Question

我一直在四处搜寻，我观看了几个视频，但是我无法得到我想要的结果。

我正在尝试使用php通过数据库填充 webname 组合框。在数据库中，我有3个字段： webid ， webname 和 weburl 。
当我从下拉列表中选择一个项目时，我想要一个输入框来显示相应 webname 的网址。
按一次按钮会在单独的浏览器窗口中打开显示的网址。

我必须使用组合框填充正确但我无法在更改下拉列表时显示正确的URL。

我的Index.php

<body id="top">

<?php
    $cn=mysql_connect("dbHost","dbUser","dbPW") or die(mysql_error());
    mysql_select_db("dbName",$cn) or die(mysql_error());
    $sql = "SELECT * FROM tblName ";
    $result = mysql_query($sql);
?>
<select id="loads" name="Loads" size=1 style="position:absolute;left:133px;top:10px;width:81px;height:22px;z-index:1;">

<?php
    while($ri = mysql_fetch_array($result))
    {
        echo "<option value=" .$ri['webname'] . ">" . $ri['webname'] . "</option>";
    }
    echo "</select> "  
?>
 <input type="text" id="ShipperName" name="ShipperName" value="">

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>   
 <script>

  $(document).ready(function(){ /* PREPARE THE SCRIPT */

    $("#loads").change(function(){ /* TRIGGER THIS WHEN USER HAS SELECTED DATA FROM THE SELECT FIELD */

      var WebName = $(this).val(); /* STORE THE SELECTED LOAD NUMBER TO THIS VARIABLE */

      $.ajax({ /* START AJAX */

        type: "POST", /* METHOD TO USE TO PASS THE DATA */
        url: "action.php", /* THE FILE WHERE WE WILL PASS THE DATA */
        data: {"WebName": WebName}, /* THE DATA WE WILL PASS TO action.php */
        dataType: 'json', /* DATA TYPE THAT WILL BE RETURNED FROM action.php */
        success: function(result){

          $("#ShipperName").val(result.shippername);
        /*  $("#ShipperAddr").val(result.shipperaddr); */
        }

      }); /* END OF AJAX */   

    }); /* END OF CHANGE #loads */

  });

</script>
</body>
</html>

my action.php

<?php

  if(!empty($_POST["WebName"])){

    /* RE-ESTABLISH YOUR CONNECTION */
    $con = new mysqli("dbHost","dbUser","dbPW","dbTable");

    /* CHECK CONNECTION */
    if (mysqli_connect_errno()) {
      printf("Connect failed: %s\n", mysqli_connect_error());
      exit();
    }

    /* PREPARE YOUR QUERY */
    $stmt = $con->prepare("SELECT weburl, webid FROM tblName WHERE webname = ?");
    $stmt->bind_param("i", $_POST["WebName"]); /* PARAMETIZE THIS VARIABLE TO YOUR QUERY */
    $stmt->execute(); /* EXECUTE QUERY */
    $stmt->bind_result($shippername, $shipperaddress); /* BIND THE RESULTS TO THESE VARIABLES */
    $stmt->fetch(); /* FETCH THE RESULTS */
    $stmt->close(); /* CLOSE THE PREPARED STATEMENT */

    /* RETURN THIS DATA TO THE MAIN FILE */
    echo json_encode(array("shippername" => $shippername, "shipperaddr" => $shipperaddress));

  } /* END OF IF NOT EMPTY loadnumber */

?>

Answer 1

我认为您希望用URL填充文本框，但是您将ShipperName分配给文本框，如下所示：

$("#ShipperName").val(result.shippername);

这应该是：

$("#ShipperName").val(result.weburl);

数据库组合框填充输入框，然后在新浏览器窗口中打开URL

1 个答案: