如何通过PHP中的MySQL从选定值和其他值绑定选择下拉列表

Question

我在php中有一个项目。 当我将选择下拉列表与选定值以及数据库中所有其他值绑定在一起时，一个选定的值将绑定两次，一个绑定为第一个选定值，另一个绑定为列表值。 下面是我的代码。

$query1 = mysql_query("select * from pincode_master where pcode_id='1'");
while ($row1 = mysql_fetch_array($query1)) 
{
    $city_id = $row1['city_id'];
    $sql1="SELECT city_name from city_master where  city_id='$city_id'";
    $result1=mysql_query($sql1);
    $row = mysql_fetch_row($result1);
    @$city_name = $row[0];
        $query11 = "select * from city_master";
        $result11 = mysql_query($query11);
        echo "<select name = 'cityname'>";
            while (($row11 = mysql_fetch_row($result11)) != null)
            {
                echo "<option value = '{$row11['city_name']}'";
                if ($city_name == $city_name)
                    echo "selected = 'selected'";
                echo ">{$row11['city_name']}</option>";
            }
        echo "</select>";
}

Answer 1

您应该在循环外编写查询。

编写如下代码：

// Check query error 
$query1 = mysql_query( "select * from pincode_master where pcode_id='1'" ) or die( mysql_error());
// Check query error
$result11 = mysql_query( "select * from city_master" ) or die( mysql_error());
while ( $row1 = mysql_fetch_assoc( $query1 ) ) {
    $city_id = $row1['city_id'];
    echo "<select name = 'cityname'>";
    while ($row11 = mysql_fetch_assoc( $result11 )) {
      $selected = $row11['city_id'] == $city_id ? "selected = 'selected'" : '';           
      echo "<option value = '{$row11['city_name']}' $selected >". $row11['city_name'] ."</option>";
    }
    echo "</select>";
 }

希望它能对您有所帮助:)

Answer 2

请检查一下-

那是你要的吗 ？

<select name="yourselection">
    <option value="">--Select--</option>
    <?php
 $msql = mysql_query("SELECT * FROM tablename");
 while($m_row = mysql_fetch_array($msql))        
        echo("<option value = '" . $m_row['table_column1'] . "'>" . $m_row['table_column2'] . "</option>");
    ?>
</select>

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