我有复选框表,我从数据库中检索:
<?php
$result = mysql_query("SELECT shop_id, name FROM shop") or die(mysql_error());
if(mysql_num_rows($result) > 0)
{
while($row = mysql_fetch_assoc($result))
{
echo '<tr>
<td>
<input type="checkbox" name="identifer[]" value="'.$row['shop_id'].'" /> <br />
</td>
<td>'.ucfirst($row['shop']).'</td>
</tr> ';
}
}
?>
我想将结果保存在数据库中:
places
places_id
,book_id
,shop_id
然而,我无法让它正常工作。在shop_id
列中,我会检查复选框的次数相同。
如果我使用它:
$identifer = $_POST['identifer'];
if( count( $identifer) > 1)
{
foreach($identifer as $x)
{
$y .= $x.",";
$val = rtrim($y,",");
$q2 = "INSERT INTO places (places_id, book_id, shop_id) VALUES (NULL, '$book_id', '$val')";
$result2 = mysql_query($q2) or die(mysql_query());
}
}
在places
表格中,无论选中多少次复选框,我都会得到一行。
那么这样做的正确方法是什么?
答案 0 :(得分:2)
我认为这就是你要找的东西:
$identifier = $_POST['identifer']; // Also, you spelled identifier wrong ;)
// There's no guarantee that you were given an array for identifier
if( is_array( $identifier) && count( $identifier) > 1)
{
$values = array();
// This creates a bunch of insert values
foreach($identifier as $x)
{
$values[] = "(NULL, '$book_id', '$x')";
}
// Join all those values together into one SQL query (this will generate multiple rows)
$q2 = "INSERT INTO places (places_id, book_id, shop_id) VALUES " . implode( ', ', $values);
$result2 = mysql_query($q2) or die(mysql_query());
}