Question

我有一个名为＆＃34的表单输入; invoiceidcopy＆＃34; ...我需要将此输入值的值与我的数据库中的值（invoiceidcopy）下的值进行匹配，然后在＆＃34; #submit时获取表单数据的相关记录/行以预填充表单-id＆＃34;单击按钮。我的表单和数据库具有以下输入/列，名为＆＃34; invoiceidcopy＆＃34;，＆＃34; location＆＃34;，＆＃34; q1＆＃34;，＆＃34; subcheck＆＃34;。以下就我而言，即便如此，我觉得我也很接近。我无法弄清楚我错过了什么或如何使这段代码有效。非常感谢并提前感谢任何人的帮助。

FORM

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Document</title>
</head>
<body>

<form action="#">

<input id="invoiceidcopy" name="invoiceidcopy" type="text" value="XXXIDXXX"/>



         <button id="submit-id">Prefill Form</button>


        <input id="location" name="location" required="" type="text">



<input type="radio" id="q1" name="q1" value="4.99" checked="checked">
<input type="radio" id="q1" name="q1" value="7.99" />

        <input id="subcheck" name="subcheck" value="0" type="hidden">
        <input id="subcheck" name="subcheck" value="1" onclick="return false" checked="" type="checkbox">Agree to Terms of Service


    <button id="btn1" type="submit" name="Submit">Submit</button>


</form>




<script src="http://code.jquery.com/jquery.min.js"></script>

<script>
     $(function(){


        $('#submit-id').on('click', function(e){  // Things to do when '#submit-id' button is clicked
            var invoiceidcopy = $('#invoiceidcopy').val(); // Grab user invoiceidcopy from text field
            e.preventDefault(); // Prevent form from submit, we are submiting form down with ajax.

             $.ajax({
              url: "/tst/orders2.php",
              data: {


invoiceidcopy:invoiceidcopy  


               }
            }).done(function(data) {
data = JSON.parse(data);
$('#location').val(data.location);
$('#q1').val(data.q1);
$('#subcheck').val(data.subcheck);
});
        });
     });
</script>
</body>
</html>

/tst/orders2.php

<?php

// Create the connection to the database
$con=mysqli_connect("xxx","xxx","xxx","xxx");


// Check if the connection failed
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  die();
}



  if (isset($_GET['invoiceidcopy']))
{
    $invoiceidcopy= $_GET['invoiceidcopy'];

   $query = "SELECT location, q1, subcheck
        FROM seguin_orders
  WHERE invoiceidcopy = '".($invoiceidcopy)."'";

    $result = mysqli_query($con,$query);

    while ($row = mysqli_fetch_assoc($result)){
 echo json_encode($row);
 die(); 
} 


    ?>

Answer 1

看起来您的表单操作无效。应该至少是“/”。

之后我会检查服务器日志，看看是否有某些东西失败了。您是否已验证该查询是否直接在数据库上运行？

使用Ajax如何将表单中的输入值与数据库中的记录进行匹配，并获取相关记录/行以预填表单？

1 个答案: