Question

如何将id="'.$row["courseid"].'"传递给ajax函数

我正在尝试data:{'courseid':deleteId}，但对如何解决此问题没有任何想法。

    <?php


echo "<table width='100%'>";
echo "<tr> 
      <th>Course name</th> 
      <th>Delete</th>
      <th>Edit</th>
      </tr>";
?>
    <?php foreach($rows as $row):  
    echo "<tr>"; 
    echo '<td><a href="#">' . htmlentities($row['coursename'], ENT_QUOTES, 'UTF-8') . '</a></td>';
    echo '<td><button onclick="deleteC(' . $row['courseid'] . ')");"><font color="#e70404"> Delete </font> </button></td>';
    echo '<td><a class="delete" id="'.$row["courseid"].'">Delette</a></td>';
    echo "</tr> ";
    endforeach; 
echo "</table>";

?>

这是同一页面中的ajax函数

<script type="text/javascript">
function deleteC(deleteId){
   $.ajax({
   type: "GET",
   url: "deleteCourse.php",
   data:{'courseid':deleteId},
   success: function(result){               
       if(result=='correct'){
           window.location='index.php';
       }else {
       window.location='coursesData.php';
       }
   }
});
}
</script>

这是deleteCourse.php

   <?php 
require("connect.php");
if (isset($_GET['courseid']) && is_numeric($_GET['courseid']))
{ 
    $id = $_GET['courseid'];

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

    mysqli_query($con,"DELETE FROM courses WHERE courseid=$id");
    echo "correct";
    mysqli_close($con);
    echo "correct";
}
else 
{
    header ("Location: ../index.php");
}


?>

Answer 1

而不是：

echo '<td><a href="finalphp/deleteCourse.php?id=' . $row['courseid'] . '" onclick="deleteC()");"><font color="#e70404"> Delete </font> </a></td>';

只需将id作为参数传递给JS函数并使用它：

PHP：

echo '<td><a href="finalphp/deleteCourse.php?id=' . $row['courseid'] . '" onclick="deleteC(' . $row['courseid'] . ')");"><font color="#e70404"> Delete </font> </a></td>';

JS

function deleteC(deleteId){
   $.ajax({
   type: "GET",
   url: "deleteCourse.php",
   data:{'courseid':deleteId},
   success: function(result){               
       if(result=='correct'){
           window.location='index.php';
       } else {

       }
   }
});
}

如何将php变量传递给ajax函数

1 个答案: