Ajax动态选择2个字段

Question

我正在使用ajax动态选择进行银行帐户验证。onkeyup在输入acc号后，它从外部文件运行检查。现在，我只能检查账号，我希望它能够检查银行和账号。我只想获得银行和帐户的值，用于在validate_acc.php中进行验证。

    <script>
function dynamic_Select3(ajax_page, account) {
$.ajax({
type: "GET",
url: ajax_page,
data: "ch=" + account,
//data: "ch2=" + bank,
dataType: "html",
//dataType: "text/html", //<--UPDATE: DELETING THIS LINE FIXES EVERYTHING
//<--UPDATE2: DON'T DELETE; REPLACE "test/html" with "html"
success: function(html){       $("#txtResult3").html(html); }
}); 
}
</script>

我的html表单

<td class="body_text_normal_npt">Select your Bank</td>
    <td><span class="body_text_normal_npt">
      <select name="bank" id="bank">
        <option value="Select Values" selected="selected">-------------------------------</option>
        <?php

$sql4="SELECT bankName FROM banks WHERE status = 'active'";
$banks = $mydb->query($sql4) or die(mysqli_error($mydb));
$row_banks = mysqli_fetch_assoc($banks);
$totalRows_banks = mysqli_num_rows($banks);

do {  

?>
        <option value="<?php echo $row_banks['bankName']?>"><?php echo $row_banks['bankName']?></option>
        <?php

} while ($row_banks = mysqli_fetch_assoc($banks));

  $rows = mysqli_num_rows($banks);

  if($rows > 0) {

      mysqli_data_seek($banks, 0);

      $row_banks = mysqli_fetch_assoc($banks);

  }

?>
      </select>
    </span></td>
  </tr>
  <tr>
    <td class="body_text_normal_npt">Acc Number</td>
    <td><p>
      <input name="account" type="text" id="account" size="25" onKeyPress="return isNumberKey(event)" onKeyUp="dynamic_Select3('validate_acc.php', this.value)" />
    </p>
</td>
  </tr>

验证页面

//$bank= $_GET['modepayment'];
$bank= $_GET['ch2'];
$accnum = $_GET['ch'];

 $query_bcode = "SELECT bankCode,abbr FROM banks WHERE bankName = '$bank'";
$bcode = $mydb->query($query_bcode) or die(mysqli_error($mydb));
$row_bcode = $bcode->fetch_assoc();

$bankCode = $row_bcode['bankCode'];
//echo $bankCode;
//echo $accnum;

$json = file_get_contents("https://api.bank.codes/ng-nuban/?format=json&api_key=2d112c21e1c5844f*******154&bank=$bankCode&nuban=$accnum");
$obj = json_decode($json);

Answer 1

可以试试这个吗？您可以通过这种方式拥有多个参数。

var request = $.ajax({
  url: "ajax_page",
  method: "POST",
  data: { ch: account, ch2: bank},
  dataType: "html"
});

代码基于http://api.jquery.com/jquery.ajax/中的一个示例