$row[betguest].'</td>';
echo '</tr>';

////////////////////////////////////////////////////////////////////////////////////


        }
        if(mysqli_query){
            //echo "all data reload !";
        }
echo    '</table>';
//  }
}
?>

Answer 1

如果你的目标是......

＆＃34;将2 VAR发送给PHP，然后使用我发送的信息加载PHP页面＆＃34;

然后听起来您想创建一个表单并在<th>点击处理程序

中提交

const form = document.createElement('form')
form.action = 'getBetsTeam.php'
form.method = 'post'

const homeInput = document.createElement('input')
homeInput.type = 'hidden'
homeInput.name = 'home'
homeInput.value = home
form.appendChild(homeInput)

// repeat for guest

document.body.appendChild(form)
form.submit()

如何将JavaScript变量传递给PHP？没有表格并提交

1 个答案: