[英]How to capture primary key from drop down and insert as foreign key of another table?
[英]Insert Primary key of Selected Drop down list value as a foreign key in another table using php mysql
我需要在另一个表中插入要作为外键插入的选定值主键,但它既没有给出错误也没有发生任何事情我已经用下面的表格填充了下拉列表
<form id="country" name="form1" method="post" action="datatodb.php">
<table border="1">
<tr>
<td>
Select Country :<select name="countryname">
<?php
include 'dblogin.php';
// Get records from database (table "name_list").
$list=mysql_query("select * from _countries");
// Show records by while loop.
while($row_list=mysql_fetch_assoc($list)){
?>
<option value="<?php echo $row_list['countryid']; ?>"><?php echo $row_list['countryname']; ?></option>
<?php
// End while loop.
}
?>
</select> </td></tr>
<tr><td>
Add Location :
<input type="text" name="locationname" value="" maxlength="40"/></td></tr>
<tr><td align="center">
<input type="submit" name="Submit" value="Add" size="20"/></td></tr></table>
</form>
使用下面的代码捕获值并尝试将所选值的主键插入另一个表中,该表是外键
<?php
include 'dblogin.php';
$countryid=$_POST['countryname'];
$locationname=$_POST['locationname'];
$countryid = stripslashes($countryid);
$locationname = stripslashes($locationname);
$countryid = mysql_real_escape_string($countryid);
$locationname = mysql_real_escape_string($locationname);
$sql="insert into _location (locationid, locationname, countryid) values ('NULL','".$locationname."','".$locationname."')";
$result=mysql_query($sql);
?>
请帮助我获取选定的值主键并插入到另一个表中,该表是外键
如果您将locationid
设为自动增量,则无需将此值添加到查询中,
喜欢
$sql="insert into _location (locationname, countryid) values
('".$locationname."','".$countryid."')";
注意:要忽略locationname
的重复值,请将此字段设置为唯一
编辑:你有一个拼写错误,你尝试插入locationname
两次,我替换它
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