如何在php中通过选定值和mysql中的其他值绑定select dropdownlist
我有一个php项目。当我将选择下拉列表与选定值以及数据库中的所有其他值绑定时,一个作为选定值的值将绑定两次,一个作为第一个选定值,另一个作为列表值。 下面是我的代码如何在php中通过选定值和mysql中的其他值绑定select dropdownlist,php,mysql,Php,Mysql,我有一个php项目。当我将选择下拉列表与选定值以及数据库中的所有其他值绑定时,一个作为选定值的值将绑定两次,一个作为第一个选定值,另一个作为列表值。 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="SEL
$query1 = mysql_query("select * from pincode_master where pcode_id='1'");
while ($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>";
}
$query1=mysql\u query(“从pincode\u master中选择*,其中pcode\u id='1'”);
while($row1=mysql\u fetch\u数组($query1))
{
$city_id=$row1['city_id'];
$sql1=“从城市主控系统中选择城市名称,其中城市id='$city\U id';
$result1=mysql\u查询($sql1);
$row=mysql\u fetch\u row($result1);
@$city_name=$row[0];
$query11=“选择*来自城市\主控”;
$result11=mysql\u查询($query11);
回声“;
while(($row11=mysql\u fetch\u row($result11))!=null)
{
echo“{$row11['city_name']}”;
}
回声“;
}
您应该在循环外部编写查询
按如下方式编写代码:-
// 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>";
}
//检查查询错误
$query1=mysql\u query(“从pcode\u id='1'”所在的pincode\u master中选择*)或die(mysql\u error());
//检查查询错误
$result11=mysql\u query(“选择*来自城市\u主控”)或die(mysql\u error());
而($row1=mysql\u fetch\u assoc($query1)){
$city_id=$row1['city_id'];
回声“;
而($row11=mysql\u fetch\u assoc($result11)){
$selected=$row11['city\u id']=$city\u id?“selected='selected'”:;
echo“$row11[“城市名称]”;
}
回声“;
}
希望它能帮助您:)请检查此-
这就是你想要的吗
<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'] . "'>" . $m_row['table_column2'] . "</option>");
?>
</select>
--挑选--
如果($city\u name==$city\u name),您将获得更多信息?嗯?