php - 如何获取所选客户的值以回显 <input> 元素？

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>

和 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“tz customer”表“vs_1”列，标记为 0 或 1。当我选择客户时，如何才能从表中获得所选客户“vs_1”值？

还有

如何获取所选客户的值来回显<input>元素？

Answer 1

嗨，据我所知，当您更改该选择时，我不完全理解您的问题。您需要将选项值更改为 id，因为它是主要的。您需要 ajax 来这样做

<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'];
  }