Question

所以我一直试图创建一个使用ajax的搜索栏来将结果获取到一个php页面然后运行一个SQL查询并显示结果，我现在已经看了一段时间并且从iv管理调试.. ajax函数有效，我相信？并在原始文件中运行相应的页面。但是我的文本输入没有加载到查询中.. 所以我的问题：我错过了什么，你怎么告诉php页面如何处理通过ajax发送的输入。是否有一种简单的方法可以将它变成一个php变量，以便在被送回之前轻松处理？

<script>
//-----------------ajax code for search by text section----(booking.php)---------
function showHint(str) {
  var xhttp;
  if (str.length == 0) { 
    document.getElementById("membersContainer").innerHTML = "";
    return;
  }
  xhttp = new XMLHttpRequest();
  xhttp.onreadystatechange = function() {
    if (xhttp.readyState == 4 && xhttp.status == 200) {
      document.getElementById("membersContainer").innerHTML = xhttp.responseText;
    }
  };
  xhttp.open("GET", "booking/searchActivity.php?q="+str, true);
  xhttp.send();   
}
</script>

-------------------- booking.php上的HTML代码--------------------- --------

<form action="">
    <td> <input type="text" name="activity" id="activity" onkeyup="showHint(this.value)"</td>
    <td></td>                      
</form>

------------------- searchActivity.php ------------------------- -----

<?php
include '../Global/DBConnect.php';


$activity=$_GET['activity'];

$query= "select * from scotiaExcursions where description like '%$activity%'";
$result=mysqli_query($db,$query);

echo "<div align='center'>";
echo "<table width='600' border='1'>";
echo "<tr>";
echo "<td>Excursion ID</td>";
echo "<td>Description</td>";
echo "<td>Price</td>";
echo "<td>Places Free</td>";
echo "<td>Order Now</td>";
echo "</tr>";

while ($row = mysqli_fetch_assoc($result))
    {
    echo "<tr>";
    echo "<td>".$row['excursionId']."</td>";
    echo "<td>".$row['description']."</td>";
    echo "<td>".$row['price']."</td>";
    echo "<td>".$row['availablePlaces']."</td>";
?>  
    <td>
        <form name="addCart" id="addCart" action="../cheakout.php" method="POST">
             <input type="hidden" name="activityID" id="activityID" value="<?php echo $row['excursionId']; ?>"/>
             <input type="hidden" name="price" id="price" value="<?php echo $row['price']; ?>"/>
             <input type="hidden" name="title" id="title" value="<?php echo $row['description']; ?>" />
             <select name="qty" id="qty">
                 <option value="1">1</option>
                 <option value="2">2</option>
                 <option value="3">3</option>
            </select>
            <input type="submit" name="submit" id="blueButton" value="Order"/>
         </form>
    </td>
    <?php
    echo "</tr>";

    }

echo "</table>";
echo "</div>";

?>

Ajax文本输入到SQL（通过php页面）

0 个答案: