Question

好的，我知道之前已经回答过这个问题（Execute PHP script on same page after selecting a dropdown list option using Ajax or JavaScript），但这些答案对以前从未使用过AJAX的人来说并不是很有帮助。如果有人从下拉菜单中选择一个选项，我如何运行创建的查询？

<head>
 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
</head>

<h3>Subject</h3>
<select name="allbooks" >
              <option value="none" ></option>
              <option value="allbooks" >All Books</option>
    </select>

<?php 
$query = "SELECT * FROM books" or die("Error in the consult.." . mysqli_error($connection)); 
    $books = $connection->query($query);
?>

Answer 1

首先，您必须使用Javascript触发AJAX请求。但我会用jQuery（一个Javascript库）来指导你。

您的HTML：

<select name="allbooks" id="allbooks">
  <option value="none" ></option>
  <option value="allbooks" >All Books</option>
</select>
<div id="show">
  <!-- ITEMS TO BE DISPLAYED HERE -->
</div>

之后，请下载jQuery。

然后让我们做脚本：

<script src="jquery-1.9.1.min.js"></script> <!-- CHANGE THE JQUERY FILE DEPENDING ON THE VERSION YOU HAVE DOWNLOADED -->
<script type="text/javascript">
  $(document).ready(function(){ /* PREPARE THE SCRIPT */
    $("#allbooks").change(function(){ /* WHEN YOU CHANGE AND SELECT FROM THE SELECT FIELD */
      var allbooks = $(this).val(); /* GET THE VALUE OF THE SELECTED DATA */
      var dataString = "allbooks="+allbooks; /* STORE THAT TO A DATA STRING */

      $.ajax({ /* THEN THE AJAX CALL */
        type: "POST", /* TYPE OF METHOD TO USE TO PASS THE DATA */
        url: "get-data.php", /* PAGE WHERE WE WILL PASS THE DATA */
        data: dataString, /* THE DATA WE WILL BE PASSING */
        success: function(result){ /* GET THE TO BE RETURNED DATA */
          $("#show").html(result); /* THE RETURNED DATA WILL BE SHOWN IN THIS DIV */
        }
      });

    });
  });
</script>

然后让我们创建get-data.php，它将接收通过AJAX发送的数据。

if(!empty($_POST["allbooks"])){
  /* DO YOUR QUERY HERE AND GET THE OUTPUT YOU WANT */
  echo $output; /* PRINT THE OUTPUT YOU WANT, IT WILL BE RETURNED TO THE ORIGINAL PAGE */
}

您可以查看此示例 - JSfiddle。

Answer 2

 var id="1";
 $.ajax({
    type: 'POST',
    url: 'yourphppage',
    dataType: "json",
    data: {
        idofrow:id
    },
    success: function(data) {
        alert(data);
    },
    error: function(data) {
        alert(data);
    }
});

这是ajax请求的一个小问题，您可以使用它，只需在查询成功时根据需要更改其他字段，您可以成功检索该数据，您可以操作您想要使用的数据，您可以返回{{1 }}，json。

在您的php页面中，您可以将ID检索为

text

你可以这个id来选择这样的

$id = ($_POST['idofrow']);

你可以回应结果。

了解更多信息，请查看此documentation

Answer 3

Check this simple tutorial希望这会有所帮助。

      <html>
    <head>
    <script>
    function showUser(str) {
        if (str == "") {
            document.getElementById("txtHint").innerHTML = "";
            return;
        } else { 
            if (window.XMLHttpRequest) {
                // code for IE7+, Firefox, Chrome, Opera, Safari
                xmlhttp = new XMLHttpRequest();
            } else {
                // code for IE6, IE5
                xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
            }
            xmlhttp.onreadystatechange = function() {
                if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                    document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
                }
            }
// getuser.php is seprate php file. q is parameter 
            xmlhttp.open("GET","getuser.php?q="+str,true);
            xmlhttp.send();
        }
    }
    </script>
    </head>
    <body>

    <form>
    <select name="users" onchange="showUser(this.value)">
      <option value="">Select a person:</option>
      <option value="1">Peter Griffin</option>
      <option value="2">Lois Griffin</option>
      <option value="3">Joseph Swanson</option>
      <option value="4">Glenn Quagmire</option>
      </select>
    </form>
    <br>
    <div id="txtHint"><b>Person info will be listed here...</b></div>

    </body>
    </html>

getuser.php文件

   <!DOCTYPE html>
    <html>
    <head>
    <style>
    table {
        width: 100%;
        border-collapse: collapse;
    }

    table, td, th {
        border: 1px solid black;
        padding: 5px;
    }

    th {text-align: left;}
    </style>
    </head>
    <body>

    <?php
// don't use intval if your select value is not numberic
    $q = $_GET['q'];

    $con = mysqli_connect('localhost','peter','abc123','my_db');
    if (!$con) {
        die('Could not connect: ' . mysqli_error($con));
    }

    mysqli_select_db($con,"ajax_demo");
    $sql="SELECT * FROM user WHERE id = '".$q."'";
    $result = mysqli_query($con,$sql);

    echo "<table>
    <tr>
    <th>Firstname</th>
    <th>Lastname</th>
    <th>Age</th>
    <th>Hometown</th>
    <th>Job</th>
    </tr>";
    while($row = mysqli_fetch_array($result)) {
        echo "<tr>";
        echo "<td>" . $row['FirstName'] . "</td>";
        echo "<td>" . $row['LastName'] . "</td>";
        echo "<td>" . $row['Age'] . "</td>";
        echo "<td>" . $row['Hometown'] . "</td>";
        echo "<td>" . $row['Job'] . "</td>";
        echo "</tr>";
    }
    echo "</table>";
    mysqli_close($con);
    ?>
    </body>
    </html>

如何使用AJAX在select选项上执行PHP查询？

3 个答案: