Question

数据库信息：

CREATE TABLE
IF NOT EXISTS tz_customer (
    id INT (10) NOT NULL,
    customervarchar (255) NOT NULL,
    NAME VARCHAR (255) NOT NULL,
    vs_1 VARCHAR (255) NOT NULL,
    stamp VARCHAR (255) NOT NULL
)

和PHP代码

这是index.php文件

<select name="nimetus" id="nimetus" onchange="chg(this.value)">
<?php $q3 = mysql_query("SELECT * FROM tz_kliendid"); while ($f3 = mysql_fetch_array($q3)) { ?>
<option value="<?php echo $f3['id']; ?>"><?php echo $f3['klient']; ?></option><?php } ?></select>

<input type="text" id="target" value="">

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>

<script type="text/javascript">function chg(str){ $.ajax({ url: "showdata.php", type: "POST", data: { "val": str }, success: function (response) { $("#target").val(response); }, error: function (xhr) { alert("Some error found!!"); } }); }</script>

AND showdata.php文件

<?php if(isset($_GET['val'])){$nimetus=$_POST['nimetus']; $q1=mysql_query("SELECT vs_1 FROM tz_kliendid WHERE id='".$nimetus."' ORDER BY id DESC LIMIT 1"); $f1=mysql_fetch_array($q1); echo $f1['vs_1']; } ?>

每个客户都有一个标记为0或1的“tz customer”表“vs_1”列。当我选择客户时，如何从表中获得所选客户的“vs_1”值？

和

如何获取所选客户的值以回显<input>元素？

Answer 1

您好我完全不了解您的问题，因为我可以理解您在更改选择时需要vs_1每个客户的价值。您需要将您的选项值更改为ID，因为它是主要的。您需要ajax到这样做

<select name="nimetus" id="nimetus" onchange="chg(this.value)">
<?php
  while ($f3 = mysql_fetch_array($q3)) {
   ?>
 <option value="<?php echo $f3['id'];  ?>">
      <?php echo $f3['customer'];  ?>
  </option>
   <?php
}
?> 
</select>
<input type="text" id="target" value="">

在java脚本中

function chg(str){
  $.ajax({
            url: "showdata.php",
            type: "POST", //send it through post method
            data: {
                "val": str
            },
            success: function (response) {
                //Do Something
                $("#target").val(response);
            },
            error: function (xhr) {
                //Do Something to handle error
                alert("Some error found!!");
            }
        });
}

在showdata.php中

if(isset($_POST['val']))//your POST method
  {
    $nimetus=$_POST['val'];
    $q1=mysql_query("SELECT vs_1 FROM tz_kliendid WHERE id='".$nimetus."' ORDER BY id DESC LIMIT 1");
   $f1=mysql_fetch_array($q1); 
   echo $f1['vs_1'];
  }

Answer 2

您可以使用data-attribute来保存v1的值，然后使用jquery访问它以显示它，通过这样做，每次更改值时都不需要额外的ajax调用

<!DOCTYPE html>
<html>
    <head>
        <title></title>
        <meta charset="UTF-8">
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
    </head>
    <body>
        <?php
            $option = '';
            $q3 = mysql_query("SELECT * FROM tz_customer");
            while ($f3 = mysql_fetch_array($q3)) {
                $option .= '
                    <option value="' . $f3['customer'] . '"' . ($f3['id'] == $f3['customer'] ? " selected" : "") . ' data-v1="'.$f3['v1'].'">
                        ' . $f3['customer'] . 
                    '</option>';
            } ?>
        <select name="nimetus" id="nimetus">
            <?= $option ?>
        </select>
        <div id="output"></div>
        <script>
        $(function() {
            $('#cbo').on('change', function() {
                var opt = $('option:selected', $(this));
                $('#output').html(opt.text()+ ' => v1 :'+ opt.data('v1'));
            });
        });
        </script>
    </body>
</html>

如何获取所选客户的值以回显<input />元素？

2 个答案: