Question

我有3列

国家/地区（id，country_name）
州（id，country_id，state_name）
files_dev（id，country_name，state_name）

Country和State我使用onchange事件处理程序来选择元素，然后每次用户选择显示相关州名的国家时，都会向我的PHP脚本发出一个AJAX请求。

问题在于POST SUBMIT我想插入国家名称而不是其值id

这是我正在使用的代码

<?php
        if ($country_result->num_rows > 0) {
        // output data of each row
        while($row = $country_result->fetch_assoc()) {
?>
    <option value="<?php echo $row["id"]; ?>">
<?php echo $row["country_name"]; ?></option><?php

我尝试将$ row [“id”]更改为$ row [“country_name”]，此处的任何帮助都将受到高度赞赏

的index.php

<form class="form-horizontal" method="post" id="signin" enctype="multipart/form-data">

                            <div class="control-group">
                                    <label class="control-label" for="inputPassword"  placeholder="Select Gram Panchayat" ><strong>Select Gram Panchayat</strong></label>
                                        <div class="controls">
                                        <?php
                                        require_once('../admin/lib/db.php');
                                        $country_result = $conn->query('select * from countries');
                                        ?>
                                            <select name="country" class="chzn-select" id="countries-list" required>

                                        <?php
                                            if ($country_result->num_rows > 0) {
                                        // output data of each row
                                        while($row = $country_result->fetch_assoc()) {
                                            ?>
                                            <option value="<?php echo $row["id"]; ?>"><?php echo $row["country_name"];?></option>
                                            <?php

                    echo "<option value='". $row['id']."'>".$row['country_name'] .'</option>';

                                        }

                                    }
                                    ?>
                                            </select>
                                    </div>
                          </div>

                                    <div class="control-group">
                                    <label for="inputPassword" class="control-label"  placeholder="Select Village" ><strong><strong>Select Village</strong></strong></label>
                                        <div class="controls">
                                            <select name="state" id="states-list" required>


                                            </select>
                                        </div>
                          </div>

                                            <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
                                            <script>
                                    $('#countries-list').on('change', function(){
                                        var country_id = this.value;
                                        $.ajax({
                                        type: "POST",
                                        url: "village_dev.php",
                                        data:'country_id='+country_id,
                                        success: function(result){
                                            $("#states-list").html(result);
                                        }
                                        });
                                    });
                                    </script>

village_dev.php

    <?php
   require_once('../admin/lib/db.php');
   error_reporting(0);
   $country_id = mysql_real_escape_string($_POST['country_id']);
   if($country_id!='')
   {
    $states_result = $conn->query('select * from states where country_id='.$country_id.'');
    $options = "<option value=''>Select Village</option>";
    while($row = $states_result->fetch_assoc()) {
    $options .= "<option value='".$row['state']."'>".$row['state']."</option>";
    }
    echo $options;
   }

   ?>

如果我在这里遗漏了任何内容，请告诉我，我想将值名称而不是id提交给数据库。

Answer 1

这里的问题是你想获得一些信息在之外输入值

这就是为什么你会获得id而不是州名！

<form class="form-horizontal" method="post" id="signin" enctype="multipart/form-data">

                            <div class="control-group">
                                    <label class="control-label" for="inputPassword"  placeholder="Select Gram Panchayat" ><strong>Select Gram Panchayat</strong></label>
                                        <div class="controls">
                                        <?php
                                        require_once('../admin/lib/db.php');
                                        $country_result = $conn->query('select * from countries');
                                        ?>
                                            <select name="country" class="chzn-select" id="countries-list" required>

                                        <?php
                                            if ($country_result->num_rows > 0) {
                                        // output data of each row
                                        while($row = $country_result->fetch_assoc()) {
                                            ?>
                                            <option value="<?php echo $row["country_name"]; ?>"><?php echo $row["country_name"];?></option>
                                            <?php

                    echo "<option value='". $row['country_name']."'>".$row['country_name'] .'</option>';

                                        }

                                    }
                                    ?>
                                            </select>
                                    </div>
                          </div>

                                    <div class="control-group">
                                    <label for="inputPassword" class="control-label"  placeholder="Select Village" ><strong><strong>Select Village</strong></strong></label>
                                        <div class="controls">
                                            <select name="state" id="states-list" required>


                                            </select>
                                        </div>
                          </div>

                                            <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
                                            <script>
                                    $('#countries-list').on('change', function(){
                                        var country_id = this.value;
                                        $.ajax({
                                        type: "POST",
                                        url: "village_dev.php",
                                        data:'country_id='+country_id,
                                        success: function(result){
                                            $("#states-list").html(result);
                                        }
                                        });
                                    });
                                    </script>

所以现在你将能够将你需要的内容发布到你的数据库中，输入的值将是状态的名称！

干杯

Answer 2

尝试html5的data-属性：

将您的国家/地区下拉代码更改为：

<?php
require_once('../admin/lib/db.php');
$country_result = $conn->query('select * from countries');
?>
<select name="country" class="chzn-select" id="countries-list" required>
<?php
if ($country_result->num_rows > 0) {
    // output data of each row
    while($row = $country_result->fetch_assoc()) {
        echo '<option value="'.$row["country_name"].'" data-countryID="'.$row["id"].'">'.$row["country_name"].'</option>';
    }
}
?>
</select>

并使用以下内容更改您的ajax脚本：

<script>
    $('#countries-list').on('change', function(){
        var country_id = $('#countries-list option:selected').data( 'countryID' );
        $.ajax({
            type: "POST",
            url: "village_dev.php",
            data:'country_id='+country_id,
            success: function(result){
                $("#states-list").html(result);
            }
        });
    });
</script>

从属下拉显示值id而不是在提交后显示名称

2 个答案: