我目前正在努力创建一个下拉框,当填充时会删除当前突出显示的任何记录。
我已经能够创建一个允许我向数据库添加新记录的表单,但从那以后我一直很难在下拉框中显示正确的记录,因为我对PHP只有模糊的理解。
<?php
$db=sqlite_open("/wwwroot/Work/bookDB.db");
if (isset($_POST['submit']))
{
$Author = $_POST['Author'];
$Title = $_POST['Title'];
$Synopsis = $_POST['Synopsis'];
$ISBN = $_POST['ISBN'];
$Publisher = $_POST['Publisher'];
sqlite_query ($db, "INSERT INTO Books (Author, Title, Synopsis, ISBN, Publisher)
VALUES ('$Author', '$Title', '$Synopsis', '$ISBN', '$Publisher')");
header("Location: /Work/Book/Book.php");
}
else
{
}
?>
这是我的提交查询,可能会对数据库的设置方式有所了解,我希望有一个下拉框,将由作者列表填充。到目前为止,我尝试使用$ row无济于事,我似乎也无法使用$ column,就像我之前说的那样,我的PHP技能非常糟糕,所以任何帮助都会非常感激。
HTML:
<form name = "delete form" action="Delete.php" method="POST">
<div class = "book">
<select name = "Title">
<option value = ""> Select </option>
</div>
PHP:
<form action = "<?php echo $_SERVER['PHP_SELF'];?>" method="get">
<select name = 'rowno' onchange = "javascript:document.forms[0].submit();">
<option> Select a Book </option>
<?php
$db = sqlite_open("/wwwroot/work/bookDB.db");
$query = sqlite_query($db,"SELECT ID, Title from Books");
$result = sqlite_fetch_all($query, SQLITE_BOTH);
$rowno = 0;
foreach($result as $entry)
{
echo "<option value = $rowno >$entry[ID] $entry[Title]</option>";
$rowno++;
}
?>
</select>
</form>
</div>